$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x. \qquad (15.73) $$

**Theorem 15.36.** If $f$ is bounded (hyperbounded) on $[a, b]$, then $f$ is integrable on $[a, b]$ if and only if for each $\varepsilon > 0$, $\varepsilon \approx 0$ there is a partition $P$ of $[a, b]$ for which

$$ S(P) - s(P) < \varepsilon. \qquad (15.74) $$

**Theorem 15.37.** If $f$ is $\#$-continuous on $[a, b]$, then $f$ is integrable on $[a, b]$.

**Proof.**

**Theorem 15.38.** If $f$ is monotonic on $[a, b]$, then $f$ is integrable on $[a, b]$.

**Proof.**

**Theorem 15.39.** (a) If $f$ and $g$ are integrable on $[a, b]$, then so is $f+g$, and

$$ \mathrm{Ext}-\int_{a}^{b} [f(x) + g(x)] d^{\#}x = \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x + \mathrm{Ext}-\int_{a}^{b} g(x)d^{\#}x \qquad (15.75) $$

(b) If $f_i, 1 \le i \le n \in \mathbb{N}^\#$ are integrable on $[a, b]$, then so is $\mathrm{Ext}-\sum_{j=1}^{n} f_i(x)$, and

$$ \mathrm{Ext}-\int_{a}^{b} \left[ \mathrm{Ext}-\sum_{j=1}^{n} f_{i}(x) \right] d^{\#}x = \mathrm{Ext}-\sum_{j=1}^{n} \left( \mathrm{Ext}-\int_{a}^{b} f_{i}(x) d^{\#}x \right) \qquad (15.76) $$

**Proof.**

**Theorem 15.40.** (a) If $f$ is integrable on $[a, b]$ and $c \in \mathbb{R}_c^\#$ is a constant, then $cf$ is integrable on $[a, b]$ and

$$ \mathrm{Ext}-\int_{a}^{b} c f(x) d^{\#}x = c \left( \mathrm{Ext}-\int_{a}^{b} f(x) d^{\#}x \right). \qquad (15.77) $$

(b) If $f_i, 1 \le i \le n \in \mathbb{N}^\#$ are integrable on $[a, b]$ and $c_i \in \mathbb{R}_c^\#$ are constants, then

$$ \mathrm{Ext}-\int_{a}^{b} \left[ \mathrm{Ext}-\sum_{j=1}^{n} c_i f_i(x) \right] d^{\#}x = \mathrm{Ext}-\sum_{j=1}^{n} c_i \left( \mathrm{Ext}-\int_{a}^{b} f_i(x) d^{\#}x \right). \qquad (15.78) $$

**Proof.**

**Theorem 15.41.** If $f$ is and $g$ integrable on $[a, b]$ and $f(x) \le g(x)$ for $x \in [a, b]$, then

$$ \mathrm{Ext}-\int_{a}^{b} f(x) d^{\#}x \le \mathrm{Ext}-\int_{a}^{b} g(x) d^{\#}x. \qquad (15.79) $$

**Proof.**

**Theorem 15.42.** If $f(x)$ is integrable on $[a, b]$, then so is $|f(x)|$, and

$$ \left| \mathrm{Ext}-\int_{a}^{b} f(x) d^{\#}x \right| \le \mathrm{Ext}-\int_{a}^{b} |f(x)| d^{\#}x. \qquad (15.80) $$

**Theorem 15.43.** If $f(x)$ and $g(x)$ are integrable on $[a, b]$, then so is the product $f(x)g(x)$.

**Proof.**
function $f(x)$.

**Lemma 15.3** Suppose that

$$|f(x)| \le M, a \le x \le b \qquad (15.66)$$

and let $P'$ be a hyperfinite partition of $[a, b]$ obtained by adding $r \in \mathbb{N}^{\#\mathbb{N}}$ points to a partition $P = \{x_0, x_1, \dots, x_n\}, n \in \mathbb{N}^{\#\mathbb{N}}$ of $[a, b]$. Then

$$S(P) \ge S(P') \ge S(P) - 2Mr \|P\| \qquad (15.67)$$

and

$$s(P) \le s(P') \le s(P) + 2Mr \|P\|. \qquad (15.68)$$

**Theorem 15.32.** If $f(x)$ is bounded on $[a, b]$, then

$$\mathrm{Ext}-\int_a^b f(x)d^\#x \le \mathrm{Ext}-\int_a^b f(x)d^\#x. \qquad (15.69)$$

**Theorem 15.33.** If $f$ is integrable on $[a, b]$, then

$$\mathrm{Ext}-\int_a^b f(x)d^\#x = \mathrm{Ext}-\int_a^b f(x)d^\#x = \mathrm{Ext}-\int_a^b f(x)d^\#x. \qquad (15.70)$$

**Theorem 15.34.** If $f$ is bounded (or hyperbounded) on $[a, b]$ and

$$\mathrm{Ext}-\int_a^b f(x)d^\#x = \mathrm{Ext}-\int_a^b f(x)d^\#x = L, \qquad (15.71)$$

then $f(x)$ is integrable on $[a, b]$ and

$$\mathrm{Ext}-\int_a^b f(x)d^\#x = L. \qquad (15.72)$$

**Theorem 15.35.** A bounded (hyperbounded) function $f$ is integrable on $[a, b]$ if and only if

$$\mathrm{Ext}-\int_a^b f(x)d^\#x = \mathrm{Ext}-\int_a^b f(x)d^\#x. \qquad (15.73)$$

**Theorem 15.36.** If $f$ is bounded (hyperbounded) on $[a, b]$, then $f$ is integrable on $[a, b]$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is a partition $P$ of $[a, b]$ for which

$$S(P) - s(P) < \varepsilon. \qquad (15.74)$$

**Theorem 15.37.** If $f$ is #-continuous on $[a, b]$, then $f$ is integrable on $[a, b]$.

**Proof.**

**Theorem 15.38.** If $f$ is monotonic on $[a, b]$, then $f$ is integrable on $[a, b]$.

**Proof.**

**Theorem 15.39.** (a) If $f$ and $g$ are integrable on $[a, b]$, then so is $f+g$, and

$$\mathrm{Ext}-\int_a^b [f(x) + g(x)]d^\#x = \mathrm{Ext}-\int_a^b f(x)d^\#x + \mathrm{Ext}-\int_a^b g(x)d^\#x \qquad (15.75)$$

(b) If $f_i, 1 \le i \le n \in \mathbb{N}^\#$ are integrable on $[a, b]$, then so is $\mathrm{Ext}-\sum_{j=1}^n f_i(x)$, and
**Theorem 15.44.** (First Mean Value Theorem for Integrals) Suppose that $u(x)$ is #-continuous and $v(x)$ is integrable and nonnegative on $[a, b]$. Then

$$ \mathrm{Ext}-\int_a^b f(x)v(x)d^\#x = v(c) \left( \mathrm{Ext}-\int_a^b f(x)d^\#x \right) \qquad (15.81) $$

for some $c \in [a, b]$.

**Proof.**

**Theorem 15.45.** If $f(x)$ is integrable on $[a, b]$ and $a \le a_1 < b_1 \le b$, then $f(x)$ is integrable on $[a_1, b_1]$.

**Proof.**

**Theorem 15.46.** If $f(x)$ is integrable on $[a, b]$ and $[b, c]$ then $f(x)$ is integrable on $[a, b]$ and

$$ \mathrm{Ext}-\int_a^c f(x)d^\#x = \mathrm{Ext}-\int_a^b f(x)d^\#x + \mathrm{Ext}-\int_b^c f(x)d^\#x. \qquad (15.82) $$

**Proof.**

**Theorem 15.47.** If $f(x)$ is integrable on $[a, b]$ and $a \le c \le b$, then the function $F(x)$ defined by

$$ F(x) = \mathrm{Ext}-\int_c^x f(t)d^\#t \qquad (15.83) $$

satisfies a Lipschitz condition on $[a, b]$, and is therefore #-continuous on $[a, b]$.

**Proof.**

**Theorem 15.48.** If $f(x)$ is integrable on $[a, b]$ and $a \le c \le b$, then $F(x) = \mathrm{Ext}-\int_c^x f(t)d^\#t$ is #-differentiable at any point $x_0 \in [a, b]$, where $f(x)$ is #-continuous, with

$$ F'(x_0) = f(x_0). \qquad (15.84) $$

If $f(x)$ is #-continuous from the right at $a$, then $F_+^\#(a) = f(a)$. If $f(x)$ is #-continuous from the left at $b$, then $F_+^\#(b) = f(b)$.

**Proof.** We consider the case where $a < x_0 < b$. From the equality

$$ \frac{1}{x-x_0} \int_{x_0}^{x} f(x_0)dt^\# = f(x_0) $$

one obtains

$$ \frac{F(x) - F(x_0)}{x - x_0} - f(x_0) = \frac{1}{x - x_0} \int_{x_0}^{x} [f(t) - f(x_0)]dt^\# $$

From this one obtains

$$ \left| \frac{F(x) - F(x_0)}{x - x_0} - f(x_0) \right| \le \frac{1}{|x - x_0|} \left| \int_{x_0}^{x} |f(t) - f(x_0)| dt^{\#} \right| \qquad (15.84') $$
$$Ext- \int_a^b \left[ Ext- \sum_{j=1}^n f_i(x) \right] d^\# x = Ext- \sum_{j=1}^n \left( Ext- \int_a^b f_i(x) d^\# x \right) \quad (15.76)$$

**Proof.**

**Theorem 15.40.** (a) If $f$ is integrable on $[a, b]$ and $c \in \mathbb{R}_c^\#$ is a constant, then $cf$ is integrable on $[a, b]$ and

$$Ext- \int_a^b c f(x) d^\# x = c \left( Ext- \int_a^b f(x) d^\# x \right). \quad (15.77)$$

(b) If $f_i, 1 \le i \le n \in \mathbb{N}^\#$ are integrable on $[a, b]$ and $c_i \in \mathbb{R}_c^\#$ are constants, then

$$Ext- \int_a^b \left[ Ext- \sum_{j=1}^n c_i f_i(x) \right] d^\# x = Ext- \sum_{j=1}^n c_i \left( Ext- \int_a^b f_i(x) d^\# x \right). \quad (15.78)$$

**Proof.**

**Theorem 15.41.** If $f$ is and $g$ integrable on $[a, b]$ and $f(x) \le g(x)$ for $x \in [a, b]$, then

$$Ext- \int_a^b f(x) d^\# x \le Ext- \int_a^b g(x) d^\# x. \quad (15.79)$$

**Proof.**

**Theorem 15.42.** If $f(x)$ is integrable on $[a, b]$, then so is $|f(x)|$, and

$$\left| Ext- \int_a^b f(x) d^\# x \right| \le Ext- \int_a^b |f(x)| d^\# x. \quad (15.80)$$

**Theorem 15.43.** If $f(x)$ and $g(x)$ are integrable on $[a, b]$, then so is the product $f(x)g(x)$.

**Proof.**

**Theorem 15.44.** (First Mean Value Theorem for Integrals) Suppose that $u(x)$ is $\#$-continuous and $v(x)$ is integrable and nonnegative on $[a, b]$. Then

$$Ext- \int_a^b f(x) v(x) d^\# x = v(c) \left( Ext- \int_a^b f(x) d^\# x \right) \quad (15.81)$$

for some $c \in [a, b]$.

**Proof.**

**Theorem 15.45.** If $f(x)$ is integrable on $[a, b]$ and $a \le a_1 < b_1 \le b$, then $f(x)$ is integrable on $[a_1, b_1]$.

**Proof.**

**Theorem 15.46.** If $f(x)$ is integrable on $[a, b]$ and $[b, c]$ then $f(x)$ is integrable on $[a, b]$ and

$$Ext- \int_a^c f(x) d^\# x = Ext- \int_a^b f(x) d^\# x + Ext- \int_b^c f(x) d^\# x. \quad (15.82)$$
Since $f$ is #-continuous at $x_0$, there is for each $\varepsilon > 0, \varepsilon \approx 0$ a $\delta > 0, \delta \approx 0$ such that $|f(t) - f(x_0)| < \varepsilon$ if $|x - x_0| < \delta$ and $x < t < x_0$. From (11.84') we get

$$ \left| \frac{F(x) - F(x_0)}{x - x_0} - f(x_0) \right| < \varepsilon \frac{|x - x_0|}{|x - x_0|} = \varepsilon, $$

where $0 < |x - x_0| < \delta$. Therefore $F'^#(x_0) = f(x_0)$.

**Theorem 15.49.** Suppose that $F(x)$ is #-continuous on the #-closed interval $[a, b]$ and #-differentiable on the #-open interval $(a, b)$, and $f^{nt}(x)$ is integrable on $[a, b]$. Suppose also that $F'^#(x) = f(x)$, $a < x < b$. Then

$$ \operatorname{Ext-} \int_a^b f(x) d^\# x = F(b) - F(a). \qquad (15.85) $$

**Proof.** Let $P$ by an partition $P = \{x_i\}_{i=0}^n$, $n \in \mathbb{N}^{\#\mathbb{N}}$ of $[a, b]$, then we get

$$ F(b) - F(a) = \operatorname{Ext-} \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})]. \qquad (15.85') $$

From Theorem 2.3.11, there is in each #-open interval $(x_{j-1}, x_j)$ a point $c_j \in (x_{j-1}, x_j)$ such that $F(x_i) - F(x_{i-1}) = f(c_j)(x_j - x_{j-1})$. Hence, Eq.(11.85') can be written as

$$ F(b) - F(a) = \operatorname{Ext-} \sum_{i=1}^{n} f(c_j)(x_j - x_{j-1}) = \sigma, \qquad (15.85'') $$

where $\sigma$ is a Riemann sum for $f$ over $P$. Since $f$ is integrable on $[a, b]$, there is for each $\varepsilon > 0, \varepsilon \approx 0$ a $\delta > 0, \delta \approx 0$ such that

$$ \left| \sigma - \int_a^b f(x) d^\# x \right| < \varepsilon $$

if $||P|| < \delta$. Therefore,

$$ \left| F(b) - F(a) - \int_a^b f(x) d^\# x \right| < \varepsilon $$

for every $\varepsilon > 0, \varepsilon \approx 0$, which implies 15.85

**Theorem 15.50.** If $f^{\#}(x)$ is integrable on $[a, b]$, then

$$ \operatorname{Ext-} \int_a^b f^{\#}(x) d^{\#} x = f(b) - f(a). \qquad (15.86) $$

**Definition 15.22.** A function $F(x)$ is an #-antiderivative of $f(x)$ on $[a, b]$ if $F(x)$ is #-continuous on $[a, b]$ and #-differentiable on $[a, b]$, with $F'^#(x) = f(x)$, $a < x < b$.

**Theorem 15.50.** If $F(x)$ is an #-antiderivative of $f(x)$ on $[a, b]$, then so is $F(x) + c$ for any constant $c$. Conversely, if $F_1(x)$ and $F_2(x)$ are #-antiderivatives of $f$ on $[a, b]$, then $F_1(x) - F_2(x)$ is constant on $[a, b]$.

**Theorem 15.51. (Fundamental Theorem of Calculus)** If $f(x)$ is #-continuous on $[a, b]$, then $f(x)$ has an #-antiderivative on $[a, b]$. Moreover, if $F(x)$ is any #-antiderivative of $f$ on $[a, b]$, then
**Proof.**

**Theorem 15.47.** If $f(x)$ is integrable on $[a, b]$ and $a \le c \le b$, then the function $F(x)$
defined by

$$F(x) = \operatorname{Ext-} \int_{c}^{x} f(t) d^{\#} t \qquad (15.83)$$

satisfies a Lipschitz condition on $[a, b]$, and is therefore #-continuous on $[a, b]$.

**Proof.**

**Theorem 15.48.** If $f(x)$ is integrable on $[a, b]$ and $a \le c \le b$, then $F(x) = \operatorname{Ext-} \int_{c}^{x} f(t) d^{\#} t$

is #-differentiable at any point $x_0 \in [a, b]$, where $f(x)$ is #-continuous, with

$$F^{\prime\#}(x_0) = f(x_0). \qquad (15.84)$$

If $f(x)$ is #-continuous from the right at $a$, then $F_{+}^{\prime\#}(a) = f(a)$. If $f(x)$ is #-continuous from the left at $b$, then $F_{-}^{\prime\#}(b) = f(b)$.

**Proof.** We consider the case where $a < x_0 < b$. From the equality

$$\frac{1}{x-x_0} \int_{x_0}^{x} f(x_0) dt^{\prime\#} = f(x_0)$$

one obtains

$$\frac{F(x) - F(x_0)}{x - x_0} - f(x_0) = \frac{1}{x - x_0} \int_{x_0}^{x} [f(t) - f(x_0)] dt^{\prime\#}.$$

From this one obtains

$$\left| \frac{F(x) - F(x_0)}{x - x_0} - f(x_0) \right| \le \frac{1}{|x - x_0|} \left| \int_{x_0}^{x} |f(t) - f(x_0)| dt^{\prime\#} \right|. \qquad (15.84')$$

Since $f$ is #-continuous at $x_0$, there is for each $\varepsilon > 0, \varepsilon \approx 0$ a $\delta > 0, \delta \approx 0$ such that $|f(t) - f(x_0)| < \varepsilon$ if $|x - x_0| < \delta$ and $x < t < x_0$. From (11.84') we get

$$\left| \frac{F(x) - F(x_0)}{x - x_0} - f(x_0) \right| < \varepsilon \left| \frac{x - x_0}{|x - x_0|} \right| = \varepsilon,$$

where $0 < |x - x_0| < \delta$. Therefore $F^{\prime\#}(x_0) = f(x_0)$.

**Theorem 15.49.** Suppose that $F(x)$ is #-continuous on the #-closed interval $[a, b]$ and #-differentiable on the #-open interval $(a, b)$, and $f^{\prime\text{int}}(x)$ is integrable on $[a, b]$. Suppose also that $F^{\prime\#}(x) = f(x)$, $a < x < b$. Then

$$\operatorname{Ext-} \int_{a}^{b} f(x) d^{\#} x = F(b) - F(a). \qquad (15.85)$$

**Proof.** Let $P$ by an partition $P = \{x_i\}_{i=0}^n$, $n \in \mathbb{N}^* \mathbb{N}$ of $[a, b]$, then we get

$$F(b) - F(a) = \operatorname{Ext-} \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})]. \qquad (15.85')$$

From Theorem 2.3.11, there is in each #-open interval $(x_{j-1}, x_j)$ a point $c_j \in (x_{j-1}, x_j)$ such that $F(x_i) - F(x_{i-1}) = f(c_j)(x_j - x_{j-1})$. Hence, Eq.(11.85') can be written as
$$ \mathrm{Ext}- \int_a^b f(x) d^\# x = F(b) - F(a). \qquad (15.87) $$

**Theorem 15.52.** (Integration by Parts) If $u^{\prime\#}(x)$ and $v^{\prime\#}(x)$ are integrable on $[a, b]$, then

$$ \mathrm{Ext}- \int_a^b u(x)v^{\prime\#}(x) d^\# x = u(x)v(x) \Big|_a^b - \mathrm{Ext}- \int_a^b u(x)^{\prime\#} v(x) d^\# x. \qquad (15.88) $$

**Theorem 15.53.** Suppose that the transformation $x = \varphi(t)$ maps the interval $c \le t \le d$ into the interval $a \le x \le b$, with $\varphi(c) = \alpha$ and $\varphi(b) = \beta$, and let $f(x)$ be \#-continuous on $[a, b]$. Let $\varphi^{\prime\#}(t)$ be integrable on $[c, d]$. Then

$$ \mathrm{Ext}- \int_{\alpha}^{\beta} f(x) d^{\#} x = \mathrm{Ext}- \int_{c}^{d} f(\varphi(t)) \varphi^{\prime\#}(t) d^{\#} t. \qquad (15.89) $$

**Theorem 15.54.** Suppose that $\varphi^{\prime\#}(t)$ is integrable and $\varphi(t)$ is monotonic on $[c, d]$, and the transformation $x = \varphi(t)$ maps $[c, d]$ onto $[a, b]$. Let $f(x)$ be bounded (hyperbounded) on $[a, b]$. Then $g(t) = f(\varphi(t))\varphi^{\prime\#}(t)$ is integrable on $[c, d]$ if and only if $f(x)$ is integrable over $[a, b]$, and in this case

$$ \mathrm{Ext}- \int_{a}^{b} f(x) d^{\#} x = \mathrm{Ext}- \int_{c}^{d} f(\varphi(t)) \varphi^{\prime\#}(t) d^{\#} t. \qquad (15.90) $$

## 15.14. Improper integrals.

**Definition 15.22.** We say $f(x)$ is locally integrable on an interval $I$ if $f(x)$ is integrable on every finite or hyperfinite \#-closed subinterval of $I$.

**Definition 15.23.** If $f$ is locally integrable on $[a, b]$, we define

$$ \mathrm{Ext}- \int_{a}^{b} f(x) d^{\#} x = \#\text{-lim}_{c \to a, c \neq b} \left( \mathrm{Ext}- \int_{a}^{c} f(x) d^{\#} x \right). \qquad (15.91) $$

**Remark 11.3.** The \#-limit in (15.91) always exists if $[a, b]$ is finite or hyperfinite and $f$ is locally integrable and bounded (hyperbounded) on $[a, b]$. In this case, Definitions 15.70 and 15.91 assign the same value to $\mathrm{Ext}-\int_{a}^{b} f(x) d^{\#} x$ no matter how $f$ is defined. However, the \#-limit may also exist in cases where $b = \infty^{\#}$ or $b < \infty^{\#}$ and $f$ is hyper unbounded as $x$ approaches $b$ from the left.

**Definition 15.24.** In the cases mentioned above, Definition 15.91 assigns a value to an integral that does not exist in the sense of Definition 15.70, and $\mathrm{Ext}-\int_{a}^{b} f(x) d^{\#} x$ is said to be an *improper integral* that \#-converges to the \#-limit in (15.91). We also say in this case that $f$ is integrable on $[a, b]$ and that $\mathrm{Ext}-\int_{a}^{b} f(x) d^{\#} x$ exists. If the \#-limit in (15.91) does not exist (finite or hyperfinite), we say that the improper integral $\mathrm{Ext}-\int_{a}^{b} f(x) d^{\#} x$ \#-diverges, and $f$ is nonintegrable on $[a, b)$. In particular, if $\#\text{-lim}_{c \to a, c \neq b} (\mathrm{Ext}-\int_{a}^{c} f(x) d^{\#} x) = \pm \infty^{\#}$ we say that \#-diverges to $\infty^{\#}$, and we write
$$F(b) - F(a) = \operatorname{Ext-} \sum_{i=1}^{n} f(c_j)(x_j - x_{j-1}) = \sigma, \qquad (15.85'')$$

where $\sigma$ is a Riemann sum for $f$ over $P$. Since $f$ is integrable on $[a, b]$, there is for each $\varepsilon > 0, \varepsilon \approx 0$ a $\delta > 0, \delta \approx 0$ such that

$$\left| \sigma - \int_{a}^{b} f(x) d^{\#} x \right| < \varepsilon$$

if $||P|| < \delta$. Therefore,

$$\left| F(b) - F(a) - \int_{a}^{b} f(x) d^{\#} x \right| < \varepsilon$$

for every $\varepsilon > 0, \varepsilon \approx 0$, which implies 15.85

**Theorem 15.50.** If $f^{\#}(x)$ is integrable on $[a, b]$, then

$$\operatorname{Ext-} \int_{a}^{b} f^{\#}(x) d^{\#} x = f(b) - f(a). \qquad (15.86)$$

**Definition 15.22.** A function $F(x)$ is an \#-antiderivative of $f(x)$ on $[a, b]$ if $F(x)$ is \#-continuous on $[a, b]$ and \#-differentiable on $[a, b]$, with $F^{\#}(x) = f(x)$, $a < x < b$.

**Theorem 15.50.** If $F(x)$ is an \#-antiderivative of $f(x)$ on $[a, b]$, then so is $F(x) + c$ for any constant $c$. Conversely, if $F_1(x)$ and $F_2(x)$ are \#-antiderivatives of $f$ on $[a, b]$, then $F_1(x) - F_2(x)$ is constant on $[a, b]$.

**Theorem 15.51. (Fundamental Theorem of Calculus)** If $f(x)$ is \#-continuous on $[a, b]$, then $f(x)$ has an \#-antiderivative on $[a, b]$. Moreover, if $F(x)$ is any \#-antiderivative of $f$ on $[a, b]$, then

$$\operatorname{Ext-} \int_{a}^{b} f(x) d^{\#} x = F(b) - F(a). \qquad (15.87)$$

**Theorem 15.52.** (Integration by Parts) If $u^{\#}(x)$ and $v^{\#}(x)$ are integrable on $[a, b]$, then

$$\operatorname{Ext-} \int_{a}^{b} u(x) v^{\#}(x) d^{\#} x = u(x) v(x) \Big|_{a}^{b} - \operatorname{Ext-} \int_{a}^{b} u(x)^{\prime \#} v(x) d^{\#} x. \qquad (15.88)$$

**Theorem 15.53.** Suppose that the transformation $x = \varphi(t)$ maps the interval $c \le t \le d$ into the interval $a \le x \le b$, with $\varphi(c) = \alpha$ and $\varphi(d) = \beta$, and let $f(x)$ be \#-continuous on $[a, b]$. Let $\varphi^{\prime}(t)$ be integrable on $[c, d]$. Then

$$\operatorname{Ext-} \int_{a}^{\beta} f(x) d^{\#} x = \operatorname{Ext-} \int_{c}^{d} f(\varphi(t)) \varphi^{\prime \#}(t) d^{\#} t. \qquad (15.89)$$

**Theorem 15.54.** Suppose that $\varphi^{\prime}(t)$ is integrable and $\varphi(t)$ is monotonic on $[c, d]$, and the transformation $x = \varphi(t)$ maps $[c, d]$ onto $[a, b]$. Let $f(x)$ be bounded (hyperbounded) on $[a, b]$. Then $g(t) = f(\varphi(t))\varphi^{\prime}(t)$ is integrable on $[c, d]$ if and only if $f(x)$ is integrable over $[a, b]$, and in this case
$$ \mathrm{Ext}-\int_a^b f(x)d^\#x = \infty^\# \qquad (15.92) $$

or

$$ \mathrm{Ext}-\int_a^b f(x)d^\#x = -\infty^\#, \qquad (15.93) $$

whichever the case may be. Similar comments apply to the next two definitions.
**Definition 15.25.** If $f(x)$ is locally integrable on $(a, b]$, we define

$$ \mathrm{Ext}-\int_a^b f(x)d^\#x = \#\lim_{c \to a^+} \left( \mathrm{Ext}-\int_c^b f(x)d^\#x \right). \qquad (15.94) $$

provided that the $\#\text{-limit}$ exists (finite or hyperfinite). To include the case where $a = -\infty^\#$, we adopt the convention that $-\infty^\# \mp = -\infty^\#$.

**Definition 15.26.** If $f(x)$ is locally integrable on $(a, b)$, we define

$$ \mathrm{Ext}-\int_a^b f(x)d^\#x = \mathrm{Ext}-\int_a^\alpha f(x)d^\#x + \mathrm{Ext}-\int_a^b f(x)d^\#x, \qquad (15.95) $$

where $a < \alpha < b$, provided that both improper integrals on the right exist i.e., finite or hyperfinite.

**Remark 15.4.** Note that the existence and value of $\mathrm{Ext}-\int_a^b f(x)d^\#x$ according to Definition 15.26 do not depend on the particular choice of $\alpha \in (a, b)$.

**Remark 15.5.** When we wish to distinguish between improper integrals and integrals in the sense of Definition 11.70, we will call the latter proper integrals.

**Theorem 15.55.** Suppose that $f_1, f_2, \dots, f_n$ are locally integrable on $[a, b)$ and that $\mathrm{Ext}-\int_a^b f_1(x)d^\#x, \dots, \mathrm{Ext}-\int_a^b f_n(x)d^\#x$ $\#\text{-converge}$. Let $c_1, c_2, \dots, c_n$ be constants. Then

$\mathrm{Ext}-\int_a^b (\mathrm{Ext}-\sum_{i=1}^n c_i f_i(x))d^\#x$ $\#\text{-converges and}$

$$ \mathrm{Ext}-\int_a^b (\mathrm{Ext}-\sum_{i=1}^n c_i f_i(x)) d^\#x = \mathrm{Ext}-\sum_{i=1}^n c_i (\mathrm{Ext}-\int_a^b f_i(x) d^\#x). \qquad (15.96) $$

## 15.15. Improper integrals of nonnegative functions
$f: D \to \mathbb{R}_c^\#$. Absolute Integrability.

**Theorem 15.56.** If $f(x)$ is nonnegative and locally integrable on $[a, b)$, then $\mathrm{Ext}-\int_a^b f(x)d^\#x$ converges if the function

$$ F(x) = \mathrm{Ext}-\int_a^x f(x)d^\#x \qquad (15.97) $$

is bounded (hyperbounded) on $[a, b)$, and $\mathrm{Ext}-\int_a^b f(x)d^\#x = \infty^\#$ if it is not.

**Theorem 15.57.** (Comparison Test) If $f$ and $g$ are locally integrable on $[a, b)$ and $0 \le f(x) \le g(x)$, $0 \le x < b$, then (a) $\mathrm{Ext}-\int_a^b f(x)d^\#x < \infty^\#$ if $\mathrm{Ext}-\int_a^b g(x)d^\#x < \infty^\#$ and (b) $\mathrm{Ext}-\int_a^b f(x)d^\#x = \infty^\#$ if $\mathrm{Ext}-\int_a^b g(x)d^\#x = \infty^\#$.

**Theorem 15.58.** Suppose that $f$ and $g$ are locally integrable on $[a, b)$, $g(x) > 0$ and
$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = \mathrm{Ext}-\int_{c}^{d} f(\varphi(t))\varphi'^{\#}(t)d^{\#}t. \qquad (15.90) $$

## 15.14. Improper integrals.

**Definition 15.22.** We say $f(x)$ is locally integrable on an interval $I$ if $f(x)$ is integrable on every finite or hyperfinite $\#$-closed subinterval of $I$.

**Definition 15.23.** If $f$ is locally integrable on $[a, b]$, we define

$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = \#\lim_{c \to \# b-} \left( \mathrm{Ext}-\int_{a}^{c} f(x)d^{\#}x \right). \qquad (15.91) $$

**Remark 11.3.** The $\#$-limit in (15.91) always exists if $[a, b]$ is finite or hyperfinite and $f$ is locally integrable and bounded (hyperbounded) on $[a, b]$. In this case, Definitions 15.70 and 15.91 assign the same value to $\mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x$ no matter how $f$ is defined. However, the $\#$-limit may also exist in cases where $b = \infty^{\#}$ or $b < \infty^{\#}$ and $f$ is hyper unbounded as $x$ approaches $b$ from the left.

**Definition 15.24.** In the cases mentioned above, Definition 15.91 assigns a value to an integral that does not exist in the sense of Definition 15.70, and $\mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x$ is said to be an **improper integral** that $\#$-converges to the $\#$-limit in (15.91). We also say in this case that $f$ is integrable on $[a, b]$ and that $\mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x$ exists. If the $\#$-limit in (15.91) does not exist (finite or hyperfinite), we say that the improper integral $\mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x$ $\#$-diverges, and $f$ is nonintegrable on $[a, b)$. In particular, if $\#\lim_{c \to \# b-} (\mathrm{Ext}-\int_{a}^{c} f(x)d^{\#}x) = \pm \infty^{\#}$ we say that $\#$-diverges to $\infty^{\#}$, and we write

$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = \infty^{\#} \qquad (15.92) $$

or

$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = -\infty^{\#}, \qquad (15.93) $$

whichever the case may be. Similar comments apply to the next two definitions.

**Definition 15.25.** If $f(x)$ is locally integrable on $(a, b]$, we define

$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = \#\lim_{c \to \# a+} \left( \mathrm{Ext}-\int_{c}^{b} f(x)d^{\#}x \right). \qquad (15.94) $$

provided that the $\#$-limit exists (finite or hyperfinite). To include the case where $a = -\infty^{\#}$, we adopt the convention that $-\infty^{\#} + = -\infty^{\#}$.

**Definition 15.26.** If $f(x)$ is locally integrable on $(a, b)$, we define

$$ \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x = \mathrm{Ext}-\int_{a}^{\alpha} f(x)d^{\#}x + \mathrm{Ext}-\int_{a}^{b} f(x)d^{\#}x, \qquad (15.95) $$

where $a < \alpha < b$, provided that both improper integrals on the right exist i.e., finite or
$f(x) \ge 0$ on some subinterval $[a_1, b) \subset [a, b)$, and

$$ \#\lim_{c \to b-} \frac{f(x)}{g(x)} = M. \qquad (15.98) $$

(a) If $0 < M < \infty^{\#}$, then $\text{Ext}-\int_a^b f(x)d^{\#}x$ and $\text{Ext}-\int_a^b g(x)d^{\#}x$ converge or diverge together.

(b) If $M = \infty^{\#}$ and $\text{Ext}-\int_a^b g(x)d^{\#}x = \infty^{\#}$, then $\text{Ext}-\int_a^b f(x)d^{\#}x = \infty^{\#}$.

(c) If $M = 0$ and $\text{Ext}-\int_a^b g(x)d^{\#}x < \infty^{\#}$, then $\text{Ext}-\int_a^b f(x)d^{\#}x < \infty^{\#}$.

**Definition 15.27.** We say that $f$ is absolutely integrable on $[a, b)$ if $f$ is locally integrable on $[a, b)$ and $\text{Ext}-\int_a^b |f(x)|d^{\#}x < \infty^{\#}$. In this case we also say that $\text{Ext}-\int_a^b f(x)d^{\#}x$ #-converges absolutely or is absolutely #-convergent.

**Theorem 11.59.** If $f$ is locally integrable on $[a, b)$ and $\text{Ext}-\int_a^b |f(x)|d^{\#}x < \infty^{\#}$, then $\text{Ext}-\int_a^b f(x)d^{\#}x$ #-converges: that is, an absolutely #-convergent integral is #-convergent.

**Theorem 15.60.** (Dirichlet's Test) Suppose that $f$ is #-continuous and its #-antiderivative $F(x) = \text{Ext}-\int_a^x f(x)d^{\#}x$ is bounded (hyperbounded) on $[a, b)$. Let $g^{\#}$ be absolutely integrable on $[a, b)$, and suppose that

$$ \#\lim_{c \to b-} g(x) = 0. \qquad (15.99) $$

Then $\text{Ext}-\int_a^x f(x)g(x)d^{\#}x$ #-converges.

**Theorem 15.61.** Suppose that $u(x)$ is #-continuous on $[a, b)$ and $\text{Ext}-\int_a^x u(x)d^{\#}x$ #-diverges. Let $v(x)$ be positive and #-differentiable on $[a, b)$, and suppose that $\#\lim_{c \to b-} v(x) = \infty^{\#}$ and $v^{\prime\#}/v^2$ is absolutely integrable on $[a, b)$. Then $\text{Ext}-\int_a^x u(x)v(x)d^{\#}x$ #-diverges.

**Theorem 15.62.** Suppose that $g(x)$ is monotonic on $[a, b)$ and $\text{Ext}-\int_a^b f(x)d^{\#}x = \infty^{\#}$. Let $f(x)$ be locally integrable on $[a, b)$ and

$$ \text{Ext}- \int_{x_j}^{x_{j+1}} |f(x)| d^{\#}x \ge \rho, j \ge 0 \qquad (15.100) $$

for some positive $\rho$ where $\{x_j\}_{j \in \mathbb{N}^\#}$ is an increasing hyper infinite sequence of points in $[a, b)$ such that $\#\lim_{j \to \infty} x_j = b$ and $x_{j+1}x_j \le M, j \ge 0$, for some $M$. Then

$$ \text{Ext}-\int_a^b |f(x)g(x)|d^{\#}x = \infty^{\#}. \qquad (15.101) $$

## 15.16. Change of Variable in an Improper Integral

**Theorem 11.63.** Suppose that $\varphi(t)$ is monotonic and $\varphi^{\prime\#}(t)$ is locally integrable on either of the half-open intervals $I = [c, d)$ or $(c, d]$, and let $x = \varphi(t)$ map $I$ onto either of the half-open intervals $J = [a, b)$ or $J = (a, b]$. Let $f$ be locally integrable on $J$. Then the improper integrals

$$ \text{Ext}-\int_a^b f(x)d^{\#}x \text{ and } \text{Ext}-\int_a^b f(\varphi)|\varphi^{\prime\#}(t)|d^{\#}t \qquad (15.102) $$

#-diverge or #-converge together, in the latter case to the same value. The same conclusion holds if $\varphi(t)$ and $\varphi^{\prime\#}(t)$ have the stated properties only on the #-open interval $(a, b)$, the transformation $x = \varphi(t)$ maps $(c, d)$ onto $(a, b)$, and $f$ is locally
hyperfinite.

**Remark 15.4.** Note that the existence and value of $Ext-\int_a^b f(x)d^\#x$ according to Definition 15.26 do not depend on the particular choice of $\alpha \in (a, b)$.

**Remark 15.5.** When we wish to distinguish between improper integrals and integrals in the sense of Definition 11.70, we will call the latter proper integrals.

**Theorem 15.55.** Suppose that $f_1, f_2, \dots, f_n$ are locally integrable on $[a, b)$ and that

$Ext-\int_a^b f_1(x)d^\#x, \dots, Ext-\int_a^b f_n(x)d^\#x$ #-converge. Let $c_1, c_2, \dots, c_n$ be constants. Then
$Ext-\int_a^b (Ext-\sum_{i=1}^n c_i f_i(x))d^\#x$ #-converges and

$$ Ext-\int_a^b (Ext-\sum_{i=1}^n c_i f_i(x)) d^\#x = Ext-\sum_{i=1}^n c_i (Ext-\int_a^b f_i(x) d^\#x). \quad (15.96) $$

## 15.15. Improper integrals of nonnegative functions
$f: D \to \mathbb{R}_c^\#$. Absolute Integrability.

**Theorem 15.56.** If $f(x)$ is nonnegative and locally integrable on $[a, b)$, then
$Ext-\int_a^b f(x)d^\#x$ converges if the function

$$ F(x) = Ext-\int_a^x f(x)d^\#x \quad (15.97) $$

is bounded (hyperbounded) on $[a, b)$, and $Ext-\int_a^b f(x)d^\#x = \infty^\#$ if it is not.

**Theorem 15.57.** (Comparison Test) If $f$ and $g$ are locally integrable on $[a, b)$ and $0 \le f(x) \le g(x)$, $0 \le x < b$, then (a) $Ext-\int_a^b f(x)d^\#x < \infty^\#$ if $Ext-\int_a^b g(x)d^\#x < \infty^\#$ and (b) $Ext-\int_a^b f(x)d^\#x = \infty^\#$ if $Ext-\int_a^b g(x)d^\#x = \infty^\#.$

**Theorem 15.58.** Suppose that $f$ and $g$ are locally integrable on $[a, b)$, $g(x) > 0$ and $f(x) \ge 0$ on some subinterval $[a_1, b) \subset [a, b)$, and

$$ \#-\lim_{c \to \infty} \lim_{b \to 0} \frac{f(x)}{g(x)} = M. \quad (15.98) $$

(a) If $0 < M < \infty^\#$, then $Ext-\int_a^b f(x)d^\#x$ and $Ext-\int_a^b g(x)d^\#x$ converge or diverge together.

(b) If $M = \infty^\#$ and $Ext-\int_a^b g(x)d^\#x = \infty^\#$, then $Ext-\int_a^b f(x)d^\#x = \infty^\#.$

(c) If $M = 0$ and $Ext-\int_a^b g(x)d^\#x < \infty^\#$, then $Ext-\int_a^b f(x)d^\#x < \infty^\#.$

**Definition 15.27.** We say that $f$ is absolutely integrable on $[a, b)$ if $f$ is locally integrable on $[a, b)$ and $Ext-\int_a^b |f(x)|d^\#x < \infty^\#$. In this case we also say that $Ext-\int_a^b f(x)d^\#x$ #-converges absolutely or is absolutely #-convergent.

**Theorem 11.59.** If $f$ is locally integrable on $[a, b)$ and $Ext-\int_a^b |f(x)|d^\#x < \infty^\#$, then
$Ext-\int_a^b f(x)d^\#x$ #-converges: that is, an absolutely #-convergent integral is #-convergent.

**Theorem 15.60.** (Dirichlet's Test) Suppose that $f$ is #-continuous and its #-antiderivative $F(x) = Ext-\int_a^x f(x)d^\#x$ is bounded (hyperbounded) on $[a, b)$.
Let $g^\#$ be absolutely integrable on $[a, b)$, and suppose that

$$ \#-\lim_{c \to \infty} \lim_{b \to 0} g(x) = 0. \quad (15.99) $$

Then $Ext-\int_a^x f(x)g(x)d^\#x$ #-converges.
integrable on $(a, b)$.

## 15.17. Generalized integrability criterion due to Lebesgue.

The main result of this section is an integrability criterion due to Lebesgue that does not require computation, but has to do with how badly #-discontinuous a function may be and still be integrable.

**Definition 15.28.** If $f(x)$ is bounded (hyperbounded) on $[a, b]$, the oscillation $W_f[a, b]$ of $f(x)$ on $[a, b]$ is defined by

$$W_f[a, b] = \sup_{a \le x, x' \le b} |f(x) - f(x')| \qquad (15.103)$$

which can also be written as

$$W_f[a, b] = \sup_{a \le x \le b} f(x) - \inf_{a \le x \le b} f(x). \qquad (15.104)$$

**Definition 15.29.** If $a < x < b$, the oscillation $w_f(x)$ of $f(x)$ at $x$ is defined by

$$w_f(x) = \#\lim_{h \to \# 0^+} W_f(x - h, x + h) \qquad (15.105)$$

The corresponding definitions for $x = a$ and $x = b$ are

$$w_f(a) = \#\lim_{h \to \# 0^+} W_f(a, a + h) \quad \text{and} \quad w_f(b) = \#\lim_{h \to \# 0^+} W_f(b - h, b). \qquad (15.106)$$

Note that for a fixed $x \in (a, b)$, $W_f(x - h, x + h)$ is a nonnegative and nondecreasing function of $h$ for $0 < h < \min\{x - a, b - x\}$, therefore, $w_f(x)$ exists and is nonnegative.

**Theorem 15.64.** Let $f$ be defined on $[a, b]$. Then $f$ is #-continuous at $x_0 \in [a, b]$ if and only if $w_f(x) = 0$; #-continuity at $a$ or $b$ means #-continuity from the right or left, respectively.

**Definition 15.30.** A subset $S$ of the $\mathbb{R}_c^\#$ is of Lebesgue measure zero if for every $\varepsilon > 0, \varepsilon \approx 0$, there is a hyperfinite or hyper infinite sequence of open intervals $I_1, I_2, \dots$ such that

$$S \subset \bigcup_{j} I_j \qquad (15.107)$$

and

$$\mathrm{Ext} - \sum_{j=1}^{n} L(I_j) < \varepsilon, n \ge 1. \qquad (15.108)$$

Note that any subset of a set of Lebesgue measure zero is also of Lebesgue measure zero.

**Example 15.1.** Any hyperfinite set $S = \{x_i\}_{i \in \mathbb{N}}$, $\mathbf{n} \in \mathbb{N}^*\mathbb{N}$ is of Lebesgue measure zero, since we can choose #-open intervals $I_1, I_2, \dots, I_n$ such that $x_j \in I_j$ and $L(I_j) < \varepsilon/\mathbf{n}$, $1 \le j \le \mathbf{n}$.

**Definition 15.31.** An infinite set $S \subset \mathbb{R}_c^\#$ is hyper denumerable if its members can be listed in a hyper infinite sequence (that is, in a one-to-one correspondence with the positive hyper integers); thus, $S = \{x_i\}_{i \in \mathbb{N}}$. An infinite set that does not have this property is hyper non hyper denumerable.

**Example 15.2.** Any denumerable set $S = \{x_i\}_{i \in \mathbb{N}}$ is of Lebesgue measure zero, since if $\varepsilon > 0, \varepsilon \approx 0$, it is possible to choose open intervals $I_1, I_2, \dots$, so that $x_j \in I_j$ and $L(I_j) < 2^{-j}\varepsilon, j \ge 1$. Then (15.108) holds since
**Theorem 15.61.** Suppose that $u(x)$ is #-continuous on $[a, b)$ and $Ext-\int_a^x u(x)d^\#x$ #-diverges. Let $v(x)$ be positive and #-differentiable on $[a, b)$, and suppose that $\#-\lim_{c \to \# b-} v(x) = \infty^\#$ and $v'^\#/v^2$ is absolutely integrable on $[a, b)$. Then $Ext-\int_a^x u(x)v(x)d^\#x$ #-diverges.

**Theorem 15.62.** Suppose that $g(x)$ is monotonic on $[a, b)$ and $Ext-\int_a^b f(x)d^\#x = \infty^\#$. Let $f(x)$ be locally integrable on $[a, b)$ and

$$ Ext- \int_{x_j}^{x_{j+1}} |f(x)| d^\#x \ge \rho, j \ge 0 \qquad (15.100) $$

for some positive $\rho$ where $\{x_j\}_{j \in \mathbb{N}^\#}$ is an increasing hyper infinite sequence of points in $[a, b)$ such that $\#-\lim_{j \to \infty} \infty^\# x_j = b$ and $x_{j+1}x_j \le M, j \ge 0$, for some $M$. Then

$$ Ext- \int_a^b |f(x)g(x)| d^\#x = \infty^\#. \qquad (15.101) $$

## 15.16. Change of Variable in an Improper Integral

**Theorem 11.63.** Suppose that $\varphi(t)$ is monotonic and $\varphi'^\#(t)$ is locally integrable on either of the half-open intervals $I = [c, d)$ or $(c, d]$, and let $x = \varphi(t)$ map $I$ onto either of the half-open intervals $J = [a, b)$ or $J = (a, b]$. Let $f$ be locally integrable on $J$. Then the improper integrals

$$ Ext- \int_a^b f(x) d^\#x \text{ and } Ext- \int_a^b f(\varphi) |\varphi'^\#(t)| d^\#t \qquad (15.102) $$

#-diverge or #-converge together, in the latter case to the same value. The same conclusion holds if $\varphi(t)$ and $\varphi'^\#(t)$ have the stated properties only on the #-open interval $(a, b)$, the transformation $x = \varphi(t)$ maps $(c, d)$ onto $(a, b)$, and $f$ is locally integrable on $(a, b)$.

## 15.17. Generalized integrability criterion due to Lebesgue.

The main result of this section is an integrability criterion due to Lebesgue that does not require computation, but has to do with how badly #-discontinuous a function may be and still be integrable.

**Definition 15.28.** If $f(x)$ is bounded (hyperbounded) on $[a, b]$, the oscillation $W_f[a, b]$ of $f(x)$ on $[a, b]$ is defined by

$$ W_f[a, b] = \sup_{a \le x, x' \le b} |f(x) - f(x')| \qquad (15.103) $$

which can also be written as

$$ W_f[a, b] = \sup_{a \le x \le b} f(x) - \inf_{a \le x \le b} f(x). \qquad (15.104) $$

**Definition 15.29.** If $a < x < b$, the oscillation $w_f(x)$ of $f(x)$ at $x$ is defined by

$$ w_f(x) = \#\lim_{h \to \infty, h \ne 0} W_f(x - h, x + h) \qquad (15.105) $$

The corresponding definitions for $x = a$ and $x = b$ are

$$ w_f(a) = \#\lim_{h \to \infty, h \ne 0} W_f(a, a + h) \text{ and } w_f(b) = \#\lim_{h \to \infty, h \ne 0} W_f(b - h, b). \qquad (15.106) $$

Note that for a fixed $x \in (a, b)$, $W_f(x - h, x + h)$ is a nonnegative and nondecreasing
$$Ext-\sum_{j=1}^{n} 2^{-j} = 1 - 2^{-n} < 1.$$

**Theorem 15.64.** If $w_f(x) < \varepsilon, \varepsilon \approx 0$, for $a \le x \le b$, then there is a $\delta > 0, \delta \approx 0$ such that $W_f[a, b] < \varepsilon$, provided that $a_1, b_1 \subset [a, b]$ and $b_1 - a_1 < \delta$.

**Theorem 15.65.** Let $f$ be bounded (hyperbounded) on $[a, b]$ and define $E_\rho = \{x \in [a, b]|w_f(x) > \rho\}$. Then $E_\rho$ is #-closed; and $f$ is integrable on $[a, b]$ if and only if for every pair of positive numbers $\rho$ and $\delta$, $E_\rho$ can be covered by hyper finitely many open intervals $I_1, I_2, \dots, I_p, \mathbf{p} \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$Ext-\sum_{j=1}^{p} L(I_j) < \delta. \qquad (15.109)$$

**Theorem 15.66.** A bounded (hyperbounded) function $f$ is integrable on a finite or hyperfinite interval $[a, b]$ if and only if the set $S$ of #-discontinuities of $f$ in $[a, b]$ is of Lebesgue measure zero.

# 16. Hyper infinite external sequences and series

## 16.1. Hyper infinite external sequences

An hyper infinite sequence (or hypersequence) of $\mathbb{R}_c^\#$-real numbers is a $\mathbb{R}_c^\#$-valued function defined on a set of hyperintegers $\{n|n \in \mathbb{N}^\# \wedge n \ge k \in \mathbb{N}\}$. We call the values of the function the terms of the hypersequence. We denote a hypersequence by listing its terms in order; thus, $\{s_n\}_k^{\infty} = \{s_k, s_{k+1}, \dots\}$. We often write $\{s_n\}_{n \in \mathbb{N}^\#}$ or simple $\{s_n\}$ for a shot.

**Definition 16.1.** A hyper infinite sequence $\{s_n\}_k^{\infty}$ converges to a limit $s \in \mathbb{R}_c^\#$ if for every $\varepsilon \approx 0, \varepsilon > 0$ there is an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$|s_n - s| < \varepsilon \text{ if } n \ge N \qquad (16.1)$$

In this case we say that $\{s_n\}$ is #-convergent and write

$$\#- \lim_{n \to \infty} \infty^\# s_n = s. \qquad (16.2)$$

A hyper infinite sequence that does not #-converge diverges, or is #-divergent.

**Theorem 16.1.** The #-limit of a #-convergent hypersequence is unique:

**Proof.** Suppose that $\#-\lim_{n \to \infty} \infty^\# s_n = s_1$ and $\#-\lim_{n \to \infty} \infty^\# s_n = s_2$. We must show that $s = s'$. Let $\varepsilon \approx 0, \varepsilon > 0$. From Definition 10.1, there are hyperintegers $N_1$ and $N_2$ such that $|s_n - s_1| < \varepsilon$ if $n \ge N_1$, and $|s_n - s_2| < \varepsilon$ if $n \ge N_2$. These inequalities both hold if $n \ge N = \max(N_1, N_2)$, which implies that: $|s_1 - s_2| < 2\varepsilon$. Since this inequality holds for every $\varepsilon \approx 0, \varepsilon > 0$ and $|s_1 - s_2|$ is independent of $\varepsilon$, we conclude that $|s_1 - s_2| = 0$; that is, $s_1 = s_2$.

**Definition 16.2.** A hypersequence $\{s_n\}$ is bounded above if there is a hyperreal number

$b \in \mathbb{R}_c^\#$ such that $s_n \le b$ for all $n \in \mathbb{N}^\#$; bounded below if there is a real number $a \in \mathbb{R}_c^\#$ such that $s_n \ge a$ for all $n \in \mathbb{N}^\#$; or bounded if there is a real number $r \in \mathbb{R}_c^\#$ such that $|s_n| \le r$ for all $n \in \mathbb{N}^\#$.

**Theorem 16.2.** Any #-convergent hypersequence $\{s_n\}$ is bounded or hyperbounded.

**Proof.** By taking $\varepsilon = 1$ in Eq.(16.1), we see that if $\#-\lim_{n \to \infty} \infty^\# s_n = s$, then there is an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that $|s_n - s| < 1$ if $n \ge N$. Therefore,
function of $h$ for $0 < h < \min\{x-a, b-x\}$, therefore, $w_f(x)$ exists and is nonnegative.

**Theorem 15.64.** Let $f$ be defined on $[a, b]$. Then $f$ is #-continuous at $x_0 \in [a, b]$ if and only if $w_f(x) = 0$; #-continuity at $a$ or $b$ means #-continuity from the right or left, respectively.

**Definition 15.30.** A subset $S$ of the $\mathbb{R}_c^\#$ is of Lebesgue measure zero if for every $\varepsilon > 0, \varepsilon \approx 0$, there is a hyperfinite or hyper infinite sequence of open intervals $I_1, I_2, \dots$ such that

$$ S \subset \bigcup_{j} I_j \qquad (15.107) $$

and

$$ \mathrm{Ext} - \sum_{j=1}^{n} L(I_j) < \varepsilon, n \ge 1. \qquad (15.108) $$

Note that any subset of a set of Lebesgue measure zero is also of Lebesgue measure zero.

**Example 15.1.** Any hyperfinite set $S = \{x_i\}_{i \in \mathbb{N}}$, $\mathbf{n} \in \mathbb{N}^*\mathbb{N}$ is of Lebesgue measure zero, since we can choose #-open intervals $I_1, I_2, \dots, I_n$ such that $x_j \in I_j$ and $L(I_j) < \varepsilon/\mathbf{n}$, $1 \le j \le \mathbf{n}$.

**Definition 15.31.** An infinite set $S \subset \mathbb{R}_c^\#$ is hyper denumerable if its members can be listed in a hyper infinite sequence (that is, in a one-to-one correspondence with the positive hyper integers); thus, $S = \{x_i\}_{i \in \mathbb{N}}$. An infinite set that does not have this property is hyper non hyper denumerable.

**Example 15.2.** Any denumerable set $S = \{x_i\}_{i \in \mathbb{N}}$ is of Lebesgue measure zero, since if $\varepsilon > 0, \varepsilon \approx 0$, it is possible to choose open intervals $I_1, I_2, \dots$, so that $x_j \in I_j$ and $L(I_j) < 2^{-j}\varepsilon, j \ge 1$. Then (15.108) holds since

$$ \mathrm{Ext} - \sum_{j=1}^{n} 2^{-j} = 1 - 2^{-\mathbf{n}} < 1. $$

**Theorem 15.64.** If $w_f(x) < \varepsilon, \varepsilon \approx 0$, for $a \le x \le b$, then there is a $\delta > 0, \delta \approx 0$ such that $W_f[a, b] < \varepsilon$, provided that $a_1, b_1 \subset [a, b]$ and $b_1 - a_1 < \delta$.

**Theorem 15.65.** Let $f$ be bounded (hyperbounded) on $[a, b]$ and define $E_\rho = \{x \in [a, b] | w_f(x) > \rho\}$. Then $E_\rho$ is #-closed; and $f$ is integrable on $[a, b]$ if and only if for every pair of positive numbers $\rho$ and $\delta$, $E_\rho$ can be covered by hyper finitely many open intervals $I_1, I_2, \dots, I_\rho, \mathbf{p} \in \mathbb{N}^*\mathbb{N}$ such that

$$ \mathrm{Ext} - \sum_{j=1}^{\mathbf{p}} L(I_j) < \delta. \qquad (15.109) $$

**Theorem 15.66.** A bounded (hyperbounded) function $f$ is integrable on a finite or hyperfinite interval $[a, b]$ if and only if the set $S$ of #-discontinuities of $f$ in $[a, b]$ is of Lebesgue measure zero.

# 16. Hyper infinite external sequences and series

## 16.1. Hyper infinite external sequences

An hyper infinite sequence (or hypersequence) of $\mathbb{R}_c^\#$-real numbers is a $\mathbb{R}_c^\#$-valued function defined on a set of hyperintegers $\{n | n \in \mathbb{N}^* \wedge n \ge k \in \mathbb{N}\}$. We call the values
$s_n = |(s_n - s) + s| \le |s_n - s| + |s| < 1 + |s|$ if $n \ge N$; and
$s_n \le \max\{(\max_{1 \le j \le N-1}\{|s_0|, |s_1|, \dots, |s_{N-1}|\}), 1 + |s|\}$ for all $n \in \mathbb{N}^\#$, so $\{s_n\}$ is bounded.

**Definition 16.3.** (Sequences Diverging to $\pm\infty$#). We say that

$$ \#\lim_{n \to \#} \infty^\# s_n = +\infty^\# $$

if for any hyperreal number $a, s_n > a$ for any $n \ge N \in \mathbb{N}^\#\mathbb{N}$. Similarly,

$$ \#\lim_{n \to \#} \infty^\# s_n = -\infty^\# $$

if for any hyperreal number $a, s_n < a$ for any $n \ge N \in \mathbb{N}^\#\mathbb{N}$. However, we do not regard $\{s_n\}$ as $\#\text{-convergent}$ unless $\#\lim_{n \to \#} \infty^\# s_n$

is finite or hyperfinite, as required by Definition 16.1. To emphasize this distinction, we say that $\{s_n\}$ diverges to $\infty^\#(-\infty^\#)$ if $\#\lim_{n \to \#} \infty^\# s_n = \infty^\#(-\infty^\#)$.

**Theorem 16.3.** Assume that a nonempty set $S \subset \mathbb{R}_c^\#$ of real $\mathbb{R}_c^\#$-numbers has a supremum $\sup(S)$, then $\sup S$ is the unique hyperreal number $\beta \in \mathbb{R}_c^\#$ such that

(a) $x \le \beta$ for all $x \in S$

(b) if $\varepsilon > 0, \varepsilon \approx 0$ (no matter how infinite small) there is an $x_0 \in S$ such that $x_0 > \beta - \varepsilon$.

**Proof.** We first show that $\beta = \sup S$ has properties (a) and (b). Since $\beta$ is an upper bound of $S$, it must satisfy (a). Since any hyperreal number $\alpha$ less than $\beta$ can be written as $\alpha = \beta - \varepsilon$ with $\varepsilon = \beta - \alpha > 0$, (b) is just another way of saying that no number less than $\beta$ is an upper bound of $S$. Hence, $\beta = \sup S$ satisfies (a) and (b). Now we show that there cannot be more than one hyperreal number with properties (a) and (b).

Suppose that $\beta_1 < \beta_2$ and $\beta_2$ has property (b); thus, if $\varepsilon > 0$, there is an $x_0 \in S$ such that $x_0 > \beta_2 - \varepsilon$. Then, by taking $\varepsilon = \beta_2 - \beta_1$, we see that there is an $x_0 \in S$ such that $x_0 > \beta_2 - (\beta_2 - \beta_1) = \beta_1$, so $\beta_1$ cannot have property (a). Therefore, there cannot be more than one hyperreal number that satisfies both (a) and (b).

**Definition 16.4.** A hypersequence $\{s_n\}_{n \in \mathbb{N}^\#}$ is nondecreasing if $s_n \ge s_{n-1}$ for all $n \in \mathbb{N}^\#$, or nonincreasing if $s_n \le s_{n-1}$ for all $n \in \mathbb{N}^\#$. A monotonic hyper infinite sequence is a hyper infinite sequence that is either nonincreasing or nondecreasing. If $s_n > s_{n-1}$ for all $n \in \mathbb{N}^\#$, then $\{s_n\}_{n \in \mathbb{N}^\#}$ is increasing, while if $s_n < s_{n-1}$ for all $n \in \mathbb{N}^\#$, $\{s_n\}_{n \in \mathbb{N}^\#}$ is decreasing.

**Theorem 16.4.** (a) If $\{s_n\}_{n \in \mathbb{N}^\#}$ is nondecreasing and there exists $\sup\{s_n|n \in \mathbb{N}^\#\}$ then
$$ \#\lim_{n \to \#} \infty^\# s_n = \sup\{s_n|n \in \mathbb{N}^\#\}. $$

(b) If $\{s_n\}_{n \in \mathbb{N}^\#}$ is nonincreasing and there exists $\inf\{s_n|n \in \mathbb{N}^\#\}$ then
$$ \#\lim_{n \to \#} \infty^\# s_n = \inf\{s_n|n \in \mathbb{N}^\#\}. $$

**Proof.** (a) Let $\beta = \sup\{s_n|n \in \mathbb{N}^\#\}$. . If $\beta < +\infty^\#$, Theorem 16.3 implies that if $\varepsilon > 0$ then $\beta - \varepsilon < s_N \le \beta$ for some hyperinteger $N \in \mathbb{N}^\#\mathbb{N}$. Since $s_N \le s_n \le \beta$ if $n \ge N$, it follows that $\beta - \varepsilon < s_n \le \beta$ if $n \ge N$. This implies that $|s_n - \beta| < \varepsilon$ if $n \ge N$, so $\#\lim_{n \to \#} \infty^\# s_n = \beta$, by definition of the $\#\text{-limit}$. If $\beta = +\infty^\#$ and $b$ is any hyperreal number, then $s_N > b$ for some hyperinteger $N$. Then $s_n > b$ for $n \ge N$, so $\#\lim_{n \to \#} \infty^\# s_n = +\infty^\#$.

**Theorem 16.5.** (Generalized Nested Intervals Theorem)

Let $\{I_n\}_{n \in \mathbb{N}^\#} = \{[a_n, b_n]\}_{n \in \mathbb{N}^\#}$, $[a_n, b_n] \subset \mathbb{R}_c^\#$ be a hyper infinite sequence of $\#\text{-closed}$ intervals satisfying each of the following conditions:

(i) $I_1 \supseteq I_2 \supseteq I_3 \supseteq \dots \supseteq I_n \supseteq \dots$,

(ii) $b_n - a_n \to \# 0$ as $n \to \infty^\#$.
of the function the terms of the hypersequence. We denote a hypersequence by listing its terms in order; thus, ${s_n}_k^\infty = \{s_k, s_{k+1}, \dots\}$. We often write ${s_n}_{n \in \mathbb{N}^\#}$ or simple ${s_n}$ for a shot.

**Definition 16.1.** A hyper infinite sequence ${s_n}_k^\infty$ converges to a limit $s \in \mathbb{R}_c^\#$ if for every $\varepsilon \approx 0, \varepsilon > 0$ there is an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$|s_n - s| < \varepsilon \text{ if } n \ge N \qquad (16.1)$$

In this case we say that ${s_n}$ is #-convergent and write

$$\#- \lim_{n \to \#} \infty^\# s_n = s. \qquad (16.2)$$

A hyper infinite sequence that does not #-converge diverges, or is #-divergent.

**Theorem 16.1.** The #-limit of a #-convergent hypersequence is unique:

**Proof.** Suppose that $\#-\lim_{n \to \#} \infty^\# s_n = s_1$ and $\#-\lim_{n \to \#} \infty^\# s_n = s_2$. We must show that $s = s'$. Let $\varepsilon \approx 0, \varepsilon > 0$. From Definition 10.1, there are hyperintegers $N_1$ and $N_2$ such that $|s_n - s_1| < \varepsilon$ if $n \ge N_1$, and $|s_n - s_2| < \varepsilon$ if $n \ge N_2$. These inequalities both hold if $n \ge N = \max(N_1, N_2)$, which implies that: $|s_1 - s_2| < 2\varepsilon$. Since this inequality holds for every $\varepsilon \approx 0, \varepsilon > 0$ and $|s_1 - s_2|$ is independent of $\varepsilon$, we conclude that $|s_1 - s_2| = 0$; that is, $s_1 = s_2$.

**Definition 16.2.** A hypersequence ${s_n}$ is bounded above if there is a hyperreal number

$b \in \mathbb{R}_c^\#$ such that $s_n \le b$ for all $n \in \mathbb{N}^\#$; bounded below if there is a real number $a \in \mathbb{R}_c^\#$ such that $s_n \ge a$ for all $n \in \mathbb{N}^\#$; or bounded if there is a real number $r \in \mathbb{R}_c^\#$ such that $|s_n| \le r$ for all $n \in \mathbb{N}^\#$.

**Theorem 16.2.** Any #-convergent hypersequence ${s_n}$ is bounded or hyperbounded.

**Proof.** By taking $\varepsilon = 1$ in Eq. (16.1), we see that if $\#-\lim_{n \to \#} \infty^\# s_n = s$, then there is an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that $|s_n - s| < 1$ if $n \ge N$. Therefore,

$s_n = |(s_n - s) + s| \le |s_n - s| + |s| < 1 + |s|$ if $n \ge N$; and
$s_n \le \max\{(\max_{1 \le i \le N-1}\{|s_0|, |s_1|, \dots, |s_{N-1}|\}), 1 + |s|\}$ for all $n \in \mathbb{N}^\#$, so ${s_n}$ is bounded.

**Definition 16.3.** (Sequences Diverging to $\pm\infty^\#$). We say that

$$\#-\lim_{n \to \#} \infty^\# s_n = +\infty^\#$$

if for any hyperreal number $a, s_n > a$ for any $n \ge N \in \mathbb{N}^\# \setminus \mathbb{N}$. Similarly,

$$\#-\lim_{n \to \#} \infty^\# s_n = -\infty^\#$$

if for any hyperreal number $a, s_n < a$ for any $n \ge N \in \mathbb{N}^\# \setminus \mathbb{N}$. However, we do not regard ${s_n}$ as #-convergent unless $\#-\lim_{n \to \#} \infty^\# s_n$

is finite or hyperfinite, as required by Definition 16.1. To emphasize this distinction, we say that ${s_n}$ diverges to $\infty^\#(-\infty^\#)$ if $\#-\lim_{n \to \#} \infty^\# s_n = \infty^\#(-\infty^\#)$.

**Theorem 16.3.** Assume that a nonempty set $S \subset \mathbb{R}_c^\#$ of real $\mathbb{R}_c^\#$-numbers has a supremum $\sup(S)$, then $\sup S$ is the unique hyperreal number $\beta \in \mathbb{R}_c^\#$ such that (a) $x \le \beta$ for all $x \in S$

(b) if $\varepsilon > 0, \varepsilon \approx 0$ (no matter how infinite small) there is an $x_0 \in S$ such that $x_0 > \beta - \varepsilon$.

**Proof.** We first show that $\beta = \sup S$ has properties (a) and (b). Since $\beta$ is an upper bound of $S$, it must satisfy (a). Since any hyperreal number $\alpha$ less than $\beta$ can be written as $\alpha = \beta - \varepsilon$ with $\varepsilon = \beta - \alpha > 0$, (b) is just another way of saying that no number less than $\beta$ is an upper bound of $S$. Hence, $\beta = \sup S$ satisfies (a) and (b). Now we show that there cannot be more than one hyperreal number with properties
Then $\bigcap_{n=1}^{\infty} I_n$ consists of exactly one hyperreal number $x \in \mathbb{R}_c^#$. Moreover both hyper infinite sequences $\{a_n\}$ and $\{b_n\}$ #-converge to $x$.

**Proof.** See proof to Theorem 8.11.

**Theorem 16.6.** (Generalized Bolzano-Weierstrass Theorem) Every bounded (hyperbounded) hyper infinite sequence $\{s_n\}_{n \in \mathbb{N}^#}$ has a #-convergent sub hyper infinite sequence.

**Proof.** Let $\{s_n\}_{n \in \mathbb{N}^#}$ be a bounded hyper infinite sequence. Then, there exists an interval $[a_1, b_1]$ such that: (i) $a_1, b_1 \in \mathbb{Q}^#$ and (ii) $a_1 \le s_n \le b_1$ for all $n \in \mathbb{N}^#$.

Either $[a_1, \frac{a_1+b_1}{2}]$ or $[\frac{a_1+b_1}{2}, b_1]$ contains hyperinfinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$.

That is, there exists hyper infinitely many $n \in \mathbb{N}^#$ such that $a_n$ is in $[a_1, \frac{a_1+b_1}{2}]$, or there exists hyper infinitely many $n \in \mathbb{N}^#$ such that $a_n$ is in $[\frac{a_1+b_1}{2}, b_1]$.

If $[a_1, \frac{a_1+b_1}{2}]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$, let $[a_2, b_2] = [a_1, \frac{a_1+b_1}{2}]$. Otherwise, let $[a_2, b_2] = [\frac{a_1+b_1}{2}, b_1]$.

Either $[a_2, \frac{a_2+b_2}{2}]$ or $[\frac{a_2+b_2}{2}, b_2]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$.

If $[a_2, \frac{a_2+b_2}{2}]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$, let $[a_3, b_3] = [a_2, \frac{a_2+b_2}{2}]$. Otherwise, let $[a_3, b_3] = [\frac{a_2+b_2}{2}, b_2]$.

By hyper infinite induction, we can continue this construction and obtain a hyper infinite sequence of intervals $\{[a_n, b_n]\}_{n \in \mathbb{N}^#}$ such that:

(i) for each $n \in \mathbb{N}^#$, interval $[a_n, b_n]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$,

(ii) for each $n \in \mathbb{N}^#$, $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$ and

(iii) for each $n \in \mathbb{N}^#$, $b_{n+1} - a_{n+1} = \frac{1}{2}(b_n - a_n)$.

The nested intervals theorem implies that the intersection $\bigcap_{n \in \mathbb{N}^#} [a_n, b_n]$ of all of the intervals $[a_n, b_n]$ is a single point $s$. We will now construct a sub hyper infinite sequence of $\{s_n\}_{n \in \mathbb{N}^#}$ which will #-converge to $s$.

Since $[a_1, b_1]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$, there exists $k_1 \in \mathbb{N}^#$ such that $s_{k_1}$ is in $[a_1, b_1]$. Since $[a_2, b_2]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$, there exists $k_2 \in \mathbb{N}^#$, $k_2 > k_1$ such that $s_{k_2}$ is in $[a_2, b_2]$. Since $[a_3, b_3]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^#}$, there exists $k_3 \in \mathbb{N}^#$, $k_3 > k_2$ such that $s_{k_3}$ is in $[a_3, b_3]$. Continuing this process by hyper infinite induction, we obtain a hyper infinite sequence $\{s_{k_n}\}_{n \in \mathbb{N}^#}$ such that $s_{k_n} \in [a_n, b_n]$ for each $n \in \mathbb{N}^#$. The hypersequence $\{s_{k_n}\}_{n \in \mathbb{N}^#}$ is a sub hyper infinite sequence of $\{s_n\}_{n \in \mathbb{N}^#}$ since $k_{n+1} > k_n$ for each $n \in \mathbb{N}^#$. Since $\#\text{-lim}_{n \to \infty} a_n = s$ and $\#\text{-lim}_{n \to \infty} b_n = s$ and $a_n \le s_n \le b_n$ for each $n \in \mathbb{N}^#$, the squeeze theorem implies that $\#\text{-lim}_{n \to \infty} s_n = s$.

## 16.2. Hyper infinite external series of constant.

**Definition 16.5.** If $\{a_n\}_k^\infty$ is an hyper infinite external sequence of Cauchy hyperreal numbers, the symbol

$$ \mathrm{Ext} - \sum_{n=k}^{\infty} a_n \qquad (16.3) $$

is an hyper infinite series, and $a_n$ is the $n$-th term of the hyper infinite series.
(a) and (b).

Suppose that $\beta_1 < \beta_2$ and $\beta_2$ has property (b); thus, if $\varepsilon > 0$, there is an $x_0 \in S$
such that $x_0 > \beta_2 - \varepsilon$. Then, by taking $\varepsilon = \beta_2 - \beta_1$, we see that there is an $x_0 \in S$
such that $x_0 > \beta_2 - (\beta_2 - \beta_1) = \beta_1$, so $\beta_1$ cannot have property (a). Therefore, there
cannot be more than one hyperreal number that satisfies both (a) and (b).

**Definition 16.4.** A hypersequence $\{s_n\}_{n \in \mathbb{N}^\#}$ is nondecreasing if $s_n \ge s_{n-1}$ for all $n \in \mathbb{N}^\#$, or nonincreasing if $s_n \le s_{n-1}$ for all $n \in \mathbb{N}^\#$. A monotonic hyper infinite sequence is a hyper infinite sequence that is either nonincreasing or nondecreasing. If $s_n > s_{n-1}$ for all $n \in \mathbb{N}^\#$, then $\{s_n\}_{n \in \mathbb{N}^\#}$ is increasing, while if $s_n < s_{n-1}$ for all $n \in \mathbb{N}^\#$, $\{s_n\}_{n \in \mathbb{N}^\#}$ is decreasing.

**Theorem 16.4. (a)** If $\{s_n\}_{n \in \mathbb{N}^\#}$ is nondecreasing and there exists $\sup\{s_n|n \in \mathbb{N}^\#\}$ then
$$ \#\text{-lim}_{n \to \#} \infty^\# s_n = \sup\{s_n|n \in \mathbb{N}^\#\}. $$

(b) If $\{s_n\}_{n \in \mathbb{N}^\#}$ is nonincreasing and there exists $\inf\{s_n|n \in \mathbb{N}^\#\}$ then
$$ \#\text{-lim}_{n \to \#} \infty^\# s_n = \inf\{s_n|n \in \mathbb{N}^\#\}. $$

**Proof.** (a) Let $\beta = \sup\{s_n|n \in \mathbb{N}^\#\}$. . If $\beta < +\infty^\#$, Theorem 16.3 implies that if $\varepsilon > 0$ then $\beta - \varepsilon < s_N \le \beta$ for some hyperinteger $N \in \mathbb{N}^\# \mathbb{N}$. Since $s_N \le s_n \le \beta$ if $n \ge N$, it follows that $\beta - \varepsilon < s_n \le \beta$ if $n \ge N$. This implies that $|s_n - \beta| < \varepsilon$ if $n \ge N$, so $\#\text{-lim}_{n \to \#} \infty^\# s_n = \beta$, by definition of the \#-limit. If $\beta = +\infty^\#$ and $b$ is any hyperreal number, then $s_N > b$ for some hyperinteger $N$. Then $s_n > b$ for $n \ge N$, so $\#\text{-lim}_{n \to \#} \infty^\# s_n = +\infty^\#$.

**Theorem 16.5. (Generalized Nested Intervals Theorem)**

Let $\{I_n\}_{n \in \mathbb{N}^\#} = \{[a_n, b_n]\}_{n \in \mathbb{N}^\#}$, $[a_n, b_n] \subset \mathbb{R}_c^\#$ be a hyper infinite sequence of \#-closed intervals satisfying each of the following conditions:

(i) $I_1 \supseteq I_2 \supseteq I_3 \supseteq \dots \supseteq I_n \supseteq \dots$,

(ii) $b_n - a_n \to \# 0$ as $n \to \infty^\#$.

Then $\cap_{n=1}^\infty I_n$ consists of exactly one hyperreal number $x \in \mathbb{R}_c^\#$. Moreover both hyper infinite sequences $\{a_n\}$ and $\{b_n\}$ \#-converge to $x$.

**Proof.** See proof to Theorem 8.11.

**Theorem 16.6. (Generalized Bolzano-Weierstrass Theorem)** Every bounded (hyperbounded) hyper infinite sequence $\{s_n\}_{n \in \mathbb{N}^\#}$ has a \#-convergent sub hyper infinite sequence.

**Proof.** Let $\{s_n\}_{n \in \mathbb{N}^\#}$ be a bounded hyper infinite sequence. Then, there exists an interval $[a_1, b_1]$ such that: (i) $a_1, b_1 \in \mathbb{Q}^\#$ and (ii) $a_1 \le s_n \le b_1$ for all $n \in \mathbb{N}^\#$.

Either $[a_1, \frac{a_1+b_1}{2}]$ or $[\frac{a_1+b_1}{2}, b_1]$ contains hyperinfinitely many terms of $\{s_n\}_{n \in \mathbb{N}^\#}$.

That is, there exists hyper infinitely many $n \in \mathbb{N}^\#$ such that $a_n$ is in $[a_1, \frac{a_1+b_1}{2}]$, or there exists hyper infinitely many $n \in \mathbb{N}^\#$ such that $a_n$ is in $[\frac{a_1+b_1}{2}, b_1]$.

If $[a_1, \frac{a_1+b_1}{2}]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^\#}$, let $[a_2, b_2] = [a_1, \frac{a_1+b_1}{2}]$. Otherwise, let $[a_2, b_2] = [\frac{a_1+b_1}{2}, b_1]$.

Either $[a_2, \frac{a_2+b_2}{2}]$ or $[\frac{a_2+b_2}{2}, b_2]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^\#}$.

If $[a_2, \frac{a_2+b_2}{2}]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^\#}$, let $[a_3, b_3] = [a_2, \frac{a_2+b_2}{2}]$. Otherwise, let $[a_3, b_3] = [\frac{a_2+b_2}{2}, b_2]$.

By hyper infinite induction, we can continue this construction and obtain a hyper infinite sequence of intervals $\{[a_n, b_n]\}_{n \in \mathbb{N}^\#}$ such that:

(i) for each $n \in \mathbb{N}^\#$, interval $[a_n, b_n]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^\#}$,
We say that $\text{Ext}-\sum_{n=k}^{\infty} a_n$ #-converges to the sum $A \in \mathbb{R}_c^\#$, and write

$$ \text{Ext}-\sum_{n=k}^{\infty} a_n = A \qquad (16.4) $$

if the hyper infinite sequence $\{A_n\}_k^\infty$ defined by

$$ A_n = \text{Ext}-\sum_{i=k}^{i=n} a_n \qquad (16.5) $$

$n \in \mathbb{N}^\#$, #-converges to A. The hyper infinite sum $A_n$ is the n-th partial sum of $\text{Ext}-\sum_{n=k}^{\infty} a_n$

If $\{A_n\}_k^\infty$ diverges, we say that $\text{Ext}-\sum_{n=k}^{\infty} a_n$ diverges; in particular, if $\lim_{n \to \infty} A_n = \infty^\#$

or $-\infty^\#$, we say that $\text{Ext}-\sum_{n=k}^{\infty} a_n$ #-diverges to $\infty^\#$ or $-\infty^\#$, and write

$$ \text{Ext}-\sum_{n=k}^{\infty} a_n = \infty^\# \text{ or } \text{Ext}-\sum_{n=k}^{\infty} a_n = -\infty^\#. \qquad (16.6) $$

A divergent hyper infinite series that does not diverge to $\pm\infty^\#$ is said to oscillate, or be oscillatory.

**Example 16.1** Consider the hyper infinite series

$$ \text{Ext}-\sum_{n=0}^{\infty} r^n, -1 < r < 1. \qquad (16.7) $$

Here $a_n = r^n, n \ge 0, n \in \mathbb{N}^\#$ and

$$ A_n = 1 + r + r^2 + \dots + r^n = \frac{1 - r^{n+1}}{1 - r} \qquad (16.8) $$

which #-converges to $1 = 1/(1-r)$ as $n \to \infty^\#$ ; thus, we write

$$ \text{Ext}-\sum_{n=0}^{\infty} r^n = 1/(1-r), -1 < r < 1. $$

An hyper infinite series can be viewed as a generalization of a hyperfinite sum

$$ A_N = \text{Ext}-\sum_{n=k}^{N} a_n \quad \text{Therefore, } \#\text{-lim}_{N \to \infty} A_N = A. $$

**Theorem 16.7.** The sum of a #-convergent hyper infinite series is unique.

**Theorem 16.8.** Let $\sum_{n=k}^{\infty} a_n = A$ and $\sum_{n=k}^{\infty} b_n = B$ where A and B are finite or hyperfinite.

Then

$$ \text{Ext}-\sum_{n=k}^{\infty} (a_n \pm b_n) = A \pm B \qquad (16.9) $$

and
(ii) for each $n \in \mathbb{N}^{\#}, [a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$ and

(iii) for each $n \in \mathbb{N}^{\#}, b_{n+1} - a_{n+1} = \frac{1}{2}(b_n - a_n)$.

The nested intervals theorem implies that the intersection $\bigcap_{n \in \mathbb{N}^{\#}} [a_n, b_n]$ of all of the intervals $[a_n, b_n]$ is a single point $s$. We will now construct a sub hyper infinite sequence of

$\{s_n\}_{n \in \mathbb{N}^{\#}}$ which will #-converge to $s$.

Since $[a_1, b_1]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^{\#}}$, there exists $k_1 \in \mathbb{N}^{\#}$ such that $s_{k_1}$ is in $[a_1, b_1]$. Since $[a_2, b_2]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^{\#}}$, there exists $k_2 \in \mathbb{N}^{\#}, k_2 > k_1$ such that $s_{k_2}$ is in $[a_2, b_2]$. Since $[a_3, b_3]$ contains hyper infinitely many terms of $\{s_n\}_{n \in \mathbb{N}^{\#}}$, there exists $k_3 \in \mathbb{N}^{\#}, k_3 > k_2$ such

that $s_{k_3}$ is in $[a_3, b_3]$. Continuing this process by hyper infinite induction, we obtain a hyper infinite sequence $\{s_{k_n}\}_{n \in \mathbb{N}^{\#}}$ such that $s_{k_n} \in [a_n, b_n]$ for each $n \in \mathbb{N}^{\#}$. The hypersequence $\{s_{k_n}\}_{n \in \mathbb{N}^{\#}}$ is a sub hyper infinite sequence of $\{s_n\}_{n \in \mathbb{N}^{\#}}$ since $k_{n+1} > k_n$ for each $n \in \mathbb{N}^{\#}$. Since $\#\text{-lim}_{n \to \infty} a_n = s$ and $\#\text{-lim}_{n \to \infty} b_n = s$ and $a_n \le s_n \le b_n$ for each $n \in \mathbb{N}^{\#}$, the squeeze theorem implies that that $\#\text{-lim}_{n \to \infty} s_n = s$.

## 16.2. Hyper infinite external series of constant.

**Definition 16.5.** If $\{a_n\}_k^{\infty}$ is an hyper infinite external sequence of Cauchy hyperreal numbers, the symbol

$$ \mathrm{Ext} - \sum_{n=k}^{\infty} a_n \qquad (16.3) $$

is an hyper infinite series, and $a_n$ is the $n$-th term of the hyper infinite series.

We say that $\mathrm{Ext} - \sum_{n=k}^{\infty} a_n$ #-converges to the sum $A \in \mathbb{R}_c^\#$, and write

$$ \mathrm{Ext} - \sum_{n=k}^{\infty} a_n = A \qquad (16.4) $$

if the hyper infinite sequence $\{A_n\}_k^{\infty}$ defined by

$$ A_n = \mathrm{Ext} - \sum_{i=k}^{i=n} a_n \qquad (16.5) $$

$n \in \mathbb{N}^{\#}$, #-converges to $A$. The hyper infinite sum $A_n$ is the $n$-th partial sum of $\mathrm{Ext} - \sum_{n=k}^{\infty} a_n$

If $\{A_n\}_k^{\infty}$ diverges, we say that $\mathrm{Ext} - \sum_{n=k}^{\infty} a_n$ diverges; in particular, if $\lim_{n \to \infty} \infty A_n = \infty$

or $-\infty$, we say that $\mathrm{Ext} - \sum_{n=k}^{\infty} a_n$ #-diverges to $\infty$ or $-\infty$, and write

$$ \mathrm{Ext} - \sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ or } \mathrm{Ext} - \sum_{n=k}^{\infty} a_n = -\infty^{\#}. \qquad (16.6) $$

A divergent hyper infinite series that does not diverge to $\pm\infty$ is said to oscillate, or be oscillatory.

**Example 16.1** Consider the hyper infinite series
$$ \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} (c \times a_n) = c \times A \qquad (16.10) $$

if $c \in \mathbb{R}_c^\#$ is a constant.

**Theorem 16.9.** (Cauchy's #-convergence criterion for hyper infinite series) A hyper infinite series $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ #-converges if and only if for every $\varepsilon > 0, \varepsilon \approx 0$ there is an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$ |\mathrm{Ext}-\sum_{n}^{m} a_n| < \varepsilon \qquad (16.11) $$

if $m \ge n \ge N$.

**Corollary 16.1.** If $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ #-converges; then $\#\lim_{N \to \infty^{\#}} a_n = 0$.

**Corollary 16.2.** If $\mathrm{Ext}-\sum_{n}^{\infty^{\#}} a_n$ #-converges; then for each $\varepsilon > 0, \varepsilon \approx 0$ there is an hyperinteger $K \in \mathbb{N}^\# \setminus \mathbb{N}$ such that $|\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n| < \varepsilon$ if $k \ge K$, that is

$$ \#\lim_{k \to \infty^{\#}} (\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n) = 0. \qquad (16.12) $$

## 16.3. Hyper Infinite Series of Nonnegative Terms.

The theory of series $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ with terms that are nonnegative for sufficiently large $n \in \mathbb{N}^\# \setminus \mathbb{N}$ is simpler than the general theory, since such a series either #-converges to a finite or hyperfinite #-limit or diverges to $\infty^{\#}$, as the next theorem shows.

**Theorem 16.10.** If $a_n \ge 0$ for $n \ge k$, then $\mathrm{Ext}-\sum_{n=1}^{\infty^{\#}} a_n$ #-converges if its partial sums are bounded or hyper bounded, or #-diverges to $\infty^{\#}$ if they are not. These are the only possibilities and, in either case, $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n = \{A_n|n \ge k\}$, where $A_n = \mathrm{Ext}-\sum_{i=k}^{n} a_i$.

**Theorem 16.11.** (The Comparison Test) Suppose that

$$ 0 \le a_n \le b_n, n \ge k. \qquad (16.13) $$

Then

$$ (a) \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n < \infty^{\#} \text{ if } \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} b_n < \infty^{\#}. \quad (b) \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n = \infty^{\#} \text{ if } \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} b_n = \infty^{\#}. $$

**Theorem 16.12.** (The Integral Test) Let

$$ c_n = f(n), n \ge k, \qquad (16.14) $$

where $f$ is positive; nonincreasing; and locally #-integrable on $[k, \infty^{\#})$. Then

$$ \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n < \infty^{\#} \qquad (16.15) $$

if and only if

$$ \mathrm{Ext}-\int_{k}^{\infty^{\#}} f(x)d^{\#}x < \infty^{\#}. \qquad (16.16) $$
$$ \mathrm{Ext}-\sum_{n=0}^{\infty^{\#}} r^n, -1 < r < 1. \qquad (16.7) $$

Here $a_n = r^n, n \ge 0, n \in \mathbb{N}^\#$ and

$$ A_n = 1 + r + r^2 + \dots + r^n = \frac{1 - r^{n+1}}{1 - r} \qquad (16.8) $$

which #-converges to $1 = 1/(1-r)$ as $n \to \infty^\#$; thus, we write

$$ \mathrm{Ext}-\sum_{n=0}^{\infty^{\#}} r^n = 1/(1-r), -1 < r < 1. $$

An hyperinfinite series can be viewed as a generalization of a hyperfinite sum

$$ A_N = \mathrm{Ext}-\sum_{n=k}^{N} a_n \quad \text{Therefore, } \#\mathrm{lim}_{N \to \infty^\#} A_N = A. $$

**Theorem 16.7.** The sum of a #-convergent hyper infinite series is unique.

**Theorem 16.8.** Let $\sum_{n=k}^{\infty^{\#}} a_n = A$ and $\sum_{n=k}^{\infty^{\#}} b_n = B$ where A and B are finite or hyperfinite.

Then

$$ \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} (a_n \pm b_n) = A \pm B \qquad (16.9) $$

and

$$ \mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} (c \times a_n) = c \times A \qquad (16.10) $$

if $c \in \mathbb{R}_c^\#$ is a constant.

**Theorem 16.9.** (Cauchy's #-convergence criterion for hyper infinite series) A hyper infinite series $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ #-converges if and only if for every $\varepsilon > 0, \varepsilon \approx 0$ there is an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$ |\mathrm{Ext}-\sum_{n=k}^{m} a_n| < \varepsilon \qquad (16.11) $$

if $m \ge n \ge N$.

**Corollary 16.1.** If $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ #-converges; then $\#\mathrm{lim}_{N \to \infty^\#} a_n = 0$.

**Corollary 16.2.** If $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ #-converges; then for each $\varepsilon > 0, \varepsilon \approx 0$ there is an

hyperinteger $K \in \mathbb{N}^\# \setminus \mathbb{N}$ such that $|\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n| < \varepsilon$ if $k \ge K$, that is

$$ \#\mathrm{lim}_{k \to \infty^{\#}} (\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n) = 0. \qquad (16.12) $$

### 16.3. Hyper Infinite Series of Nonnegative Terms.

The theory of series $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n$ with terms that are nonnegative for sufficiently large $n \in \mathbb{N}^\# \setminus \mathbb{N}$ is simpler than the general theory, since such a series either #-converges to a finite or hyperfinite #-limit or diverges to $\infty^\#$, as the next theorem shows.
**Example 16.2.** The integral test implies that the hyper infinite series $Ext-\sum_{n=k}^{\infty} n^{-p}$ converge if $p > 1$ and diverge if $0 < p \le 1$, because the same is true of the integral $Ext-\int_{a}^{\infty} x^{-p}d^{\#}x, a > 1$.

The next theorem is often applicable where the integral test is not.

**Theorem 16.13.** Suppose that $a_n \ge 0$ and $b_n > 0$ for $n \ge k$. Then

(a) $Ext-\sum_{n=k}^{\infty} a_n < \infty^{\#}$ if $Ext-\sum_{n=k}^{\infty} b_n < \infty^{\#}$ and $\overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_n}{b_n} < \infty^{\#}$.

(b) $Ext-\sum_{n=k}^{\infty} a_n = \infty^{\#}$ if $Ext-\sum_{n=k}^{\infty} b_n = \infty^{\#}$ and $\overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_n}{b_n} > 0$.

**Corollary 16.3.** Suppose that $a_n \ge 0$ and $b_n > 0$ for $n \ge k$, and $\#\lim_{n \to \infty^{\#}} \frac{a_n}{b_n} = L$.

where $0 < L < \infty^{\#}$. Then $Ext-\sum_{n=k}^{\infty} a_n$ and $Ext-\sum_{n=k}^{\infty} b_n$ $\#$-converge or $\#$-diverge together.

**Theorem 16.14.** Suppose that $a_n > 0, b_n > 0$, and

$$ \frac{a_{n+1}}{a_n} \le \frac{b_{n+1}}{b_n}. \qquad (16.17) $$

Then (a) $Ext-\sum_{n=k}^{\infty} a_n < \infty^{\#}$ if $Ext-\sum_{n=k}^{\infty} b_n < \infty^{\#}$. (b) $Ext-\sum_{n=k}^{\infty} a_n = \infty^{\#}$ if $Ext-\sum_{n=k}^{\infty} b_n = \infty^{\#}$.

**Theorem 16.15.** (The Ratio Test) Suppose that $a_n > 0$ for $n \ge k$. Then

(a) $Ext-\sum_{n=k}^{\infty} a_n < \infty^{\#}$ if $\overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_{n+1}}{a_n} < 1$. (b) $Ext-\sum_{n=k}^{\infty} a_n = \infty^{\#}$ if $\overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_{n+1}}{a_n} > 1$. If

$$ \overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_{n+1}}{a_n} \le 1 \le \overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_{n+1}}{a_n} \qquad (16.18) $$

then the test is inconclusive; that is, $Ext-\sum_{n=k}^{\infty} a_n$ may $\#$-converge or $\#$-diverge.

**Proof.** (a) If $\overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_{n+1}}{a_n} < 1$, there is a number $r$ such that $0 < r < 1$ and $\frac{a_{n+1}}{a_n} < r$ for $n \in \mathbb{N}^{\#}$ sufficiently large. This can be rewritten as $\frac{a_{n+1}}{a_n} < \frac{r^{n+1}}{r^n}$. Since $Ext-\sum_{n=k}^{\infty} r^n < \infty^{\#}$ Theorem 16.14 (a) with $b_n = r^n$ implies that $Ext-\sum_{n=k}^{\infty} a_n < \infty^{\#}$.

(b) If $\overline{\#-\lim}_{n \to \infty^{\#}} \frac{a_{n+1}}{a_n} > 1$, there is a number $r$ such that $r > 1$ and $\frac{a_{n+1}}{a_n} > r$ for $n \in \mathbb{N}^{\#}$ sufficiently large. This can be rewritten as $\frac{a_{n+1}}{a_n} > \frac{r^{n+1}}{r^n}$. Since

$$ \text{Ext-}\sum_{n=k}^{\infty} r^n = \infty^{\#} \text{Theorem 16.14 (b) with } b_n = r^n \text{ implies that } \text{Ext-}\sum_{n=k}^{\infty} a_n = \infty^{\#}. $$

To see that no conclusion can be drawn if (12.18) holds, consider hyper infinite series

$$ \text{Ext-}\sum_{n=k}^{\infty} a_n = \text{Ext-}\sum_{n=k}^{\infty} n^{-p}. \qquad (16.19) $$
**Theorem 16.10.** If $a_n \ge 0$ for $n \ge k$, then $Ext-\sum_{n=1}^{\infty} a_n$ #-converges if its partial sums are bounded or hyper bounded, or #-diverges to $\infty^#$ if they are not. These are the only possibilities and, in either case, $Ext-\sum_{n=k}^{\infty} a_n = \{A_n|n \ge k\}$, where $A_n = Ext-\sum_{i=k}^{n} a_i$.

**Theorem 16.11.** (The Comparison Test) Suppose that

$$0 \le a_n \le b_n, n \ge k. \qquad (16.13)$$

Then

$$(a) \ Ext-\sum_{n=k}^{\infty} a_n < \infty^# \text{ if } Ext-\sum_{n=k}^{\infty} b_n < \infty^# . \quad (b) \ Ext-\sum_{n=k}^{\infty} a_n = \infty^# \text{ if } Ext-\sum_{n=k}^{\infty} b_n = \infty^# .$$

**Theorem 16.12.** (The Integral Test) Let

$$c_n = f(n), n \ge k, \qquad (16.14)$$

where $f$ is positive; nonincreasing; and locally #-integrable on $[k, \infty^#)$. Then

$$Ext-\sum_{n=k}^{\infty} a_n < \infty^# \qquad (16.15)$$

if and only if

$$Ext-\int_{k}^{\infty} f(x)d^{\#}x < \infty^{\#}. \qquad (16.16)$$

**Example 16.2.** The integral test implies that the hyper infinite series $Ext-\sum_{n=k}^{\infty} n^{-p}$ converge if $p > 1$ and diverge if $0 < p \le 1$, because the same is true of the integral $Ext-\int_{a}^{\infty} x^{-p}d^{\#}x, a > 1$.

The next theorem is often applicable where the integral test is not.

**Theorem 16.13.** Suppose that $a_n \ge 0$ and $b_n > 0$ for $n \ge k$. Then

$$(a) \ Ext-\sum_{n=k}^{\infty} a_n < \infty^# \text{ if } Ext-\sum_{n=k}^{\infty} b_n < \infty^# \text{ and } \overline{\#-\lim}_{n \to \infty^#} \frac{a_n}{b_n} < \infty^# .$$

$$(b) \ Ext-\sum_{n=k}^{\infty} a_n = \infty^# \text{ if } Ext-\sum_{n=k}^{\infty} b_n = \infty^# \text{ and } \underline{\#-\lim}_{n \to \infty^#} \frac{a_n}{b_n} > 0 .$$

**Corollary 16.3.** Suppose that $a_n \ge 0$ and $b_n > 0$ for $n \ge k$, and $\#-\lim_{n \to \infty^#} \frac{a_n}{b_n} = L$.

where $0 < L < \infty^#$. Then $Ext-\sum_{n=k}^{\infty} a_n$ and $Ext-\sum_{n=k}^{\infty} b_n$ #-converge or #-diverge together.

**Theorem 16.14.** Suppose that $a_n > 0, b_n > 0$, and

$$\frac{a_{n+1}}{a_n} \le \frac{b_{n+1}}{b_n}. \qquad (16.17)$$

$$(a) \ Ext-\sum_{n=k}^{\infty} a_n < \infty^# \text{ if } Ext-\sum_{n=k}^{\infty} b_n < \infty^# . \quad (b) \ Ext-\sum_{n=k}^{\infty} a_n = \infty^# \text{ if } Ext-\sum_{n=k}^{\infty} b_n = \infty^# .$$
This series #-converges if $p > 1$ or #-diverges if $p \le 1$, however,

$$ \overline{\#-\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = \overline{\#-\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = 1. \qquad (16.20) $$

**Corollary 16.4.** Suppose that $a_n > 0$ for $n \ge k$ and $\overline{\#-\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = L$. Then

$$ (a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } L < 1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } L > 1. $$

The test is inconclusive if $L = 1$.

**Theorem 16.16.** (Generalized Raabe's Test) Suppose that $a_n > 0$ for large $n \in \mathbb{N}^*\mathbb{N}$. Let $M = \overline{\#-\lim}_{n \to \infty} \left(\frac{a_{n+1}}{a_n} - 1\right)$ and $m = \overline{\#-\lim}_{n \to \infty} \left(\frac{a_{n+1}}{a_n} - 1\right)$. Then

$$ (a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } M < -1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } m > -1. $$

The test is inconclusive if $m \le -1 \le M$.

**Theorem 16.17.** (Generalized Cauchy's Root Test) Suppose that $a_n \ge 0$ for $n \ge k \in \mathbb{N}^*\mathbb{N}$, then

$$ (a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } \overline{\#-\lim}_{n \to \infty} \sqrt[n]{a_n} < 1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } \overline{\#-\lim}_{n \to \infty} \sqrt[n]{a_n} > 1. $$

The test is inconclusive if $\overline{\#-\lim}_{n \to \infty} \sqrt[n]{a_n} = 1$.

## 16.4. Absolute and Conditional #-Convergence.

**Definition 16.6.** A series $\operatorname{Ext}-\sum_{n=k}^{\infty} a_n$ #-converges absolutely, or is absolutely #-convergent if $\operatorname{Ext}-\sum_{n=k}^{\infty} |a_n| < \infty^{\#}$.

**Theorem 12.18.** If $\operatorname{Ext}-\sum_{n=k}^{\infty} a_n$ #-converges absolutely; then $\operatorname{Ext}-\sum_{n=k}^{\infty} a_n$ #-converges.

**Theorem 12.19.** (Dirichlet's Test for Hyper Infinite Series) The hyper infinite series $\operatorname{Ext}-\sum_{n=k}^{\infty} a_n b_n$ is #-converges if the following conditions are satisfied

(i) $\overline{\#-\lim}_{n \to \infty} a_n = 0$,

(ii)

$$ \operatorname{Ext}-\sum_{n=k}^{\infty} |a_{n+1} - a_n| < \infty^{\#} \qquad (16.21) $$

and

(iii) for all $n \ge k$

$$ \operatorname{Ext}-\sum_{i=k}^{n} b_n \le M \qquad (16.22) $$

for some constant $M$.

**Proof.** Let $B_n, n \ge k$ be the partial sum
**Theorem 16.15.** (The Ratio Test) Suppose that $a_n > 0$ for $n \ge k$. Then

$$
(a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } \overline{\operatorname{lim}}_{n \to \infty} \# \frac{a_{n+1}}{a_n} < 1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } \overline{\operatorname{lim}}_{n \to \infty} \# \frac{a_{n+1}}{a_n} > 1. \text{ If } \\
\underline{\operatorname{lim}}_{n \to \infty} \# \frac{a_{n+1}}{a_n} \le 1 \le \overline{\operatorname{lim}}_{n \to \infty} \# \frac{a_{n+1}}{a_n} \qquad (16.18)
$$

then the test is inconclusive; that is, $\operatorname{Ext}-\sum_{n=k}^{\infty} a_n$ may #-converge or #-diverge.

**Proof.**(a) If $\overline{\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} < 1$, there is a number $r$ such that $0 < r < 1$ and $\frac{a_{n+1}}{a_n} < r$ for $n \in \mathbb{N}^\#$ sufficiently large. This can be rewritten as $\frac{a_{n+1}}{a_n} < \frac{r^{n+1}}{r^n}$.
Since $\operatorname{Ext}-\sum_{n=k}^{\infty} r^n < \infty^\#$ Theorem 16.14 (a) with $b_n = r^n$ implies that $\operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^\#$.

(b) If $\overline{\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} > 1$, there is a number $r$ such that $r > 1$ and $\frac{a_{n+1}}{a_n} > r$ for $n \in \mathbb{N}^\#$ sufficiently large. This can be rewritten as $\frac{a_{n+1}}{a_n} > \frac{r^{n+1}}{r^n}$. Since

$$
\operatorname{Ext}-\sum_{n=k}^{\infty} r^n = \infty^{\#} \text{Theorem 16.14 (b) with } b_n = r^n \text{ implies that } \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#}.
$$

To see that no conclusion can be drawn if (12.18) holds, consider hyper infinite series

$$
\operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \operatorname{Ext}-\sum_{n=k}^{\infty} n^{-p}. \qquad (16.19)
$$

This series #-converges if $p > 1$ or #-diverges if $p \le 1$, however,

$$
\overline{\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = \overline{\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = 1. \qquad (16.20)
$$

**Corollary 16.4.** Suppose that $a_n > 0$ for $n \ge k$ and $\overline{\lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = L$. Then

$$
(a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } L < 1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } L > 1.
$$

The test is inconclusive if $L = 1$.

**Theorem 16.16.** (Generalized Raabe's Test) Suppose that $a_n > 0$ for large $n \in \mathbb{N}^\# \setminus \mathbb{N}$.
Let $M = \overline{\lim}_{n \to \infty} \left( \frac{a_{n+1}}{a_n} - 1 \right)$ and $m = \overline{\lim}_{n \to \infty} \left( \frac{a_{n+1}}{a_n} - 1 \right)$. Then

$$
(a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } M < -1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } m > -1.
$$

The test is inconclusive if $m \le -1 \le M$.

**Theorem 16.17.** (Generalized Cauchy's Root Test) Suppose that $a_n \ge 0$ for $n \ge k \in \mathbb{N}^* \setminus \mathbb{N}$, then

$$
(a) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n < \infty^{\#} \text{ if } \overline{\lim}_{n \to \infty} \# \sqrt{a_n} < 1. \quad (b) \operatorname{Ext}-\sum_{n=k}^{\infty} a_n = \infty^{\#} \text{ if } \overline{\lim}_{n \to \infty} \# \sqrt{a_n} > 1.
$$

The test is inconclusive if $\overline{\lim}_{n \to \infty} \# \sqrt{a_n} = 1$.

16.4 Absolute and Conditional #-Convergence.
$$B_n = \mathrm{Ext}-\sum_{i=k}^{n} b_n \qquad (16.23)$$

Let us consider the partial sums $S_n, n \ge k$ of $\mathrm{Ext}-\sum_{n=k}^{\infty} a_n b_n$, where

$$S_n = \mathrm{Ext}-\sum_{i=k}^{n} a_n b_n \qquad (16.24)$$

By substituting $b_k = B_k$ and $b_n = B_n - B_{n-1}, n \ge k+1$, into (16.24), we obtain

$$S_n = a_k b_k + \mathrm{Ext}-\sum_{i=k+1}^{n} a_i (B_i - B_{i-1}), \qquad (16.25)$$

which we rewrite as

$$S_n = a_n B_n + \mathrm{Ext}-\sum_{i=k}^{n-1} (a_i - a_{i+1}) B_i. \qquad (16.26)$$

Now (16.26) can be viewed as

$$S_n = T_{n-1} + a_n B_n, \qquad (16.27)$$

where $T_{n-1} = \mathrm{Ext}-\sum_{i=k}^{n-1} (a_i - a_{i+1}) B_i$; that is, $\{T_n\}$ is the hyper infinite sequence of partial sums of the hyper infinite series

$$\mathrm{Ext}-\sum_{i=k}^{\infty} (a_i - a_{i+1}) B_i. \qquad (16.28)$$

Since $|(a_i - a_{i+1}) B_i| \le M|a_i - a_{i+1}|$ from (16.22), the comparison test and (16.21) imply that the series (16.28) #-converges absolutely. Theorem 12.18 now implies that $\{T_n\}_{n \in \mathbb{N}^\#}$ #-converges. Let $T = \#\mathrm{lim}_{n \to \infty^\#} T_n$. Since $B_n$ is bounded (hyperbounded) and $\#\mathrm{lim}_{n \to \infty^\#} a_n = 0$, we infer from (16.27) that

$$\#\mathrm{lim}_{n \to \infty^\#} S_n = \#\mathrm{lim}_{n \to \infty^\#} T_{n-1} + \#\mathrm{lim}_{n \to \infty^\#} a_n B_n = T. \qquad (16.29)$$

Therefore, $\mathrm{Ext}-\sum_{n=k}^{\infty} a_n b_n$ is #-converges.

**Corollary 16.4. (Abel's Test for Hyper Infinite Series)** The series $\mathrm{Ext}-\sum_{n=k}^{\infty} a_n b_n$ #-converges if $a_{n+1} \le a_n$ for $n \ge k$, $\#\mathrm{lim}_{n \to \infty^\#} a_n = 0$ and $\mathrm{Ext}-\sum_{i=k}^{n} b_n \le M$, for some constant $M$.

**Corollary 16.5. (Alternating Hyper Infinite Series Test)** The series $\mathrm{Ext}-\sum_{n=0}^{\infty} (-1)^n a_n$ #-converges if $0 \le a_{n+1} \le a_n$ and $\#\mathrm{lim}_{n \to \infty^\#} a_n = 0$.

**Proof.** Let $b_n = (-1)^n$, then $\{|B_n|\}_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence of zeros and ones and therefore bounded. The conclusion now follows from Abel's test.
**Definition 16.6.** A series $Ext-\sum_{n=k}^{\infty} a_n$ #-converges absolutely, or is absolutely

#-convergent if $Ext-\sum_{n=k}^{\infty}|a_n| < \infty^#$.

**Theorem 12.18.** If $Ext-\sum_{n=k}^{\infty} a_n$ #-converges absolutely; then $Ext-\sum_{n=k}^{\infty} a_n$ #-converges.

**Theorem 12.19.** (Dirichlet's Test for Hyper Infinite Series) The hyper infinite series $Ext-\sum_{n=k}^{\infty} a_n b_n$ is #-converges if the following conditions are satisfied

(i) $\#\lim_{n\to\infty} a_n = 0$,

(ii)

$$Ext-\sum_{n=k}^{\infty}|a_{n+1}-a_n| < \infty^# \qquad (16.21)$$

and

(iii) for all $n \ge k$

$$Ext-\sum_{i=k}^{n} b_n \le M \qquad (16.22)$$

for some constant $M$.

**Proof.** Let $B_n, n \ge k$ be the partial sum

$$B_n = Ext-\sum_{i=k}^{n} b_n \qquad (16.23)$$

Let us consider the partial sums $S_n, n \ge k$ of $Ext-\sum_{n=k}^{\infty} a_n b_n$, where

$$S_n = Ext-\sum_{i=k}^{n} a_n b_n \qquad (16.24)$$

By substituting $b_k = B_k$ and $b_n = B_n - B_{n-1}, n \ge k+1$, into (16.24), we obtain

$$S_n = a_k b_k + Ext-\sum_{i=k+1}^{n} a_i (B_i - B_{i-1}), \qquad (16.25)$$

which we rewrite as

$$S_n = a_n B_n + Ext-\sum_{i=k}^{n-1} (a_i - a_{i+1}) B_i. \qquad (16.26)$$

Now (16.26) can be viewed as

$$S_n = T_{n-1} + a_n B_n, \qquad (16.27)$$

where $T_{n-1} = Ext-\sum_{i=k}^{n-1} (a_i - a_{i+1}) B_i$; that is, $\{T_n\}$ is the hyper infinite sequence of partial

sums of the hyper infinite series
$$ \mathrm{Ext}-\sum_{i=k}^{\infty^{\#}} (a_i - a_{i+1})B_i. \qquad (16.28) $$

Since $|(a_i - a_{i+1})B_i| \le M|a_i - a_{i+1}|$ from (16.22), the comparison test and (16.21) imply that the series (16.28) #-converges absolutely. Theorem 12.18 now implies that $\{T_n\}_{n \in \mathbb{N}^\#}$ #-converges. Let $T = \#\lim_{n \to \infty^\#} T_n$. Since $B_n$ is bounded (hyperbounded) and $\#\lim_{n \to \infty^\#} a_n = 0$, we infer from (16.27) that

$$ \#\lim_{n \to \infty^\#} S_n = \#\lim_{n \to \infty^\#} T_{n-1} + \#\lim_{n \to \infty^\#} a_n B_n = T. \qquad (16.29) $$

Therefore, $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n b_n$ is #-converges.

**Corollary 16.4.** (Abel's Test for Hyper Infinite Series) The series $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n b_n$ #-converges if $a_{n+1} \le a_n$ for $n \ge k$, $\#\lim_{n \to \infty^\#} a_n = 0$ and $\mathrm{Ext}-\sum_{i=k}^{n} b_n \le M$, for some constant $M$.

**Corollary 16.5.** (Alternating Hyper Infinite Series Test) The series $\mathrm{Ext}-\sum_{n=0}^{\infty^{\#}} (-1)^n a_n$ #-converges if $0 \le a_{n+1} \le a_n$ and $\#\lim_{n \to \infty^\#} a_n = 0$.

**Proof.** Let $b_n = (-1)^n$, then $\{|B_n|\}_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence of zeros and ones and therefore bounded. The conclusion now follows from Abel's test.

## 16.5. Grouping Terms in a Hyper Infinite Series.

The terms of a hyper finite sum can be grouped arbitrarily by it hyper finite (but not by countable set of it finite subsets) subsets by inserting corresponding parentheses, see Appendix C. According to the next theorem, the same is true of an hyper infinite series that #-converges or #-diverges to $\pm\infty^\#$.

**Theorem 16.20.** Suppose that $\mathrm{Ext}-\sum_{n=k}^{\infty^{\#}} a_n = A$, where $-\infty^{\#} \le A \le \infty^{\#}$. Let $\{n_j\}_{n \in \mathbb{N}^\#}$ be an increasing hyper infinite sequence of integers, with $n_1 \ge k$. Define

$$ \begin{align} b_1 &= \mathrm{Ext}-\sum_{n=k}^{n_1} a_n, \notag \\ b_2 &= \mathrm{Ext}-\sum_{n=n_1+1}^{n_2} a_n, \tag{16.30} \end{align} $$

............

$$ b_r = \mathrm{Ext}-\sum_{n=n_{r-1}+1}^{n_r} a_n $$

Then

$$ \mathrm{Ext}-\sum_{j=1}^{\infty^{\#}} b_{n_j} = A. \qquad (12.31) $$
$Ext-\sum_{n=1}^{\infty} a_n$ can be rearranged to form a series $Ext-\sum_{n=1}^{\infty} b_n$ with partial sums

$B_n = Ext-\sum_{i=1}^{n} b_i$ such that

$$ \lim_{n \to \infty} B_n = v \quad \text{and} \quad \lim_{n \to \infty} B_n = \mu. \qquad (16.33) $$

## 16.7. Multiplication of hyper infinite Series.

Given two hyper infinite series $Ext-\sum_{n=0}^{\infty} a_n$ and $Ext-\sum_{n=0}^{\infty} b_n$ we can arrange all possible products $a_ib_j$, $i,j \ge 0$ in a two-dimensional array:

$$ \begin{array}{l} a_0 b_0 \quad a_0 b_1 \quad a_0 b_2 \quad a_0 b_3 \quad \dots \\ a_1 b_0 \quad a_1 b_1 \quad a_1 b_2 \quad a_1 b_3 \quad \dots \\ a_2 b_0 \quad a_2 b_1 \quad a_2 b_2 \quad a_2 b_3 \quad \dots \\ \vdots \quad \vdots \quad \vdots \quad \vdots \quad \dots \end{array} \qquad (16.34) $$

where the subscript on a is constant in each row and the subscript on b is constant in each column. Any sensible definition of the product

$$ \left( Ext-\sum_{n=0}^{\infty} a_n \right) \left( Ext-\sum_{n=0}^{\infty} b_n \right) \qquad (12.35) $$

clearly must involve every product in this array exactly once; thus, we might define the product of the two series to be the series $Ext-\sum_{n=0}^{\infty} p_n$, where $\{p_n\}_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence obtained by ordering the products in (12.34) according to some method that chooses every product exactly once.

**Theorem 16.24.** Let $Ext-\sum_{n=0}^{\infty} a_n = A$ and $Ext-\sum_{n=0}^{\infty} b_n = B$, where A and B are finite or hyperfinite, and at least one term of each series is nonzero. Then $Ext-\sum_{n=0}^{\infty} p_n = A \times B$ for every hyper infinite sequence $\{p_n\}_{n \in \mathbb{N}^\#}$ obtained by ordering the products in (16.34) if and only if $Ext-\sum_{n=0}^{\infty} a_n$ and $Ext-\sum_{n=0}^{\infty} b_n$ #-converge absolutely:

Moreover, in this case, $Ext-\sum_{n=0}^{\infty} p_n$ #-converges absolutely.

**Definition 16.7.** The Cauchy product of $Ext-\sum_{n=0}^{\infty} a_n$ and $Ext-\sum_{n=0}^{\infty} b_n$ is $Ext-\sum_{n=0}^{\infty} c_n$, where
## 16.6. Rearrangement of hyper infinite series.

A hyperfinite sum is not changed by rearranging its terms, see Appendix C. According to the next theorem, we see that every rearrangement of an absolutely #-convergent hyper infinite series has the same sum, but that conditionally #-convergent series fail, spectacularly, to have this property.

**Theorem 16.21.** If $Ext-\sum_{n=1}^{\infty} b_n$ is a rearrangement of an absolutely #-convergent series $Ext-\sum_{n=1}^{\infty} a_n$ then $Ext-\sum_{n=1}^{\infty} b_n$ also #-converges absolutely, and to the same sum.

**Theorem 16.22.** If $\{a_{n_i}\}_{i \in \mathbb{N}^\#}$ and $\{a_{m_j}\}_{j \in \mathbb{N}^\#}$ are respectively the subsequences of all positive and negative terms in a conditionally #-convergent series $Ext-\sum_{n=1}^{\infty} a_n$ then

$$ Ext-\sum_{i=1}^{\infty} a_{n_i} = \infty^{\#} \text{ and } Ext-\sum_{j=1}^{\infty} a_{m_j} = -\infty^{\#}. \qquad (16.32) $$

**Theorem 16.23.** Suppose that $Ext-\sum_{n=1}^{\infty} a_n$ is conditionally #-convergent and $\mu$ and $\nu$ are arbitrarily given in the extended hyperreals; with $\mu \le \nu$. Then the terms of

$Ext-\sum_{n=1}^{\infty} a_n$ can be rearranged to form a series $Ext-\sum_{n=1}^{\infty} b_n$ with partial sums

$B_n = Ext-\sum_{i=1}^{n} b_i$ such that

$$ \varlimsup_{n \to \infty} B_n = \nu \text{ and } \varliminf_{n \to \infty} B_n = \mu. \qquad (16.33) $$

## 16.7. Multiplication of hyper infinite Series.

Given two hyper infinite series $Ext-\sum_{n=0}^{\infty} a_n$ and $Ext-\sum_{n=0}^{\infty} b_n$ we can arrange all possible products $a_i b_j$, $i, j \ge 0$ in a two-dimensional array:

$$ \begin{array}{l} a_0 b_0 \quad a_0 b_1 \quad a_0 b_2 \quad a_0 b_3 \quad \cdots \\ a_1 b_0 \quad a_1 b_1 \quad a_1 b_2 \quad a_1 b_3 \quad \cdots \\ a_2 b_0 \quad a_2 b_1 \quad a_2 b_2 \quad a_2 b_3 \quad \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ a_m b_0 \quad a_m b_1 \quad a_m b_2 \quad a_m b_3 \quad \cdots \\ \cdot & \cdot & \cdot & \cdot & \ddots \end{array} \qquad (16.34) $$

where the subscript on a is constant in each row and the subscript on b is constant in each column. Any sensible definition of the product
$$c_n = \operatorname{Ext} - \sum_{j=0}^{n} a_j b_{n-j}. \qquad (16.36)$$

Thus, $c_n$ is the external sum of all products $a_l b_k$, where $i \ge 0, j \ge 0$, and $i + j = n$; thus,

$$c_n = \operatorname{Ext} - \sum_{j=0}^{n} a_j b_{n-j} = \operatorname{Ext} - \sum_{j=0}^{n} b_j a_{n-j}. \qquad (16.37)$$

**Theorem 16.25.** If $\operatorname{Ext} - \sum_{n=0}^{\infty} a_n$ and $\operatorname{Ext} - \sum_{n=0}^{\infty} b_n$ #-converge absolutely to sums A and B, then the Cauchy product $\operatorname{Ext} - \sum_{j=0}^{n} a_j b_{n-j}$ #-converges absolutely to AB.

**Theorem 16.26.** Let $f(\alpha) = \operatorname{Ext} - \sum_{n=0}^{\infty} \frac{\alpha^n}{n!}$ and $f(\beta) = \operatorname{Ext} - \sum_{n=0}^{\infty} \frac{\beta^n}{n!}$, then
$$f(\alpha)f(\beta) = f(\alpha + \beta). \qquad (16.38)$$

**Proof.** From Eq. (16.37) we obtain

$$c_n = \operatorname{Ext} - \sum_{n=0}^{m} \frac{\alpha^{n-m} \beta^m}{(n-m)!m!} = \frac{1}{n!} \left( \operatorname{Ext} - \sum_{n=0}^{m} \binom{n}{m} \alpha^{n-m} \beta^m \right) = \operatorname{Ext} - \sum_{n=0}^{\infty} \frac{(\alpha + \beta)^n}{n!} \qquad (16.39)$$

Thus

$$f(\alpha)f(\beta) = \operatorname{Ext} - \sum_{n=0}^{\infty} \frac{(\alpha + \beta)^n}{n!} = f(\alpha + \beta). \qquad (16.40)$$

## 16.8. Double Hyper Infinite Sequences.

**Definition 16.8.** A double hyper infinite sequence of hyperreal numbers $\mathbb{R}_c^\#$ (complex numbers $\mathbb{C}_c^\# = \mathbb{R}_c^\# + i\mathbb{R}_c^\#$) is a $\mathbb{R}_c^\#$-valued ($\mathbb{C}_c^\#$-valued) function $s: \mathbb{N}^\# \times \mathbb{N}^\# \to \mathbb{R}_c^\#$ or $s: \mathbb{N}^\# \times \mathbb{N}^\# \to \mathbb{C}_c^\#$. We shall use the notation $\{s_{n,m}\}_{n,m \in \mathbb{N}^\#}$ or simply $s_{nm}$.

**Definition 16.9.** We say that a double hyper infinite sequence $s_{n,m}$ #-converges to $a \in \mathbb{C}_c^\#$ and we write $\#-\lim_{n,m \to \infty} s_{n,m} = a$, if the following condition is satisfied: for every $\varepsilon > 0$, $\varepsilon \approx 0$, there exists $N \in \mathbb{N}^\#$ such that $|s_{n,m} - a| < \varepsilon$ if $n, m \ge N$.

**Theorem 16.27.** (Uniqueness of Double #-Limits). A double hyper infinite $\mathbb{C}_c^\#$-valued sequence has at most one #-limit.

**Definition 16.10.** A double hyper infinite sequence $s_{n,m}$ is called bounded (hyper bounded) if there exists finite (hyperfinite) number $M \in \mathbb{R}_c^\#$, $M > 0$ such that $|s_{n,m}| \le M, \forall n, m \in \mathbb{N}^\#$.

**Theorem 16.28.** A #-convergent double $\mathbb{C}_c^\#$-valued hyper infinite sequence is bounded or hyper bounded.

**Definition 16.11.** A double $\mathbb{C}_c^\#$-valued hyper infinite sequence $s_{n,m}$ is called a Cauchy sequence if and only if for every $\varepsilon > 0$, $\varepsilon \approx 0$, there exists a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that $|s_{p,q} - s_{n,m}| < \varepsilon, \forall p \ge n \ge N$ and $\forall q(q \ge m \ge N)$.

**Theorem 16.29.** (Cauchy Convergence Criterion for Double hyper infinite Sequences). A double $\mathbb{C}_c^\#$-valued hyper infinite sequence $s_{n,m}, n, m \in \mathbb{N}^\#$ #-converges if and only if it is a Cauchy sequence.
$$ \left( \operatorname{Ext} - \sum_{n=0}^{\infty^{\#}} a_n \right) \left( \operatorname{Ext} - \sum_{n=0}^{\infty^{\#}} b_n \right) \qquad (12.35) $$

clearly must involve every product in this array exactly once; thus, we might define the product of the two series to be the series $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} p_n$, where $\{p_n\}_{i \in \mathbb{N}^{\#}}$ is a hyper infinite sequence obtained by ordering the products in (12.34) according to some method that chooses every product exactly once.

**Theorem 16.24.** Let $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} a_n = A$ and $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} b_n = B$, where A and B are finite or hyperfinite, and at least one term of each series is nonzero. Then $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} p_n = A \times B$ for every hyper infinite sequence $\{p_n\}_{i \in \mathbb{N}^{\#}}$ obtained by ordering the products in (16.34) if and only if $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} a_n$ and $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} b_n$ #-converge absolutely:

Moreover, in this case, $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} p_n$ #-converges absolutely.

**Definition 16.7.** The Cauchy product of $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} a_n$ and $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} b_n$ is $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} c_n$, where

$$ c_n = \operatorname{Ext}-\sum_{j=0}^{n} a_j b_{n-j}. \qquad (16.36) $$

Thus, $c_n$ is the external sum of all products $a_j b_{n-j}$, where $i \ge 0, j \ge 0$, and $i + j = n$; thus,

$$ c_n = \operatorname{Ext}-\sum_{j=0}^{n} a_j b_{n-j} = \operatorname{Ext}-\sum_{j=0}^{n} b_j a_{n-j}. \qquad (16.37) $$

**Theorem 16.25.** If $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} a_n$ and $\operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} b_n$ #-converge absolutely to sums A and B, then the Cauchy product $\operatorname{Ext}-\sum_{j=0}^{n} a_j b_{n-j}$ #-converges absolutely to AB.

**Theorem 16.26.** Let $f(\alpha) = \operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} \frac{\alpha^n}{n!}$ and $f(\beta) = \operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} \frac{\beta^n}{n!}$, then
$$ f(\alpha)f(\beta) = f(\alpha + \beta). \qquad (16.38) $$

**Proof.** From Eq. (16.37) we obtain

$$ c_n = \operatorname{Ext}-\sum_{n=0}^{m} \frac{\alpha^{n-m} \beta^m}{(n-m)!m!} = \frac{1}{n!} \left( \operatorname{Ext}-\sum_{n=0}^{m} \binom{n}{m} \alpha^{n-m} \beta^m \right) = \operatorname{Ext}-\sum_{n=0}^{\infty^{\#}} \frac{(\alpha + \beta)^n}{n!} \qquad (16.39) $$

Thus
**Definition 16.12.** Let $s_{n,m}$ be a double $\mathbb{R}_c^\#$-valued hyper infinite sequence.

(i) If $s_{n,m} \le s_{j,k}, \forall n \forall j \forall m \forall k (n \le j \land m \le k)$, $n,m,j,k \in \mathbb{N}^\#$, we say the sequence $s_{n,m}$ is increasing.

(ii) $s_{n,m} \ge s_{j,k}, \forall n \forall j \forall m \forall k (n \le j \land m \le k)$, $n,m,j,k \in \mathbb{N}^\#$, we say the sequence $s_{n,m}$ is decreasing.

(ii) If $s_{n,m}$ is either increasing or decreasing, then we say it is monotone.

**Definition 16.13.** For a double sequence $s_{n,m}$, the #-limits

$\#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right)$ and $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right)$ are called repeated #-limits.

**Theorem 16.30.** Let $\#\lim_{n,m \to \#} \infty^{\#} s_{n,m} = a$. Then $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = a$ if and only if $\#\lim_{n \to \#} \infty^{\#} s_{n,m}$ exists for each $m \in \mathbb{N}^\#$.

**Theorem 16.31.** Let $\#\lim_{n,m \to \#} \infty^{\#} s_{n,m} = a$. Then the repeated #-limits

$\#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right)$ and $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right)$ exist and both are equal to a if and only if (i) $\#\lim_{n \to \#} \infty^{\#} s_{n,m}$ exists for each $m \in \mathbb{N}^\#$, and (ii) $\#\lim_{m \to \#} \infty^{\#} s_{n,m}$ exists for each $n \in \mathbb{N}^\#$.

**Theorem 16.32.** If $s_{n,m}$ is a double sequence such that the repeated #-limit $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = a$ and the #-limit $\#\lim_{n \to \#} \infty^{\#} s_{n,m}$ exists uniformly in $m \in \mathbb{N}^\#$, then the double #-limit $\#\lim_{n,m \to \#} \infty^{\#} s_{n,m} = a$.

**Theorem 16.33.** (Monotone Convergence Theorem). A monotone double $\mathbb{R}_c^\#$-valued hyper infinite sequence is #-convergent if and only if it is bounded (hyper bounded).

Further: (i) If $s_{n,m}$ is increasing and bounded (hyper bounded) above, then
$$ \#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n,m \to \#} \infty^{\#} s_{n,m}. $$

(ii) If $s_{n,m}$ is decreasing and bounded (hyper bounded) below, then
$$ \#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n,m \to \#} \infty^{\#} s_{n,m}. $$

**Theorem 16.34.** (The Sandwich Theorem). Suppose that $x_{n,m}, s_{n,m}$, and $y_{n,m}$ are double $\mathbb{R}_c^\#$-valued hyper infinite sequences such that $x_{n,m} \le s_{n,m} \le y_{n,m}, \forall n, m \in \mathbb{N}^\#$, and $\#\lim_{n,m \to \#} \infty^{\#} x_{n,m} = \#\lim_{n,m \to \#} \infty^{\#} y_{n,m}$.

Then $s_{n,m}$ is #-convergent and $\#\lim_{n,m \to \#} \infty^{\#} x_{n,m} = \#\lim_{n,m \to \#} \infty^{\#} y_{n,m} = \#\lim_{n,m \to \#} \infty^{\#} s_{n,m}$.

**Definition 16.14.** Let $s_{n,m}$ be a double $\mathbb{C}_c^\#$-valued hyper infinite sequence and let $(k_1, r_1) < (k_2, r_2) < \dots < (k_n, r_n) < \dots$ be a strictly increasing sequences of pairs of hypernatural numbers. Then the sequence $s_{k_n, r_m}$ is called a subsequence of $s_{n,m}$.

**Theorem 16.35.** If a double $\mathbb{C}_c^\#$-valued hyper infinite sequence $s_{n,m}$ #-converges to number $a \in \mathbb{C}_c^\#$, then any hyper infinite subsequence of $s_{n,m}$ also #-converges to $a$.

**Theorem 16.36.** If the repeated #-limits of a double sequence $s_{n,m}$ exist and satisfy $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right) = a$, then the

repeated #-limits for any subsequence $s_{p_n, q_m}$ exist and satisfy
$$ \#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{p_n, q_m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{p_n, q_m} \right) = a. $$

**Theorem 16.37.** Every double $\mathbb{R}_c^\#$-valued hyper infinite sequence has a monotone hyper infinite subsequence.

**Theorem 16.38.** (Bolzano-Weierstrass Theorem). A bounded (hyper bounded) double $\mathbb{R}_c^\#$-valued hyper infinite sequence has a #-convergent monotone subsequence.
$$f(\alpha)f(\beta) = \mathrm{Ext}-\sum_{n=0}^{\infty} \frac{(\alpha + \beta)^n}{n!} = f(\alpha + \beta). \qquad (16.40)$$

## 16.8. Double Hyper Infinite Sequences.

**Definition 16.8.** A double hyper infinite sequence of hyperreal numbers $\mathbb{R}_c^\#$ (complex numbers $\mathbb{R}_c^\# = \mathbb{R}_c^\# + i\mathbb{R}_c^\#$) is a $\mathbb{R}_c^\#$-valued ($\mathbb{C}_c^\#$-valued) function $s: \mathbb{N}^\# \times \mathbb{N}^\# \to \mathbb{R}_c^\#$ or $s: \mathbb{N}^\# \times \mathbb{N}^\# \to \mathbb{C}_c^\#$. We shall use the notation $\{s_{n,m}\}_{n,m\in\mathbb{N}^\#}$ or simply $s_{nm}$.

**Definition 16.9.** We say that a double hyper infinite sequence $s_{n,m}$ #-converges to $a \in \mathbb{C}_c^\#$ and we write $\#\lim_{n,m\to\#} \infty^\# s_{n,m} = a$, if the following condition is satisfied: for every $\varepsilon > 0$, $\varepsilon \approx 0$, there exists $N \in \mathbb{N}^\#$ such that $|s_{n,m} - a| < \varepsilon$ if $n, m \ge N$.

**Theorem 16.27.** (Uniqueness of Double #-Limits). A double hyper infinite $\mathbb{C}_c^\#$-valued sequence has at most one #-limit.

**Definition 16.10.** A double hyper infinite sequence $s_{n,m}$ is called bounded (hyper bounded) if there exists finite (hyperfinite) number $M \in \mathbb{R}_c^\#$, $M > 0$ such that $|s_{n,m}| \le M, \forall n, m \in \mathbb{N}^\#$.

**Theorem 16.28.** A #-convergent double $\mathbb{C}_c^\#$-valued hyper infinite sequence is bounded or hyper bounded.

**Definition 16.11.** A double $\mathbb{C}_c^\#$-valued hyper infinite sequence $s_{n,m}$ is called a Cauchy sequence if and only if for every $\varepsilon > 0$, $\varepsilon \approx 0$, there exists a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that $|s_{p,q} - s_{n,m}| < \varepsilon$, $\forall p(p \ge n \ge N)$ and $\forall q(q \ge m \ge N)$.

**Theorem 16.29.** (Cauchy Convergence Criterion for Double hyper infinite Sequences). A double $\mathbb{C}_c^\#$-valued hyper infinite sequence $s_{n,m}, n, m \in \mathbb{N}^\#$ #-converges if and only if it is a Cauchy sequence.

**Definition 16.12.** Let $s_{n,m}$ be a double $\mathbb{R}_c^\#$-valued hyper infinite sequence.

(i) If $s_{n,m} \le s_{j,k}, \forall n, j, m, k \in \mathbb{N}^\#$, we say the sequence $s_{n,m}$ is increasing.

(ii) $s_{n,m} \ge s_{j,k}, \forall n, j, m, k \in \mathbb{N}^\#$, we say the sequence $s_{n,m}$ is decreasing.

(ii) If $s_{n,m}$ is either increasing or decreasing, then we say it is monotone.

**Definition 16.13.** For a double sequence $s_{n,m}$, the #-limits

$\#\lim_{n \to \#} \infty^\# (\#\lim_{m \to \#} \infty^\# s_{n,m})$ and $\#\lim_{n \to \#} \infty^\# (\#\lim_{m \to \#} \infty^\# s_{n,m})$ are called repeated #-limits.

**Theorem 16.30.** Let $\#\lim_{n,m \to \#} \infty^\# s_{n,m} = a$. Then $\#\lim_{m \to \#} \infty^\# (\#\lim_{n \to \#} \infty^\# s_{n,m}) = a$ if and only if $\#\lim_{n \to \#} \infty^\# s_{n,m}$ exists for each $m \in \mathbb{N}^\#$.

**Theorem 16.31.** Let $\#\lim_{n,m \to \#} \infty^\# s_{n,m} = a$. Then the repeated #-limits $\#\lim_{n \to \#} \infty^\# (\#\lim_{m \to \#} \infty^\# s_{n,m})$ and $\#\lim_{m \to \#} \infty^\# (\#\lim_{n \to \#} \infty^\# s_{n,m})$ exist and both are equal to a if and only if (i) $\#\lim_{n \to \#} \infty^\# s_{n,m}$ exists for each $m \in \mathbb{N}^\#$, and (ii) $\#\lim_{m \to \#} \infty^\# s_{n,m}$ exists for each $n \in \mathbb{N}^\#$.

**Theorem 16.32.** If $s_{n,m}$ is a double sequence such that the repeated #-limit $\#\lim_{m \to \#} \infty^\# (\#\lim_{n \to \#} \infty^\# s_{n,m}) = a$ and the #-limit $\#\lim_{n \to \#} \infty^\# s_{n,m}$ exists uniformly in $m \in \mathbb{N}^\#$, then the double #-limit $\#\lim_{n,m \to \#} \infty^\# s_{n,m} = a$.

**Theorem 16.33.** (Monotone Convergence Theorem). A monotone double $\mathbb{R}_c^\#$-valued hyper infinite sequence is #-convergent if and only if it is bounded (hyper bounded). Further: (i) If $s_{n,m}$ is increasing and bounded (hyper bounded) above, then
## 16.9. External Double Hyper Infinite Series.

**Definition 16.15.** Let $z: \mathbb{N}^{\#} \times \mathbb{N}^{\#} \to \mathbb{C}_c^{\#}$ be external hyper infinite double sequence of complex numbers $\mathbb{C}_c^{\#}$ and let $s_{n,m}$ be the double hyper infinite sequence defined by the equation

$$s_{n,m} = \mathrm{Ext} - \sum_{i=1}^{n} \left( \mathrm{Ext} - \sum_{j=1}^{m} z_{ij} \right). \qquad (16.41)$$

The pair $(z, s)$ is called a double hyper infinite series and is denoted by the symbol

$$\mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m} \qquad (16.42)$$

or, more briefly by $\mathrm{Ext} - \sum_{n,m=1}^{\infty^{\#}} z_{n,m}$. Each number $z_{n,m}$ is called a term of the double series and each $s_{n,m}$ is called a partial sum.

**Definition 16.16.** We say that the double series $\mathrm{Ext} - \sum_{n,m=1}^{\infty^{\#}} z_{n,m}$ is #-convergent to the sum $s$ if $\#\lim_{n,m \to \infty} s_{n,m} = s$. If no such #-limit exists, we say that the double series $\mathrm{Ext} - \sum_{n,m=1}^{\infty^{\#}} z_{n,m}$ is #-divergent.

**Definition 16.17.** The hyper infinite series

$$\mathrm{Ext} - \sum_{n=1}^{\infty^{\#}} \left( \mathrm{Ext} - \sum_{m=1}^{\infty^{\#}} z_{n,m} \right) \qquad (16.43)$$

and

$$\mathrm{Ext} - \sum_{m=1}^{\infty^{\#}} \left( \mathrm{Ext} - \sum_{n=1}^{\infty^{\#}} z_{n,m} \right) \qquad (16.44)$$

are called repeated hyper infinite series.

**Theorem 16.38.** If the double hyper infinite series $\mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m}$ is #-convergent, then
$$\#\lim_{n,m \to \infty} z_{n,m} = 0. \qquad (12.45)$$

**Theorem 16.39.** (Cauchy #-Convergence Criterion for Double hyper infinite Series.)

A double hyper infinite series $\mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m}$ #-converges if and only if its sequence of partial sums $s_{n,m}$ is Cauchy.

**Theorem 16.40.** If the double series $\mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m}$ #-converges to $s_1$ and $\mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} u_{n,m}$ #-converges to $s_2$, then: (i) $\mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m} + \mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} u_{n,m} = s_1 + s_2$.

$$(ii) \mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} c \times z_{n,m} = c \times \left( \mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m} \right).$$
$$ \#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n,m \to \#} \infty^{\#} s_{n,m}. $$

(ii) If $s_{n,m}$ is decreasing and bounded (hyper bounded) below, then

$$ \#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n,m \to \#} \infty^{\#} s_{n,m}. $$

**Theorem16.34.** (The Sandwich Theorem). Suppose that $x_{n,m}, s_{n,m}$, and $y_{n,m}$ are double $\mathbb{R}^\#$-valued hyper infinite sequences such that $x_{n,m} \le s_{n,m} \le y_{n,m}, \forall n,m \in \mathbb{N}^\#$, and $\#\lim_{n,m \to \#} \infty^{\#} x_{n,m} = \#\lim_{n,m \to \#} \infty^{\#} y_{n,m}$.

Then $s_{n,m}$ is #-convergent and $\#\lim_{n,m \to \#} \infty^{\#} x_{n,m} = \#\lim_{n,m \to \#} \infty^{\#} y_{n,m} = \#\lim_{n,m \to \#} \infty^{\#} s_{n,m}$.

**Definition 16.14.** Let $s_{n,m}$ be a double $\mathbb{C}^\#$-valued hyper infinite sequence and let $(k_1, r_1) < (k_2, r_2) < \dots < (k_n, r_n) < \dots$ be a strictly increasing sequences of pairs of hypernatural numbers. Then the sequence $s_{k_n, r_m}$ is called a subsequence of $s_{n,m}$.

**Theorem16.35.** If a double $\mathbb{C}^\#$-valued hyper infinite sequence $s_{n,m}$ #-converges to number $a \in \mathbb{C}^\#$, then any hyper infinite subsequence of $s_{n,m}$ also #-converges to $a$.

**Theorem16.36.** If the repeated #-limits of a double sequence $s_{n,m}$ exist and satisfy $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{n,m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{n,m} \right) = a$, then the

repeated #-limits for any subsequence $s_{p_n, q_m}$ exist and satisfy $\#\lim_{m \to \#} \infty^{\#} \left( \#\lim_{n \to \#} \infty^{\#} s_{p_n, q_m} \right) = \#\lim_{n \to \#} \infty^{\#} \left( \#\lim_{m \to \#} \infty^{\#} s_{p_n, q_m} \right) = a$.

**Theorem16.37.** Every double $\mathbb{R}^\#$-valued hyper infinite sequence has a monotone hyper infinite subsequence.

**Theorem16.38.** (Bolzano-Weierstrass Theorem). A bounded (hyper bounded) double $\mathbb{R}^\#$-valued hyper infinite sequence has a #-convergent monotone subsequence.

## 16.9. External Double Hyper Infinite Series.

**Definition 16.15.** Let $z : \mathbb{N}^\# \times \mathbb{N}^\# \to \mathbb{C}^\#$ be external hyper infinite double sequence of complex numbers $\mathbb{C}^\#$ and let $s_{n,m}$ be the double hyper infinite sequence defined by the equation

$$ s_{n,m} = \mathrm{Ext} - \sum_{i=1}^{n} \left( \mathrm{Ext} - \sum_{j=1}^{m} z_{i,j} \right). \qquad (16.41) $$

The pair $(z, s)$ is called a double hyper infinite series and is denoted by the symbol

$$ \mathrm{Ext} - \sum_{n=1, m=1}^{\infty^{\#}} z_{n,m} \qquad (16.42) $$

or, more briefly by $\mathrm{Ext} - \sum_{n,m=1}^{\infty^{\#}} z_{n,m}$. Each number $z_{n,m}$ is called a term of the double series and each $s_{n,m}$ is called a partial sum.

**Definition 16.16.** We say that the double series $\mathrm{Ext} - \sum_{n,m=1}^{\infty^{\#}} z_{n,m}$ is #-convergent to the sum $s$ if $\#\lim_{n,m \to \#} \infty^{\#} s_{n,m} = s$. If no such #-limit exists, we say that the double series $\mathrm{Ext} - \sum_{n,m=1}^{\infty^{\#}} z_{n,m}$ is #-divergent.

**Definition 16.17.** The hyper infinite series

$$ \mathrm{Ext} - \sum_{n=1}^{\infty^{\#}} \left( \mathrm{Ext} - \sum_{m=1}^{\infty^{\#}} z_{n,m} \right) \qquad (16.43) $$
**Theorem16.41.** Suppose that the double series $\text{Ext}-\sum_{n=1, m=1}^{\infty} z_{n,m}$ is #-convergent, with sum $s$. Then the repeated series $\text{Ext}-\sum_{n=1}^{\infty} \left( \text{Ext}-\sum_{m=1}^{\infty} z_{n,m} \right)$ and $\text{Ext}-\sum_{m=1}^{\infty} \left( \text{Ext}-\sum_{n=1}^{\infty} z_{n,m} \right)$ are both #-convergent with sum $s$ if and only if for every $m \in \mathbb{N}^{\#}$, the series $\text{Ext}-\sum_{n=1}^{\infty} z_{n,m}$ is #-convergent, and for every $n \in \mathbb{N}^{\#}$, the series $\text{Ext}-\sum_{m=1}^{\infty} z_{n,m}$ is #-convergent.

## 16.10. Interchanging the order of summation of hyper infinite sum.

**Theorem 12.** Assum that

$$ \text{Ext}-\sum_{i=1}^{\infty} \left( \text{Ext}-\sum_{k=1}^{\infty} |a_{jk}| \right) < \infty^{\#}. \qquad (16.46) $$

Then

$$ \text{Ext}-\sum_{i=1}^{\infty} \left( \text{Ext}-\sum_{k=1}^{\infty} |a_{jk}| \right) = \text{Ext}-\sum_{k=1}^{\infty} \left( \text{Ext}-\sum_{j=1}^{\infty} |a_{jk}| \right) \qquad (16.47) $$

## 17. Hyper infinite sequences and series of $\mathbb{R}^\#_c$-valued functions.

### 17.1. Uniform #-Convergence

If $f_1, \dots, f_k, f_{k+1}, \dots, f_n, \dots, n \in \mathbb{N}^\#$ are $\mathbb{R}^\#$-valued functions defined on a subset $D \subset \mathbb{R}^\#$ of the hyperreals, we say that $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ is an hyper infinite sequence of functions on $D$. If the sequence of values $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ #-converges for each $x$ in some subset $S$ of $D$, then $\{f_n\}_{n \in \mathbb{N}^\#}$ defines a #-limit function on $S$. The formal definition is as follows.

**Definition 17.1.** Suppose that $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence of functions on $D \subset \mathbb{R}^\#$ and the hyper infinite sequence of values $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ #-converges for each $x$ in some subset $S$ of $D$. Then we say that $\{f_n\}_{n \in \mathbb{N}^\#}$ #-converges pointwise on $S$ to the #-limit function $f$, defined by

$$ f(x) = \#\lim_{n \to \infty} f_n(x), x \in S. \qquad (17.1) $$

**Definition 17.2.** Let $f$ be a function defined on $S \subset \mathbb{R}^\#$ and there exist $\sup_{x \in S} |f(x)|$, then we set

$$ \|f\|_S = \sup_{x \in S} |f(x)|. \qquad (17.2) $$
and

$$ \operatorname{Ext}-\sum_{m=1}^{\infty} \left( \operatorname{Ext}-\sum_{n=1}^{\infty} z_{n,m} \right) \qquad (16.44) $$

are called repeated hyper infinite series.

**Theorem 16.38.** If the double hyper infinite series $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} z_{n,m}$ is #-convergent, then

$$ \lim_{n,m \to \infty} z_{n,m} = 0. \qquad (12.45) $$

**Theorem 16.39.** (Cauchy #-Convergence Criterion for Double hyper infinite Series.)

A double hyper infinite series $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} z_{n,m}$ #-converges if and only if its sequence of partial sums $s_{n,m}$ is Cauchy.

**Theorem 16.40.** If the double series $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} z_{n,m}$ #-converges to $s_1$ and $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} u_{n,m}$ #-converges to $s_2$, then: (i) $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} z_{n,m} + \operatorname{Ext}-\sum_{n=1,m=1}^{\infty} u_{n,m} = s_1 + s_2$.

(ii) $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} c \times z_{n,m} = c \times (\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} z_{n,m})$.

**Theorem 16.41.** Suppose that the double series $\operatorname{Ext}-\sum_{n=1,m=1}^{\infty} z_{n,m}$ is #-convergent, with sum $s$. Then the repeated series $\operatorname{Ext}-\sum_{n=1}^{\infty} (\operatorname{Ext}-\sum_{m=1}^{\infty} z_{n,m})$ and

$\operatorname{Ext}-\sum_{m=1}^{\infty} (\operatorname{Ext}-\sum_{n=1}^{\infty} z_{n,m})$ are both #-convergent with sum $s$ if and only if for every $m \in \mathbb{N}^{\#}$, the series $\operatorname{Ext}-\sum_{n=1}^{\infty} z_{n,m}$ is #-convergent, and for every $n \in \mathbb{N}^{\#}$, the

series $\operatorname{Ext}-\sum_{m=1}^{\infty} z_{n,m}$ is #-convergent.

## 16.10. Interchanging the order of summation of hyper infinite sum.

**Theorem 12.** Assume that

$$ \operatorname{Ext}-\sum_{i=1}^{\infty} \left( \operatorname{Ext}-\sum_{k=1}^{\infty} |a_{jk}| \right) < \infty^{\#}. \qquad (16.46) $$

Then
**Lemma 17.1.** If $g$ and $h$ are defined on $S$, then $\|g+h\|_S \le \|g\|_S + \|h\|_S$ and $\|g \times h\|_S \le \|g\|_S \times \|h\|_S$. Moroever if either $g$ or $h$ is bounded on $S$, then $\|g-h\|_S \ge \|g\|_S - \|h\|_S$.

**Definition 17.2.** A hyper infinite sequence $\{f_n\}_{n \in \mathbb{N}^\#}$ of functions defined on a set $S$ #-converges uniformly to the #-limit function $f$ on $S$ if $\#\lim_{n \to \infty^\#} \|f_n - f\|_S = 0$. Thus, $f_n$ #-converges uniformly to $f$ on $S$ if for each $\varepsilon > 0$, $\varepsilon \approx 0$, there is an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$ \|f_n - f\| < \varepsilon \text{ if } n \ge N. \qquad (17.3) $$

**Theorem 17.1.** Let $f_n, n \in \mathbb{N}^\#$ be hyper infinite sequence defined on $S$. Then (a) $f_n$ #-converges pointwise to $f$ on $S$ if and only if there is, for each $\varepsilon > 0$, $\varepsilon \approx 0$, and $x \in S$, an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ which may depend on $x$ as well as $\varepsilon$ such that $|f_n(x) - f(x)| < \varepsilon$ if $n \ge N$;

(b) $f_n$ #-converges uniformly to $f$ on $S$ if and only if there is for each $\varepsilon > 0$, $\varepsilon \approx 0$, an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ which depends only on and not on any particular $x$ in $S$ such that $|f_n(x) - f(x)| < \varepsilon$ for all $x \in S$ if $n \ge N$.

**Theorem 17.2.** If $f_n$ #-converges uniformly to $f$ on $S$, then $f_n$ #-converges pointwise to $f$ on $S$. The converse is false; that is pointwise #-convergence does not imply uniform #-convergence.

**Theorem 17.3.** (Cauchy's Uniform #-Convergence Criterion) A sequence of functions $f_n$ #-converges uniformly on a set $S$ if and only if for each $\varepsilon > 0$, $\varepsilon \approx 0$, there is an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$ \|f_n - f_m\|_S < \varepsilon \text{ if } n, m \ge N. \qquad (17.4) $$

**Theorem 17.4.** If $f_n$ #-converges uniformly to $f$ on $S$ and each $f_n$ is #-continuous at a point $x_0 \in S$; then so is $f$. Similar statements hold for #-continuity from the right and left.

**Theorem 17.5.** Suppose that $f_n$ #-converges uniformly to $f$ on $S = [a, b]$. Assume that $f$ and all $f_n$ are #-integrable on $[a, b]$. Then

$$ \mathrm{Ext}- \int_a^b f(x) d^\# x = \#\lim_{n \to \infty^\#} \left( \mathrm{Ext}- \int_a^b f_n(x) d^\# x \right). \qquad (17.5) $$

**Proof.** Since

$$ \left| \mathrm{Ext}- \int_a^b f(x) d^\# x - \mathrm{Ext}- \int_a^b f_n(x) d^\# x \right| \le \mathrm{Ext}- \int_a^b |f(x) - f_n(x)| d^\# x \le (b-a) \|f - f_n\|_S \quad (17.6) $$

and $\#\lim_{n \to \infty^\#} \|f - f_n\|_S = 0$, the Eq. (17.5) follows.

**Theorem 17.6.** Suppose that $f_n(x)$ #-converges pointwise to $f$ and each $f_n(x)$ is #-integrable on $[a, b]$. Then

(a) If the #-convergence is uniform, then $f(x)$ is #-integrable on $[a, b]$ and (13.5) holds.

(b) If the sequence $\|f_n\|_{[a,b]}$ is bounded and $f(x)$ is #-integrable on $[a, b]$, then (13.5) holds.

**Theorem 17.7.** Suppose that $f_n^{\prime\#}(x)$ is #-continuous on $[a, b]$ for all $n \in \mathbb{N}^\#$ and $\{f_n^{\prime\#}\}_{n \in \mathbb{N}^\#}$ #-converges uniformly on $[a, b]$ Suppose also that $\{f_n(x_0)\}_{n \in \mathbb{N}^\#}$ #-converges for some $x_0 \in [a, b]$. Then $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ #-converges uniformly on $[a, b]$ to
$$ Ext-\sum_{i=1}^{\infty} \left( Ext-\sum_{k=1}^{\infty} |a_{jk}| \right) = Ext-\sum_{k=1}^{\infty} \left( Ext-\sum_{j=1}^{\infty} |a_{jk}| \right) \quad (16.47) $$

# 17. Hyper infinite sequences and series of $\mathbb{R}^\#$-valued functions.

## 17.1. Uniform #-Convergence

If $f_1, \dots, f_k, f_{k+1}, \dots, f_n, \dots, n \in \mathbb{N}^\#$ are $\mathbb{R}^\#$-valued functions defined on a subset $D \subset \mathbb{R}^\#$ of the hyperreals, we say that $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ is an hyper infinite sequence of functions on $D$. If the sequence of values $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ #-converges for each $x$ in some subset $S$ of $D$, then $\{f_n\}_{n \in \mathbb{N}^\#}$ defines a #-limit function on $S$. The formal definition is as follows.

**Definition 17.1.** Suppose that $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence of functions on $D \subset \mathbb{R}^\#$ and the hyper infinite sequence of values $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ #-converges for each $x$ in some subset $S$ of $D$. Then we say that $\{f_n\}_{n \in \mathbb{N}^\#}$ #-converges pointwise on $S$ to the #-limit function $f$, defined by

$$ f(x) = \#-\lim_{n \to \infty} f_n(x), x \in S. \quad (17.1) $$

**Definition 17.2.** Let $f$ be a function defined on $S \subset \mathbb{R}^\#$ and there exist $\sup_{x \in S}|f(x)|$, then we set

$$ \|f\|_S = \sup_{x \in S} |f(x)|. \quad (17.2) $$

**Lemma 17.1.** If $g$ and $h$ are defined on $S$, then $\|g+h\|_S \le \|g\|_S + \|h\|_S$ and $\|g \times h\|_S \le \|g\|_S \times \|h\|_S$. Moroever if either $g$ or $h$ is bounded on $S$, then $\|g-h\|_S \ge \|g\|_S - \|h\|_S$.

**Definition 17.2.** A hyper infinite sequence $\{f_n\}_{n \in \mathbb{N}^\#}$ of functions defined on a set $S$ #-converges uniformly to the #-limit function $f$ on $S$ if $\#-\lim_{n \to \infty} \|f_n - f\|_S = 0$. Thus, $f_n$ #-converges uniformly to $f$ on $S$ if for each $\varepsilon > 0$, $\varepsilon \approx 0$, there is an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$ \|f_n - f\| < \varepsilon \text{ if } n \ge N. \quad (17.3) $$

**Theorem 17.1.** Let $f_n, n \in \mathbb{N}^\#$ be hyper infinite sequence defined on $S$. Then (a) $f_n$ #-converges pointwise to $f$ on $S$ if and only if there is, for each $\varepsilon > 0$, $\varepsilon \approx 0$, and $x \in S$, an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ which may depend on $x$ as well as $\varepsilon$ such that $|f_n(x) - f(x)| < \varepsilon$ if $n \ge N$; (b) $f_n$ #-converges uniformly to $f$ on $S$ if and only if there is for each $\varepsilon > 0$, $\varepsilon \approx 0$, an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ which depends only on and not on any particular $x$ in $S$ such that $|f_n(x) - f(x)| < \varepsilon$ for all $x \in S$ if $n \ge N$.

**Theorem 17.2.** If $f_n$ #-converges uniformly to $f$ on $S$, then $f_n$ #-converges pointwise to $f$ on $S$. The converse is false; that is pointwise #-convergence does not imply uniform #-convergence.

**Theorem 17.3.** (Cauchy's Uniform #-Convergence Criterion) A sequence of functions $f_n$ #-converges uniformly on a set $S$ if and only if for each $\varepsilon > 0$, $\varepsilon \approx 0$, there is an integer $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that
a #-differentiable #-limit function $f(x)$ and

$$f'^{\#}(x) = \#\lim_{n \to \infty} f_n^{\#}(x), x \in (a, b), \qquad (17.7)$$

while

$$f_{+}^{\prime\#}(a) = \#\lim_{n \to \infty} f_{n}^{\prime\#}(a+), f_{-}^{\prime\#}(b) = \#\lim_{n \to \infty} f_{n}^{\prime\#}(b-). \qquad (17.8)$$

## 17.2. Hyper Infinite Series of Functions.

**Definition 17.3.** If {$f_j(x)$}$_{j=k}^{\infty}$ is a hyper infinite sequence of $\mathbb{R}_c^\#$-valued functions defined on a set $D \subset \mathbb{R}_c^\#$ of hyperreals, then

$$\mathrm{Ext}-\sum_{j=k}^{\infty} f_j(x) \qquad (17.9)$$

is an hyper infinite series of functions on $D$. The partial sums of, $\mathrm{Ext}-\sum_{j=k}^{\infty} f_j(x)$ are defined by

$$F_n(x) = \mathrm{Ext}-\sum_{j=k}^{n} f_j(x), n \in \mathbb{N}^{\#}. \qquad (17.10)$$

If $F_n(x)$ #-converges pointwise to a function $F$ on a subset $S \subset D$, we say that $\mathrm{Ext}-\sum_{j=k}^{n} f_j(x)$ #-converges pointwise to the sum $F(x)$ on $S$, and write

$$F(x) = \mathrm{Ext}-\sum_{j=k}^{\infty} f_j(x). \qquad (17.11)$$

If $F_n(x)$ #-converges uniformly to $F(x)$ on $S$, we say that $\mathrm{Ext}-\sum_{j=k}^{n} f_j(x)$ #-converges uniformly to $F(x)$ on $S$.

**Example 17.1.** The functions $f_j(x) = x^j, j \in \mathbb{N}^\#$ define the hyper infinite series $\mathrm{Ext}-\sum_{j=0}^{\infty} x^j$ on $D = (-\infty^\#, \infty^\#)$. The n-th partial sum of the series is $F_n(x) = \mathrm{Ext}-\sum_{j=0}^{n} x^j$, or, in closed form,

$$F_n(x) = \begin{cases} \frac{1-x^n}{1-x} & x \neq 1 \\ n+1 & x=1 \end{cases} \qquad (17.12)$$

Therefore {$F_n$} #-converges pointwise to $\mathrm{Ext}|_{x|<1}$ and #-diverges if $\mathrm{Ext}|_{x|>1}$, hence, we get $F(x) = \mathrm{Ext}-\sum_{j=0}^{\infty} x^j = (1-x)^{-1}, -1 < x < 1$. Since the difference $F(x) - F_n(x) = \frac{x^n}{1-x}$ can be made arbitrarily infinite large by taking $x$ infinite close to 1, $\|F - F_n\|_{(-1,1)} = \infty^\#$ so the #-convergence is not uniform on $(-1,1)$. Neither is it uniform on any interval $(-1, r]$ with $1 < r < 1$, since $\|F - F_n\|_{[-r,r]} = r^n/(1-r)$ and $\#\lim_{n \to \infty} r^n = \infty^\#$. Put another way, the series #-converges uniformly on #-closed subsets of $(-1,1)$.

**Theorem 17.8.** (Cauchy's Uniform #-Convergence Criterion) A hyper infinite series
$$ \|f_n - f_m\|_S < \varepsilon \text{ if } n, m \ge N. \qquad (17.4) $$

**Theorem 17.4.** If $f_n$ #-converges uniformly to $f$ on $S$ and each $f_n$ is #-continuous at a point $x_0 \in S$; then so is $f$. Similar statements hold for #-continuity from the right and left.

**Theorem 17.5.** Suppose that $f_n$ #-converges uniformly to $f$ on $S = [a, b]$. Assume that $f$ and all $f_n$ are #-integrable on $[a, b]$. Then

$$ \mathrm{Ext}-\int_a^b f(x)d^\#x = \#\lim_{n\to\infty}\left(\mathrm{Ext}-\int_a^b f_n(x)d^\#x\right). \qquad (17.5) $$

**Proof.** Since

$$ \left| \mathrm{Ext}-\int_a^b f(x)d^\#x - \mathrm{Ext}-\int_a^b f_n(x)d^\#x \right| \le \mathrm{Ext}-\int_a^b |f(x)-f_n(x)|d^\#x \le (b-a)\|f-f_n\|_S \quad (17.6) $$

and $\#\lim_{n\to\infty}\|f-f_n\|_S = 0$, the Eq. (17.5) follows.

**Theorem 17.6.** Suppose that $f_n(x)$ #-converges pointwise to $f$ and each $f_n(x)$ is #-integrable on $[a, b]$. Then

(a) If the #-convergence is uniform, then $f(x)$ is #-integrable on $[a, b]$ and (13.5) holds.

(b) If the sequence $\|f_n\|_{[a,b]}$ is bounded and $f(x)$ is #-integrable on $[a, b]$, then (13.5) holds.

**Theorem 17.7.** Suppose that $f_n^{\prime\#}(x)$ is #-continuous on $[a, b]$ for all $n \in \mathbb{N}^\#$ and

$\{f_n^{\prime\#}\}_{n \in \mathbb{N}^\#}$ #-converges uniformly on $[a, b]$. Suppose also that $\{f_n(x_0)\}_{n \in \mathbb{N}^\#}$

#-converges for some $x_0 \in [a, b]$. Then $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ #-converges uniformly on $[a, b]$ to

a #-differentiable #-limit function $f(x)$ and

$$ f^{\prime\#}(x) = \#\lim_{n\to\infty} f_n^{\prime\#}(x), x \in (a, b), \qquad (17.7) $$

while

$$ f_{+}^{\prime\#}(a) = \#\lim_{n\to\infty} f_n^{\prime\#}(a+), f_{-}^{\prime\#}(b) = \#\lim_{n\to\infty} f_n^{\prime\#}(b-). \qquad (17.8) $$

## 17.2. Hyper Infinite Series of Functions.

**Definition 17.3.** If $\{f_j(x)\}_{j=k}^{\infty}$ is a hyper infinite sequence of $\mathbb{R}_c^\#$-valued functions defined on a set $D \subset \mathbb{R}_c^\#$ of hyperreals, then

$$ \mathrm{Ext}-\sum_{j=k}^{\infty} f_j(x) \qquad (17.9) $$

is an hyper infinite series of functions on $D$. The partial sums of, $\mathrm{Ext}-\sum_{j=k}^{\infty} f_j(x)$ are defined by

$$ F_n(x) = \mathrm{Ext}-\sum_{j=k}^{n} f_j(x), n \in \mathbb{N}^\#. \qquad (17.10) $$

If $F_n(x)$ #-converges pointwise to a function $F$ on a subset $S \subset D$, we say that $\mathrm{Ext}-\sum_{j=k}^{n} f_j(x)$ #-converges pointwise to the sum $F(x)$ on $S$, and write
Ext-$\sum_{i=0}^{\infty} f_i(x)$ #-converges uniformly on a set $S \subset \mathbb{R}_c^\#$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$
there is an hyperinteger $N \in \mathbb{N}^\#$ such that

$$ \left\| \mathrm{Ext} - \sum_{n}^{m} f_{i}(x) \right\|_{S} < \varepsilon \qquad (17.13) $$

if $m \ge n \ge N$.

**Corollary 17.1.** If $\mathrm{Ext}-\sum_{i=0}^{\infty} f_i(x)$ #-converges uniformly on $S$, then $\#\lim_{n \to \infty} \|f_n\|_S = 0$.

**Theorem 17.9.** (Weierstrass's Test) The hyper infinite series $\mathrm{Ext}-\sum_{i=0}^{\infty} f_i(x)$ #-converges uniformly on $S$ if

$$ \|f_n\|_S \le M_n, n \ge k, \qquad (17.14) $$

where $\mathrm{Ext}-\sum_{n=k}^{\infty} M_n < \infty^\#$.

**Theorem 17.10.** (Dirichlet's Test for Uniform #-Convergence) The hyper infinite series

$\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)g_n(x)$ #-converges uniformly on $S$ if $f_n$ #-converges uniformly to zero on $S$,

$\mathrm{Ext}-\sum_{n=k}^{\infty} (f_{n+1}(x) - f_n(x))$ #-converges absolutely uniformly on $S$, and

$$ \left\| \mathrm{Ext} - \sum_{i=k}^{n} g_i(x) \right\|_{S} \le M, \qquad (17.14) $$

where $n \ge k$, for some constant $M$.

**Corollary 17.2.** The hyper infinite series $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)g_n(x)$ #-converges uniformly on $S$ if $f_{n+1}(x) \le f_n(x), x \in S, n \ge k, \{f_n\}$ #-converges uniformly to zero on $S$, and

$$ \left\| \mathrm{Ext} - \sum_{i=k}^{n} g_i(x) \right\|_{S} \le M, \qquad (17.15) $$

for some constant $M$.

## 17.3. #-Continuity, #-Differentiability, and Integrability of hyper infinite Series.

**Theorem 17.11.** If $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)$ #-converges uniformly to $F(x)$ on $S$ and each $f_n$ is #-continuous at a point $x_0$ in $S$, then so is $F(x)$. Similar statements hold for #-continuity from the right and left.

**Theorem 17.12.** Suppose that $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)$ #-converges uniformly to $F(x)$ on $S = [a, b]$ Assume that $F(x)$ and $f_n(x), n \ge k$, are integrable on $[a, b]$. Then

$$ \mathrm{Ext} - \int_{a}^{b} F(x) d^{\#}x = \mathrm{Ext} - \sum_{n=k}^{\infty} \left( \mathrm{Ext} - \int_{a}^{b} f_n(x) d^{\#}x \right). \qquad (17.16) $$
$$F(x) = \operatorname{Ext}-\sum_{j=k}^{\infty} f_j(x). \qquad (17.11)$$

If $F_n(x)$ #-converges uniformly to $F(x)$ on $S$, we say that $\operatorname{Ext}-\sum_{j=k}^{n} f_j(x)$ #-converges uniformly to $F(x)$ on $S$.

**Example 17.1.** The functions $f_j(x) = x^j, j \in \mathbb{N}^\#$ define the hyper infinite series $\operatorname{Ext}-\sum_{j=0}^{\infty} x^j$ on $D = (-\infty^\#, \infty^\#)$. The n-th partial sum of the series is $F_n(x) = \operatorname{Ext}-\sum_{j=0}^{n} x^j$, or, in closed form,

$$F_n(x) = \begin{cases} \frac{1-x^n}{1-x} & x \neq 1 \\ n+1 & x=1 \end{cases} \qquad (17.12)$$

Therefore $\{F_n\}$ #-converges pointwise toif $|x| < 1$ and #-diverges if $|x| > 1$, hence, we get $F(x) = \operatorname{Ext}-\sum_{j=0}^{\infty} x^j = (1-x)^{-1}, -1 < x < 1$. Since the difference $F(x) - F_n(x) =$ $= \frac{x^n}{1-x}$ can be made arbitrarily infinite large by taking $x$ infinite close to 1, $\|F - F_n\|_{(-1,1)} = \infty^\#$ so the #-convergence is not uniform on $(-1,1)$. Neither is it uniform on any interval $(-1, r]$ with $1 < r < 1$, since $\|F - F_n\|_{[-r,r]} = r^n/(1-r)$ and $\#-\lim_{n \to \infty^\#} r^n = \infty^\#$. Put another way, the series #-converges uniformly on #-closed subsets of $(-1,1)$.

**Theorem 17.8.** (Cauchy's Uniform #-Convergence Criterion) A hyper infinite series $\operatorname{Ext}-\sum_{i=0}^{\infty} f_i(x)$ #-converges uniformly on a set $S \subset \mathbb{R}^\#$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is an hyperinteger $N \in \mathbb{N}^\#$ such that

$$\left\| \operatorname{Ext}-\sum_{n}^{m} f_{i}(x) \right\|_{S} < \varepsilon \qquad (17.13)$$

if $m \ge n \ge N$.

**Corollary 17.1.** If $\operatorname{Ext}-\sum_{i=0}^{\infty} f_i(x)$ #-converges uniformly on $S$, then $\#-\lim_{n \to \infty^\#} \|f_n\|_S = 0$.

**Theorem 17.9.** (Weierstrass's Test) The hyper infinite series $\operatorname{Ext}-\sum_{i=0}^{\infty} f_i(x)$ #-converges uniformly on $S$ if

$$\|f_n\|_S \le M_n, n \ge k, \qquad (17.14)$$

where $\operatorname{Ext}-\sum_{n=k}^{\infty} M_n < \infty^\#$.

**Theorem 17.10.** (Dirichlet's Test for Uniform #-Convergence) The hyper infinite series $\operatorname{Ext}-\sum_{n=k}^{\infty} f_n(x)g_n(x)$ #-converges uniformly on $S$ if $f_n$ #-converges uniformly to zero on $S$, $\operatorname{Ext}-\sum_{n=k}^{\infty} (f_{n+1}(x) - f_n(x))$ #-converges absolutely uniformly on $S$, and
**Theorem 17.13.** Suppose that $f_n$ is #-continuously #-differentiable on $[a, b]$ for each $n \ge k$, $Ext-\sum_{n=k}^{\infty} f_n(x_0)$ #-converges for some $x_0 \in [a, b]$ and $Ext-\sum_{n=k}^{\infty} f_n^{\#}(x)$ #-converges uniformly on $[a, b]$. $Ext-\sum_{n=k}^{\infty} f_n(x)$ #-converges uniformly on $[a, b]$ to a #-differentiable function $F(x)$, and $F^{\#}(x) = Ext-\sum_{n=k}^{\infty} f_n^{\#}(x)$, $a < x < b$, while $F^{\#}(a+) = Ext-\sum_{n=k}^{\infty} f_n^{\#}(a+)$ and $F^{\#}(b-) = Ext-\sum_{n=k}^{\infty} f_n^{\#}(b-)$.

# 18. Hyper Infinite Power Series.

## 18.1. The convergence properties of hyper infinite power series.

**Definition 18.1.** A hyper infinite series of the form

$$ Ext-\sum_{n=0}^{\infty} a_n(x - x_0)^n \qquad (18.1) $$

where $x_0 \in \mathbb{R}_c^\#$ and $a_n \in \mathbb{R}_c^\#$, $n \in \mathbb{N}^\#$ is called a hyper infinite power series in $(x - x_0)$. The following theorem summarizes the #-convergence properties of hyper infinite power series.

**Theorem 18.1.** In connection with the hyper infinite power series (14.1) define $R$ in the extended hyperreals by

$$ \frac{1}{R} = \overline{\#-\lim}_{n \to \infty} \sqrt[n]{|a_n|} \qquad (18.2) $$

In particular, $R = 0$ if $\overline{\#-\lim}_{n \to \infty} \sqrt[n]{|a_n|} = \infty^\#$, and $R = \infty^\#$ if $\overline{\#-\lim}_{n \to \infty} \sqrt[n]{|a_n|} = 0$. Then the hyper infinite power series #-converges:

(a) only for $x = x_0$ if $R = 0$

(b) for all $x \in \mathbb{R}_c^\#$ if $R = \infty^\#$, and absolutely uniformly in every bounded set;

(c) for $x \in (x_0 - R, x_0 + R)$ if $0 < R < 1$, and absolutely uniformly in every closed subset of this interval.

The series #-diverges if $|x - x_0| > R$. No general statement can be made concerning #-convergence at the endpoints $x = x_0 + R$ and $x = x_0 - R$: the series may #-converge absolutely or conditionally at both; #-converge conditionally at one and #-diverge at the other; or #-diverge at both.

**Theorem 18.2.** The radius of #-convergence of $Ext-\sum_{n=0}^{\infty} a_n(x - x_0)^n$ is given by

$$ \frac{1}{R} = \#-\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \qquad (18.3) $$

if the #-limit exists in the extended hyperreals.

**Example 18.1.** For the hyper infinite power series

$$ Ext-\sum_{n=0}^{\infty} \frac{x^n}{n!} \qquad (18.4) $$

one obtains that
$$ \left\| \mathrm{Ext} - \sum_{i=k}^{n} g_i(x) \right\|_{S} \leq M, \qquad (17.14) $$

where $n \geq k$, for some constant $M$.

**Corollary 17.2.** The hyper infinite series $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)g_n(x)$ #-converges uniformly on $S$ if $f_{n+1}(x) \leq f_n(x), x \in S, n \geq k, \{f_n\}$ #-converges uniformly to zero on $S$, and

$$ \left\| \mathrm{Ext} - \sum_{i=k}^{n} g_i(x) \right\|_{S} \leq M, \qquad (17.15) $$

for some constant $M$.

## 17.3. #-Continuity, #-Differentiability, and Integrability of hyper infinite Series.

**Theorem 17.11.** If $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)$ #-converges uniformly to $F(x)$ on $S$ and each $f_n$ is #-continuous at a point $x_0$ in $S$, then so is $F(x)$. Similar statements hold for #-continuity from the right and left.

**Theorem 17.12.** Suppose that $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)$ #-converges uniformly to $F(x)$ on $S = [a, b]$ Assume that $F(x)$ and $f_n(x), n \geq k$, are integrable on $[a, b]$. Then

$$ \mathrm{Ext} - \int_{a}^{b} F(x) d^{\#} x = \mathrm{Ext} - \sum_{n=k}^{\infty} \left( \mathrm{Ext} - \int_{a}^{b} f_{n}(x) d^{\#} x \right). \qquad (17.16) $$

**Theorem 17.13.** Suppose that $f_n$ is #-continuously #-differentiable on $[a, b]$ for each $n \geq k$, $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x_0)$ #-converges for some $x_0 \in [a, b]$ and $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n^{\#}(x)$ #-converges uniformly on $[a, b]$. $\mathrm{Ext}-\sum_{n=k}^{\infty} f_n(x)$ #-converges uniformly on $[a, b]$ to a #-differentiable function $F(x)$, and $F^{\prime\#}(x) = \mathrm{Ext}-\sum_{n=k}^{\infty} f_n^{\#}(x), a < x < b$, while $F^{\prime\#}(a+) = \mathrm{Ext}-\sum_{n=k}^{\infty} f_n^{\#}(a+)$ and $F^{\prime\#}(b-) = \mathrm{Ext}-\sum_{n=k}^{\infty} f_n^{\#}(b-)$.

# 18. Hyper Infinite Power Series.

## 18.1. The convergence properties of hyper infinite power series.

**Definition 18.1.** A hyper infinite series of the form

$$ \mathrm{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n \qquad (18.1) $$

where $x_0 \in \mathbb{R}_c^\#$ and $a_n \in \mathbb{R}_c^\#$, $n \in \mathbb{N}^\#$ is called a hyper infinite power series in $(x - x_0)$. The following theorem summarizes the #-convergence properties of hyper infinite power series.

**Theorem 18.1.** In connection with the hyper infinite power series (14.1) define $R$ in
$$ \#\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \#\lim_{n \to \infty} \frac{n!}{(n+1)!} = \#\lim_{n \to \infty} \frac{1}{n+1} = 0. \quad (18.4') $$

Therefore, $R = \infty^{\#}$; that is, the series #-converges for all $x \in \mathbb{R}^{\#}$, and absolutely uniformly in every bounded set.

**Theorem 18.3.** A hyper infinite power series

$$ f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n \quad (18.5) $$

with positive radius of #-convergence $R$ is #-continuous and #-differentiable in its interval of #-convergence; and its #-derivative can be obtained by #-differentiating term by term; that is;

$$ f'^{\#}(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} n a_n (x - x_0)^{n-1} \quad (18.6) $$

which can also be written as

$$ f'^{\#}(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} (n+1) a_{n+1} (x - x_0)^n \quad (18.7) $$

This hyper infinite series also has radius of #-convergence $R$.

**Theorem 18.4.** A hyper infinite power series

$$ f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n \quad (18.8) $$

with positive radius of #-convergence $R$ has #-derivatives of all orders in its interval of #-convergence, which can be obtained by repeated term by term #-differentiation thus,

$$ \begin{aligned} f^{(n)^{\#}}(x) &= \operatorname{Ext} - \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)a_n(x-x_0)^n = \\ &= \operatorname{Ext} - \sum_{n=k}^{\infty} \left[ \left( \operatorname{Ext} - \prod_{j=n-k+1}^{n} j \right) a_n(x-x_0)^n \right]. \end{aligned} \quad (18.9) $$

The radius of #-convergence of each of these hyper infinite series is $R$.

**Corollary 18.1.** (Uniqueness of hyper infinite Power Series) If

$$ \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n = \operatorname{Ext} - \sum_{n=0}^{\infty} b_n (x - x_0)^n \quad (18.10) $$

for all $x$ in some interval $(x_0 - r, x_0 + r)$ then

$$ a_n = b_n, n \ge 0. \quad (18.11) $$

**Corollary 18.2.** If

$$ f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n, |x - x_0| < R \quad (18.12) $$

then

$$ a_n = \frac{f^{(n)^{\#}}(x)}{n!} \quad (18.13) $$

**Theorem 18.5.** If $x_1$ and $x_2$ are in the interval of #-convergence of
the extended hyperreals by

$$ \frac{1}{R} = \overline{\text{-lim}_{n \to \infty}} \sqrt[n]{|a_n|} \qquad (18.2) $$

In particular, $R = 0$ if $\overline{\text{-lim}_{n \to \infty}} \sqrt[n]{|a_n|} = \infty^\#$, and $R = \infty^\#$ if $\overline{\text{-lim}_{n \to \infty}} \sqrt[n]{|a_n|} = 0$. Then the hyper infinite power series #-converges:

(a) only for $x = x_0$ if $R = 0$

(b) for all $x \in \mathbb{R}_c^\#$ if $R = \infty^\#$, and absolutely uniformly in every bounded set;

(c) for $x \in (x_0 - R, x_0 + R)$ if $0 < R < 1$, and absolutely uniformly in every closed subset of this interval.

The series #-diverges if $|x - x_0| > R$. No general statement can be made concerning #-convergence at the endpoints $x = x_0 + R$ and $x = x_0 - R$: the series may #-converge absolutely or conditionally at both; #-converge conditionally at one and #-diverge at the other; or #-diverge at both.

**Theorem 18.2.** The radius of #-convergence of $Ext-\sum_{n=0}^{\infty} a_n(x - x_0)^n$ is given by

$$ \frac{1}{R} = \overline{\text{-lim}_{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \qquad (18.3) $$

if the #-limit exists in the extended hyperreals.

**Example 18.1.** For the hyper infinite power series

$$ \mathrm{Ext}-\sum_{n=0}^{\infty} \frac{x^n}{n!} \qquad (18.4) $$

one obtains that

$$ \overline{\text{-lim}_{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \overline{\text{-lim}_{n \to \infty}} \frac{n!}{(n+1)!} = \overline{\text{-lim}_{n \to \infty}} \frac{1}{n+1} = 0. \qquad (18.4') $$

Therefore, $R = \infty^\#$; that is, the series #-converges for all $x \in \mathbb{R}_c^\#$, and absolutely uniformly

in every bounded set.

**Theorem 18.3.** A hyper infinite power series

$$ f(x) = \mathrm{Ext}-\sum_{n=0}^{\infty} a_n (x - x_0)^n \qquad (18.5) $$

with positive radius of #-convergence $R$ is #-continuous and #-differentiable in its interval of #-convergence; and its #-derivative can be obtained by #-differentiating term by term; that is;

$$ f^{\prime\#}(x) = \mathrm{Ext}-\sum_{n=0}^{\infty} n a_n (x - x_0)^{n-1} \qquad (18.6) $$

which can also be written as

$$ f^{\prime\#}(x) = \mathrm{Ext}-\sum_{n=0}^{\infty} (n+1) a_{n+1} (x - x_0)^n \qquad (18.7) $$

This hyper infinite series also has radius of #-convergence $R$.

**Theorem 18.4.** A hyper infinite power series

$$ f(x) = \mathrm{Ext}-\sum_{n=0}^{\infty} a_n (x - x_0)^n \qquad (18.8) $$

with positive radius of #-convergence $R$ has #-derivatives of all orders in its interval of
$$f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n \qquad (18.14)$$

Then

$$\operatorname{Ext} - \int_{x_1}^{x_2} f(x) d^\# x = \operatorname{Ext} - \sum_{n=0}^{\infty} \frac{a_n}{n+1} \left[ (x_2 - x_0)^{n+1} - (x_1 - x_0)^{n+1} \right] \quad (18.15)$$

that is, a hyper infinite power series may be integrated term by term between any two points in its interval of #-convergence.

**Theorem 18.6.** Suppose that $f(x)$ is hyper infinitely #-differentiable on an interval $I$ and

$$\#\operatorname{lim}_{n \to \infty} \frac{r^n}{n!} \left\| f^{(n)}(x) \right\|_I = 0. \qquad (18.16)$$

Then, if $x_0 \in I^0$, the hyper infinite Taylor series

$$\operatorname{Ext} - \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} (x - x_0)^n \qquad (18.17)$$

#-converges uniformly to $f(x)$ on $I_r = I \cap [x_0 - r, x_0 + r]$.

**Theorem 18.7.** If

$$f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n, |x - x_0| < R_1 \qquad (18.18)$$

and

$$g(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} b_n (x - x_0)^n, |x - x_0| < R_2 \qquad (18.19)$$

and $\alpha$ and $\beta$ are constants, then

$$\alpha f(x) + \beta g(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} (\alpha a_n + \beta b_n) (x - x_0)^n, |x - x_0| < R, \qquad (18.20)$$

where $R \ge \min\{R_1, R_2\}$.

**Theorem 18.8.** If $f(x)$ and $g(x)$ are given by Eq.(18.19) and Eq.(18.20) correspondingly, then

$$f(x)g(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} c_n (x - x_0)^n, |x - x_0| < R, \qquad (18.21)$$

where

$$c_n = \operatorname{Ext} - \sum_{j=0}^{n} a_j b_{n-j} = \sum_{j=0}^{n} a_{n-j} b_j, \qquad (18.22)$$

$n \in \mathbb{N}^\#$ and $R \ge \min\{R_1, R_2\}$.

**Theorem 18.9.** (Generalized Abel's Theorem) Let $f(x)$ be defined by a hyper infinite power series

$$f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n, |x - x_0| < R \qquad (18.23)$$

with finite or hyperfinite radius of #-convergence $R \in \mathbb{R}^\#_c$.

(a) If $\operatorname{Ext} - \sum_{n=0}^{\infty} a_n R^n$ #-converges, then
#-convergence, which can be obtained by repeated term by term #-differentiation thus,

$$
\begin{align}
f^{(n)\#}(x) &= \operatorname{Ext-}\sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1) a_n(x-x_0)^n = \nonumber \\
&= \operatorname{Ext-}\sum_{n=k}^{\infty} \left[ \left( \operatorname{Ext-}\prod_{j=n-k+1}^{n} j \right) a_n(x-x_0)^n \right]. \tag{18.9}
\end{align}
$$

The radius of #-convergence of each of these hyper infinite series is R.

**Corollary 18.1.** (Uniqueness of hyper infinite Power Series) If

$$
\operatorname{Ext-}\sum_{n=0}^{\infty} a_n(x - x_0)^n = \operatorname{Ext-}\sum_{n=0}^{\infty} b_n(x - x_0)^n \quad (18.10)
$$

for all $x$ in some interval $(x_0 - r, x_0 + r)$ then

$$
a_n = b_n, n \ge 0. \tag{18.11}
$$

**Corollary 18.2.** If

$$
f(x) = \operatorname{Ext-} \sum_{n=0}^{\infty} a_n (x - x_0)^n, |x - x_0| < R \quad (18.12)
$$

then

$$
a_n = \frac{f^{(n)\#}(x)}{n!}. \tag{18.13}
$$

**Theorem 18.5.** If $x_1$ and $x_2$ are in the interval of #-convergence of

$$
f(x) = \operatorname{Ext-} \sum_{n=0}^{\infty} a_n (x - x_0)^n \qquad (18.14)
$$

Then

$$
\operatorname{Ext-} \int_{x_1}^{x_2} f(x) d^\# x = \operatorname{Ext-} \sum_{n=0}^{\infty} \frac{a_n}{n+1} \left[ (x_2 - x_0)^{n+1} - (x_1 - x_0)^{n+1} \right] \quad (18.15)
$$

that is, a hyper infinite power series may be integrated term by term between any
two points in its interval of #-convergence.

**Theorem 18.6.** Suppose that $f(x)$ is hyper infinitely #-differentiable on an interval $I$
and

$$
\#\text{-lim}_{n\to\infty} \frac{r^n}{n!} \left\| f^{(n)\#}(x) \right\|_I = 0. \quad (18.16)
$$

Then, if $x_0 \in I^0$, the hyper infinite Taylor series

$$
\operatorname{Ext-} \sum_{n=0}^{\infty} \frac{f^{(n)\#}(x)}{n!} (x - x_0)^n \quad (18.17)
$$

#-converges uniformly to $f(x)$ on $I_r = I \cap [x_0 - r, x_0 + r]$.

**Theorem 18.7.** If

$$
f(x) = \operatorname{Ext-} \sum_{n=0}^{\infty} a_n (x - x_0)^n, |x - x_0| < R_1 \quad (18.18)
$$

and
$$ \#- \lim_{x \to \# (x_0+R)-} f(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} a_n R^n. \qquad (18.24) $$

(b) If $\mathrm{Ext}-\sum_{n=0}^{\infty} (-1)^n a_n R^n$ #-converges, then

$$ \#- \lim_{x \to \# (x_0-R)+} f(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} (-1)^n a_n R^n. \qquad (18.25) $$

## 18.2. The $\mathbb{R}_c^\#$-valued #-exponential $\mathrm{Ext}\text{-exp}(x)$

We define the #-exponential $\mathrm{Ext}\text{-exp}(x)$ function as the solution of the differential equation

$$ f^{\prime\#}(x) = f(x), f(0) = 1. \qquad (18.26) $$

We solve it by setting

$$ f(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} a_n x^n, f^{\prime\#}(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} n a_n x^n. \qquad (18.27) $$

Therefore

$$ \mathrm{Ext}\text{-exp}(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} \frac{x^n}{n!} \qquad (18.28) $$

From Eq. (18.40) and Eq. (18.28) we get

$$ (\mathrm{Ext}\text{-exp}(x))(\mathrm{Ext}\text{-exp}(y)) = \mathrm{Ext}\text{-exp}(x+y), \qquad (18.29) $$

for any $x, y \in \mathbb{R}_c^\#$. We often denote #-exponential $\mathrm{Ext}\text{-exp}(x)$ by $\mathrm{Ext}\text{-e}^x$

$$ \mathrm{Ext}\text{-e}^x. \qquad (18.30) $$

## 18.3. The $\mathbb{R}_c^\#$-valued Trigonometric Functions $\mathrm{Ext}\text{-sin}(x)$ and $\mathrm{Ext}\text{-cos}(x)$.

We define the $\mathbb{R}_c^\#$-valued Trigonometric Functions $\mathrm{Ext}\text{-sin}(x)$ and $\mathrm{Ext}\text{-cos}(x)$ by

$$ \mathrm{Ext}\text{-sin}(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \qquad (18.31) $$

and

$$ \mathrm{Ext}\text{-cos}(x) = \mathrm{Ext}- \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}. \qquad (18.32) $$

It can be shown that the series (18.30) and (18.31) #-converge for all $x \in \mathbb{R}_c^\#$ and #-differentiating yields

$$ [\mathrm{Ext}\text{-sin}(x)]^{\prime\#} = \mathrm{Ext}- \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = \mathrm{Ext}\text{-cos}(x) \qquad (18.33) $$

and

$$ \begin{aligned} [\mathrm{Ext}\text{-cos}(x)]^{\prime\#} &= \mathrm{Ext}- \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n-1)!} = -\mathrm{Ext}- \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = \\ &= -[\mathrm{Ext}\text{-sin}(x)]. \end{aligned} \qquad (18.34) $$
$$g(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} b_n (x - x_0)^n, |x - x_0| < R_2 \quad (18.19)$$

and $\alpha$ and $\beta$ are constants, then

$$\alpha f(x) + \beta g(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} (\alpha a_n + \beta b_n)(x - x_0)^n, |x - x_0| < R, \quad (18.20)$$

where $R \ge \min\{R_1, R_2\}$.

**Theorem 18.8.** If $f(x)$ and $g(x)$ are given by Eq.(18.19) and Eq.(18.20) correspondingly, then

$$f(x)g(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} c_n (x - x_0)^n, |x - x_0| < R, \quad (18.21)$$

where

$$c_n = \operatorname{Ext} - \sum_{j=0}^{n} a_j b_{n-j} = \sum_{j=0}^{n} a_{n-j} b_j, \quad (18.22)$$

$n \in \mathbb{N}^\#$ and $R \ge \min\{R_1, R_2\}$.

**Theorem 18.9.** (Generalized Abel's Theorem) Let $f(x)$ be defined by a hyper infinite power series

$$f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n (x - x_0)^n, |x - x_0| < R \quad (18.23)$$

with finite or hyperfinite radius of #-convergence $R \in \mathbb{R}^\#_c$.

(a) If $\operatorname{Ext} - \sum_{n=0}^{\infty} a_n R^n$ #-converges, then

$$\#- \lim_{x \to \# (x_0+R)-} f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n R^n. \quad (18.24)$$

(b) If $\operatorname{Ext} - \sum_{n=0}^{\infty} (-1)^n a_n R^n$ #-converges, then

$$\#- \lim_{x \to \# (x_0-R)+} f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} (-1)^n a_n R^n. \quad (18.25)$$

## 18.2. The $\mathbb{R}^\#$-valued #-exponential $\operatorname{Ext}-\exp(x)$

We define the #-exponential $\operatorname{Ext}-\exp(x)$ function as the solution of the differential equation

$$f^{\#}(x) = f(x), f(0) = 1. \quad (18.26)$$

We solve it by setting

$$f(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} a_n x^n, f^{\#}(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} n a_n x^n. \quad (18.27)$$

Therefore

$$\operatorname{Ext}-\exp(x) = \operatorname{Ext} - \sum_{n=0}^{\infty} \frac{x^n}{n!} \quad (18.28)$$

From Eq.(18.40) and Eq.(18.28) we get
### 18.4. $\mathbb{R}_c^\#$-valued functions of several variables.

In this subsection we study $\mathbb{R}_c^\#$-valued functions defined on subsets of the $n$-dimensional external linear space $\mathbb{R}_c^{\#n}$, $n \in \mathbb{N}^\#$ which consists of all external and internal hyperfinite (or finite) sequences (called a vector) $\mathbf{X} = \{x_i\}_{i=1}^{i=n} = \{x_i\}_{i \in n}$ of hyperreal numbers, called the coordinates or components of vector $\mathbf{X}$.

**Definition 18.2.** The vector sum of $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ and $\mathbf{Y} = \{y_i\}_{i=1}^{i=n}$ is

$$ \mathbf{X} + \mathbf{Y} = \{x_i + y_i\}_{i=1}^{i=n}. \qquad (18.35) $$

If $a \in \mathbb{R}_c^\#$ is a hyperreal number, the scalar multiple of $\mathbf{X}$ by $a$ is

$$ a \cdot \mathbf{X} = \{ax_i\}_{i=1}^{i=n}. \qquad (18.36) $$

**Theorem 18.10.** If $\mathbf{X}, \mathbf{Y}$, and $\mathbf{Z}$ are in $\mathbb{R}_c^{\#n}$ and $a, b \in \mathbb{R}_c^\#$ are hyperreal numbers, then

(i) $\mathbf{X} + \mathbf{Y} = \mathbf{Y} + \mathbf{X}$ - vector addition is commutative

(ii) $(\mathbf{X} + \mathbf{Y}) + \mathbf{Z} = \mathbf{X} + (\mathbf{Y} + \mathbf{Z})$ - vector addition is associative

(iii) There is a unique vector $\mathbf{0}$, called the zero vector, such that $\mathbf{X} + \mathbf{0} = \mathbf{X}$ for all $\mathbf{X} \in \mathbb{R}_c^{\#n}$

(iv) For each $\mathbf{X} \in \mathbb{R}_c^{\#n}$ there is a unique vector $-\mathbf{X}$ such that $\mathbf{X} + (-\mathbf{X}) = \mathbf{0}$

(v) $a \cdot (b \cdot \mathbf{X}) = (ab) \cdot \mathbf{X}$

(vi) $(a+b) \cdot \mathbf{X} = a \cdot \mathbf{X} + b \cdot \mathbf{X}$

(vii) $a \cdot (\mathbf{X} + \mathbf{Y}) = a \cdot \mathbf{X} + a \cdot \mathbf{Y}$

(viii) $1 \cdot \mathbf{X} = \mathbf{X}$.

Clearly, $\mathbf{0} = \{0\}_{i=1}^{i=n}$ and, if $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$, then $-\mathbf{X} = \{-x_i\}_{i=1}^{i=n}$.

We write $\mathbf{X} + (-\mathbf{Y})$ as $\mathbf{X} - \mathbf{Y}$. The point $\mathbf{0}$ is called the origin.

**Definition 18.3.** The length of the vector $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ is

$$ \|\mathbf{X}\| = (\mathrm{Ext} - \sum_{i=1}^{n} x_i^2)^{1/2}. \qquad (18.37) $$

The distance between points $\mathbf{X}$ and $\mathbf{Y}$ is $\|\mathbf{X} - \mathbf{Y}\|$; in particular, $\|\mathbf{X}\|$ is the distance between $\mathbf{X}$ and the origin. If $\|\mathbf{X}\| = 1$, then $\mathbf{X}$ is a unit vector.

**Definition 18.4.** The inner product $\mathbf{X} \cdot \mathbf{Y}$ of $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ and $\mathbf{Y} = \{y_i\}_{i=1}^{i=n}$ is

$$ \mathbf{X} \cdot \mathbf{Y} = \mathrm{Ext} - \sum_{i=1}^{n} x_i y_i. \qquad (18.38) $$

**Theorem 18.11.** (Schwarz's Inequality) If $\mathbf{X}, \mathbf{Y} \in \mathbb{R}_c^{\#n}$ then

$$ \|\mathbf{X} \cdot \mathbf{Y}\| \le \|\mathbf{X}\| \|\mathbf{Y}\|, \qquad (18.39) $$

with equality if and only if one of the vectors is a scalar multiple of the other:

**Theorem 18.12.** (Triangle Inequality) If $\mathbf{X}, \mathbf{Y} \in \mathbb{R}_c^{\#n}$ then

$$ \|\mathbf{X} + \mathbf{Y}\| \le \|\mathbf{X}\| + \|\mathbf{Y}\|, \qquad (18.40) $$

with equality if and only if one of the vectors is a nonnegative multiple of the other.

**Corollary 18.3.** If $\mathbf{X}, \mathbf{Y}, \mathbf{Z} \in \mathbb{R}_c^{\#n}$, then

$$ \|\mathbf{X} - \mathbf{Z}\| \le \|\mathbf{X} - \mathbf{Y}\| + \|\mathbf{Y} - \mathbf{Z}\|. \qquad (18.41) $$

**Corollary 18.4.** If $\mathbf{X}, \mathbf{Y} \in \mathbb{R}_c^{\#n}$, then

$$ \|\mathbf{X} - \mathbf{Y}\| \ge \|\mathbf{X}\| - \|\mathbf{Y}\|. \qquad (18.42) $$

**Theorem 18.13.** If $\mathbf{X}, \mathbf{Y}, \mathbf{Z} \in \mathbb{R}_c^{\#n}$ and $a \in \mathbb{R}_c^\#$ is a scalar, then

(i) $\|a\mathbf{X}\| = |a|\|\mathbf{X}\|$

(ii) $\|\mathbf{X}\| \ge 0$, with equality if and only if $\mathbf{X} = \mathbf{0}$
$$ (\mathrm{Ext}-\exp(x))(\mathrm{Ext}-\exp(y)) = \mathrm{Ext}-\exp(x+y), \qquad (18.29) $$

for any $x, y \in \mathbb{R}_c^\#$. We often denote $\#\text{-exponential } \mathrm{Ext}-\exp(x)$ by $\mathrm{Ext}-e^x$

$$ \mathrm{Ext}-e^x. \qquad (18.30) $$

### 18.3. The $\mathbb{R}_c^\#$-valued Trigonometric Functions $\mathrm{Ext}-\sin(x)$ and $\mathrm{Ext}-\cos(x)$.

We define the $\mathbb{R}_c^\#$-valued Trigonometric Functions $\mathrm{Ext}-\sin(x)$ and $\mathrm{Ext}-\cos(x)$ by

$$ \mathrm{Ext}-\sin(x) = \mathrm{Ext}-\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \qquad (18.31) $$

and

$$ \mathrm{Ext}-\cos(x) = \mathrm{Ext}-\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}. \qquad (18.32) $$

It can be shown that the series (18.30) and (18.31) $\#\text{-converge for all } x \in \mathbb{R}_c^\#$ and $\#\text{-differentiating yields}$

$$ [\mathrm{Ext}-\sin(x)]^{\prime\#} = \mathrm{Ext}-\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = \mathrm{Ext}-\cos(x) \qquad (18.33) $$

and

$$ \begin{aligned} [\mathrm{Ext}-\cos(x)]^{\prime\#} &= \mathrm{Ext}-\sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n-1)!} = -\mathrm{Ext}-\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = \\ &= -[\mathrm{Ext}-\sin(x)]. \end{aligned} \qquad (18.34) $$

### 18.4. $\mathbb{R}_c^\#$-valued functions of several variables.

In this subsection we study $\mathbb{R}_c^\#$-valued functions defined on subsets of the $n$-dimensional external linear space $\mathbb{R}_c^{\#n}, n \in \mathbb{N}^\#$ which consists of all external and internal hyperfinite (or finite) sequences (called a vector) $\mathbf{X} = \{x_i\}_{i=1}^{i=n} = \{x_i\}_{i \in n}$ of hyperreal numbers, called the coordinates or components of vector $\mathbf{X}$.

**Definition 18.2.** The vector sum of $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ and $\mathbf{Y} = \{y_i\}_{i=1}^{i=n}$ is

$$ \mathbf{X} + \mathbf{Y} = \{x_i + y_i\}_{i=1}^{i=n}. \qquad (18.35) $$

If $a \in \mathbb{R}_c^\#$ is a hyperreal number, the scalar multiple of $\mathbf{X}$ by $a$ is

$$ a \cdot \mathbf{X} = \{ax_i\}_{i=1}^{i=n}. \qquad (18.36) $$

**Theorem 18.10.** If $\mathbf{X}, \mathbf{Y}$, and $\mathbf{Z}$ are in $\mathbb{R}_c^{\#n}$ and $a, b \in \mathbb{R}_c^\#$ are hyperreal numbers, then

(i) $\mathbf{X} + \mathbf{Y} = \mathbf{Y} + \mathbf{X}$ - vector addition is commutative

(ii) $(\mathbf{X} + \mathbf{Y}) + \mathbf{Z} = \mathbf{X} + (\mathbf{Y} + \mathbf{Z})$ - vector addition is associative

(iii) There is a unique vector $\mathbf{0}$, called the zero vector, such that $\mathbf{X} + \mathbf{0} = \mathbf{X}$ for all $\mathbf{X} \in \mathbb{R}_c^{\#n}$

(iv) For each $\mathbf{X} \in \mathbb{R}_c^{\#n}$ there is a unique vector $-\mathbf{X}$ such that $\mathbf{X} + (-\mathbf{X}) = \mathbf{0}$

(v) $a \cdot (b \cdot \mathbf{X}) = (ab) \cdot \mathbf{X}$

(vi) $(a + b) \cdot \mathbf{X} = a \cdot \mathbf{X} + b \cdot \mathbf{X}$

(vii) $a \cdot (\mathbf{X} + \mathbf{Y}) = a \cdot \mathbf{X} + a \cdot \mathbf{Y}$

(viii) $1 \cdot \mathbf{X} = \mathbf{X}$.
(iii) $||X - Y|| \ge 0$, with equality if and only if $X = Y$

(iv) $X \cdot Y = Y \cdot X$

(v) $X \cdot (Y + Z) = X \cdot Y + X \cdot Z$

(vi) $(cX) \cdot Y = X \cdot (cY) = c(X \cdot Y)$

**Definition 18.5.** Non-Archimedean metric space $(X, d)$ is a set $X$ together with a $\mathbb{R}_c^\#$-valued function $d: X \times X \to \mathbb{R}_c^\#$ (called a metric or non-Archimedean distance function) which assigns a hyperreal number $d(x, y)$ to every pair $x, y$ belongs $X$ satisfying the properties:

1. $d(x, y) \ge 0$ and $d(x, y) = 0$ iff $x = y$,

2. $d(x, y) = d(y, x)$,

3. $d(x, y) + d(y, z) \ge d(x, z)$.

**Remark 18.1.** Note that external linear space $\mathbb{R}_c^{\#n}$ endowed with distance function $d(X, Y) = ||X - Y||$ satisfying the properties 1-3 mentioned above in Definition 14.5.

## 18.5. Line Segments in $\mathbb{R}_c^{\#n}, n \in \mathbb{N}^\#$.

**Definition 18.6.** Suppose that $X_0, U \in \mathbb{R}_c^{\#n}$ and $U \ne 0$. Then the line through $X_0$ in the direction of $U$ is the set of all points in $\mathbb{R}_c^{\#n}$ of the form

$$X(X_0, U) = X_0 + tU, -\infty < t < \infty. \quad (18.43)$$

A set of points of the form

$$X = X_0 + tU, t_1 \le t \le t_2 \quad (18.44)$$

is called a line segment. In particular, the line segment from $X_0$ to $X_1$ is the set of points of the form

$$X = X_0 + t(X_1 - X_0) = tX_1 + (1-t)X_0, 0 \le t \le 1. \quad (18.45)$$

**Definition 18.7.** A hyper infinite sequence of points $X_n, n \in \mathbb{N}^\#$ in $\mathbb{R}_c^{\#n}$ #-converges to the #-limit $\bar{X}$ if

$$\#\lim_{n \to \infty} \# ||X_n - \bar{X}|| = 0. \quad (18.46)$$

In this case we write $\#\lim_{n \to \infty} \# X_n = \bar{X}$.

**Theorem 18.14.** Let $\bar{X} = \{x_i\}_{i=1}^{i=n}$ and $X_m = \{x_{im}\}_{i=1}^{i=n}, m \ge 1$. Then $\#\lim_{m \to \infty} \# X_m = \bar{X}$ if and only if $\#\lim_{m \to \infty} \# x_{im} = \bar{x}_i, 1 \le i \le n$; that is a hyper infinite sequence $\{X_m\}$ of points in $\mathbb{R}_c^{\#n}$ #-converges to a #-limit $\bar{X}$ if and only if the hyper infinite sequences of components of $\{X_m\}$ #-converge to the respective components of $\bar{X}$.

**Theorem 18.15.** (Cauchy's #-Convergence Criterion) A hyper infinite sequence $\{X_m\}$ in $\mathbb{R}_c^{\#n}$ #-converges if and only if for each $\varepsilon > 0, \varepsilon \approx 0$, there is an hyperinteger $N \in \mathbb{N}^\# \mathbb{N}$ such that

$$||X_n - X_m|| < \varepsilon \quad (18.47)$$

if $n, m \ge N$.

**Definition 18.8.** If $A$ is a subset of a metric space $\mathbb{R}_c^{\#n}$ then $x$ is a #-limit point of $A$ if it is the #-limit of an eventually non-constant hyper infinite sequence $\{a_i\}_{i \in \mathbb{N}^\#}$ of points of $A$.

**Definition 18.9.** A subset $A$ is said to be a #-closed subset of $\mathbb{R}_c^{\#n}$ if it contains all its #-limit points.

**Example 18.1.** (i) $\mathbb{R}_c^\#$ with the canonical metric $d(x, y) = |x - y|$, since in $\mathbb{R}_c^\#$ every hyperreal number is a #-limit point of the hyper infinite sequence $\{q_i\}_{i \in \mathbb{N}^\#}$ of
Clearly, $\mathbf{0} = \{0\}_{i=1}^{i=n}$ and, if $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$, then $-\mathbf{X} = \{-x_i\}_{i=1}^{i=n}$.

We write $\mathbf{X} + (-\mathbf{Y})$ as $\mathbf{X} - \mathbf{Y}$. The point $\mathbf{0}$ is called the origin.

**Definition 18.3.** The length of the vector $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ is

$$ \|\mathbf{X}\| = (\mathrm{Ext} - \sum_{i=1}^{n} x_i^2)^{1/2}. \qquad (18.37) $$

The distance between points $\mathbf{X}$ and $\mathbf{Y}$ is $\|\mathbf{X} - \mathbf{Y}\|$; in particular, $\|\mathbf{X}\|$ is the distance between $\mathbf{X}$ and the origin. If $\|\mathbf{X}\| = 1$, then $\mathbf{X}$ is a unit vector.

**Definition 18.4.** The inner product $\mathbf{X} \cdot \mathbf{Y}$ of $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ and $\mathbf{Y} = \{y_i\}_{i=1}^{i=n}$ is

$$ \mathbf{X} \cdot \mathbf{Y} = \mathrm{Ext} - \sum_{i=1}^{n} x_i y_i. \qquad (18.38) $$

**Theorem 18.11.** (Schwarz's Inequality) If $\mathbf{X}, \mathbf{Y} \in \mathbb{R}_c^{\#n}$ then

$$ \|\mathbf{X} \cdot \mathbf{Y}\| \le \|\mathbf{X}\| \|\mathbf{Y}\|, \qquad (18.39) $$

with equality if and only if one of the vectors is a scalar multiple of the other:

**Theorem 18.12.** (Triangle Inequality) If $\mathbf{X}, \mathbf{Y} \in \mathbb{R}_c^{\#n}$ then

$$ \|\mathbf{X} + \mathbf{Y}\| \le \|\mathbf{X}\| + \|\mathbf{Y}\|, \qquad (18.40) $$

with equality if and only if one of the vectors is a nonnegative multiple of the other.

**Corollary 18.3.** If $\mathbf{X}, \mathbf{Y}, \mathbf{Z} \in \mathbb{R}_c^{\#n}$, then

$$ \|\mathbf{X} - \mathbf{Z}\| \le \|\mathbf{X} - \mathbf{Y}\| + \|\mathbf{Y} - \mathbf{Z}\|. \qquad (18.41) $$

**Corollary 18.4.** If $\mathbf{X}, \mathbf{Y} \in \mathbb{R}_c^{\#n}$, then

$$ \|\mathbf{X} - \mathbf{Y}\| \ge \|\|\mathbf{X}\| - \|\mathbf{Y}\|\|. \qquad (18.42) $$

**Theorem 18.13.** If $\mathbf{X}, \mathbf{Y}, \mathbf{Z} \in \mathbb{R}_c^{\#n}$ and $a \in \mathbb{R}^\#$ is a scalar, then

(i) $\|a\mathbf{X}\| = |a|\|\mathbf{X}\|$

(ii) $\|\mathbf{X}\| \ge 0$, with equality if and only if $\mathbf{X} = \mathbf{0}$

(iii) $\|\mathbf{X} - \mathbf{Y}\| \ge 0$, with equality if and only if $\mathbf{X} = \mathbf{Y}$

(iv) $\mathbf{X} \cdot \mathbf{Y} = \mathbf{Y} \cdot \mathbf{X}$

(v) $\mathbf{X} \cdot (\mathbf{Y} + \mathbf{Z}) = \mathbf{X} \cdot \mathbf{Y} + \mathbf{X} \cdot \mathbf{Z}$

(vi) $(c\mathbf{X}) \cdot \mathbf{Y} = \mathbf{X} \cdot (c\mathbf{Y}) = c(\mathbf{X} \cdot \mathbf{Y})$

**Definition 18.5.** Non-Archimedean metric space $(X, d)$ is a set $X$ together with a $\mathbb{R}_c^\#$-valued function $d: X \times X \to \mathbb{R}_c^\#$ (called a metric or non-Archimedean distance function) which assigns a hyperreal number $d(x, y)$ to every pair $x, y$ belongs $X$ satisfying the properties:

1. $d(x, y) \ge 0$ and $d(x, y) = 0$ iff $x = y$,

2. $d(x, y) = d(y, x)$,

3. $d(x, y) + d(y, z) \ge d(x, z)$.

**Remark 18.1.** Note that external linear space $\mathbb{R}_c^{\#n}$ endroved with distance function $d(\mathbf{X}, \mathbf{Y}) = \|\mathbf{X} - \mathbf{Y}\|$ satisfying the properties 1-3 mentioned above in Definition 14.5.

## 18.5. Line Segments in $\mathbb{R}_c^{\#n}, n \in \mathbb{N}^\#$.

**Definition 18.6.** Suppose that $X_0, U \in \mathbb{R}_c^{\#n}$ and $U \ne \mathbf{0}$. Then the line through $X_0$ in the direction of $U$ is the set of all points in $\mathbb{R}_c^{\#n}$ of the form

$$ \mathbf{X}(X_0, U) = X_0 + tU, -\infty^\# < t < \infty^\#. \qquad (18.43) $$

A set of points of the form

$$ \mathbf{X} = X_0 + t\mathbf{U}, t_1 \le t \le t_2 \qquad (18.44) $$

is called a line segment. In particular, the line segment from $X_0$ to $X_1$ is the set of
hyperrationals $q_i \in \mathbb{Q}^\#; i \in \mathbb{N}^\#.$

(ii) The empty set is #-closed.

(iii) Any finite set is #-closed.

(iv) Any hyperfinite set is #-closed.

(v) The closed interval $[a, b]$, where $a, b \in \mathbb{R}_c^\#$, is #-closed subset of $\mathbb{R}_c^\#$ with its canonical metric.

(vi) Let $\Delta$ be a set $\Delta = \{\varepsilon \parallel \varepsilon| \approx 0\}$. A set $\Delta$ is #-closed subset of $\mathbb{R}_c^\#$, since in $\Delta$ every hyperreal number $\delta \in \Delta$ is a #-limit point of the hyper infinite sequence $\{q_i\}_{i \in \mathbb{N}^\#}$ of hyperrationals $q_i \in \Delta \cap \mathbb{Q}^\#; i \in \mathbb{N}^\#.$

**Definition 18.10.** An #-neighbourhood of a point $p$ in a metric space $(X, d)$ is the set $N_\varepsilon(p) = \{x \in X | d(x, p) < \varepsilon, \varepsilon \approx 0\}$

**Definition 18.11.** A subset $A$ of a metric space $(X, d)$ is called #-open in $X$ if every point of $A$ has an #-neighbourhood which lies completely in $A$.

**Example 18.2.** (i) Any open interval $(a, b)$ is an #-open set in $\mathbb{R}_c^\#$ with its canonical metric $d(x, y) = |x - y|$.

(ii) A set $\Delta = \{\varepsilon \parallel \varepsilon| \approx 0\}$ is #-open subset of $\mathbb{R}_c^\#$, since every point of $\Delta$ obviously has a #-neighbourhood which lies completely in $\Delta$.

**Remark 18.2.** Note that a set $\Delta = \{\varepsilon \parallel \varepsilon| \approx 0\}$ are #-open and #-closed simultaneously.

**Definition 18.12.** A subset $A$ of a non-Archimedean metric space $X$ is *admissible* if $A$ is exactly #-closed or exactly #-open but not #-open and #-closed simultaneously.

**Theorem 18.16.** (i) The union (of an arbitrary number) of #-open admissible sets is #-open. (ii) The intersection of finitely or hyper finitely many #-open admissible sets is #-open.

**Proof.** (i) Let $x \in \bigcup A_i = A$. Then $x \in A_i$ for some $i$. Since this is #-open, $x$ has an #-neighbourhood lying completely inside $A_i$ and this is also inside $A$.

(ii) It is enough to show this for just two #-open sets $A$ and $B$. So suppose $x \in A \cap B$. Then $x \in A$ and so has an #-neighbourhood $N_{\varepsilon_1}(p), \varepsilon_1 \approx 0$ lying in $A$. Similarly $x$ has an #-neighbourhood $N_{\varepsilon_2}(p), \varepsilon_2 \approx 0$ lying in $B$. So if $\varepsilon = \min\{\varepsilon_1, \varepsilon_2\}$ the #-neighbourhood $N_\varepsilon(p)$ lies in both $A$ and $B$ and hence in $A \cap B$. By hyper infinite induction statement (ii) holds in general.

**Theorem 18.17.** Any admissible subset $A$ of a metric space $X$ is #-closed if and only if its complement $X\setminus A$ is admissible and is #-open subset of a metric space $X$.

**Proof.** 1. Suppose $A$ is admissible and $A$ is #-closed. We need to show that $X\setminus A$ is #-open.

So suppose that $x$ belongs $X\setminus A$. Then some #-neighbourhood of $x$ does not meet $A$ (otherwise $x$ would be a #-limit point of $A$ and hence in $A$). Thus this #-neighbourhood of $x$ lies completely in $X\setminus A$ which is what we needed to prove.

2. Conversely, suppose that $X\setminus A$ is #-open. We need to show that $A$ contains all its #-limit points. So suppose $x$ is a #-limit point of $A$ and that $x \notin A$. Then $x \in X\setminus A$ and hence

has an #-neighbourhood subset $X\setminus A$. But this is an #-neighbourhood that does not meet $A$

and we have a contradiction.

**Definition 18.13.** If $S$ is a nonempty subset of $\mathbb{R}_c^\#$, then

$$d(S) = \sup\{\|\mathbf{X} - \mathbf{Y}\|\mid \mathbf{X}, \mathbf{Y} \in S\} \qquad (18.48)$$
points of the form

$$ \mathbf{X} = \mathbf{X}_0 + t(\mathbf{X}_1 - \mathbf{X}_0) = t\mathbf{X}_1 + (1-t)\mathbf{X}_0, 0 \le t \le 1. \quad (18.45) $$

**Definition 18.7.** A hyper infinite sequence of points $\mathbf{X}_n, n \in \mathbb{N}^\# \text{ in } \mathbb{R}_c^{\#n}$ #-converges to the #-limit $\bar{\mathbf{X}}$ if

$$ \#\lim_{n \to \infty} \#\# \|\mathbf{X}_n - \bar{\mathbf{X}}\| = 0. \quad (18.46) $$

In this case we write $\#\lim_{n \to \infty} \#\# \mathbf{X}_n = \bar{\mathbf{X}}$.

**Theorem 18.14.** Let $\bar{\mathbf{X}} = \{x_i\}_{i=1}^{i=n}$ and $\mathbf{X}_m = \{x_{i_m}\}_{i=1}^{i=n}, m \ge 1$. Then $\#\lim_{m \to \infty} \#\# \mathbf{X}_m = \bar{\mathbf{X}}$ if and only if $\#\lim_{m \to \infty} \#\# x_{i_m} = \bar{x}_i, 1 \le i \le n$; that is a hyper infinite sequence $\{\mathbf{X}_m\}$ of points in $\mathbb{R}_c^{\#n}$ #-converges to a #-limit $\bar{\mathbf{X}}$ if and only if the hyper infinite sequences of components of $\{\mathbf{X}_m\}$ #-converge to the respective components of $\bar{\mathbf{X}}$.

**Theorem 18.15.** (Cauchy's #-Convergence Criterion) A hyper infinite sequence $\{\mathbf{X}_m\}$ in $\mathbb{R}_c^{\#n}$ #-converges if and only if for each $\varepsilon > 0, \varepsilon \approx 0$, there is an hyperinteger $N \in \mathbb{N}^\# \mathbb{N}$ such that

$$ \|\mathbf{X}_n - \mathbf{X}_m\| < \varepsilon \quad (18.47) $$

if $n, m \ge N$.

**Definition 18.8.** If $A$ is a subset of a metric space $\mathbb{R}_c^{\#n}$ then $x$ is a #-limit point of $A$ if it is the #-limit of an eventually non-constant hyper infinite sequence $\{a_i\}_{i \in \mathbb{N}^\#}$ of points of $A$.

**Definition 18.9.** A subset $A$ is said to be a #-closed subset of $\mathbb{R}_c^{\#n}$ if it contains all its #-limit points.

**Example 18.1.** (i) $\mathbb{R}_c^\#$ with the canonical metric $d(x, y) = |x - y|$, since in $\mathbb{R}_c^\#$ every hyperreal number is a #-limit point of the hyper infinite sequence $\{q_i\}_{i \in \mathbb{N}^\#}$ of hyperrationals $q_i \in \mathbb{Q}^\#$, $i \in \mathbb{N}^\#$.

(ii) The empty set is #-closed.

(iii) Any finite set is #-closed.

(iv) Any hyperfinite set is #-closed.

(v) The closed interval $[a, b]$, where $a, b \in \mathbb{R}_c^\#$, is #-closed subset of $\mathbb{R}_c^\#$ with its canonical metric.

(vi) Let $\Delta$ be a set $\Delta = \{\varepsilon \mid \varepsilon \approx 0\}$. A set $\Delta$ is #-closed subset of $\mathbb{R}_c^\#$, since in $\Delta$ every hyperreal number $\delta \in \Delta$ is a #-limit point of the hyper infinite sequence $\{q_i\}_{i \in \mathbb{N}^\#}$ of hyperrationals $q_i \in \Delta \cap \mathbb{Q}^\#$, $i \in \mathbb{N}^\#$.

**Definition 18.10.** An #-neighbourhood of a point $p$ in a metric space $(X, d)$ is the set $N_\varepsilon(p) = \{x \in X | d(x, p) < \varepsilon, \varepsilon \approx 0\}$

**Definition 18.11.** A subset $A$ of a metric space $(X, d)$ is called #-open in $X$ if every point of $A$ has an #-neighbourhood which lies completely in $A$.

**Example 18.2.** (i) Any open interval $(a, b)$ is an #-open set in $\mathbb{R}_c^\#$ with its canonical metric $d(x, y) = |x - y|$.

(ii) A set $\Delta = \{\varepsilon \mid \varepsilon \approx 0\}$ is #-open subset of $\mathbb{R}_c^\#$, since every point of $\Delta$ obviously has a #-neighbourhood which lies completely in $\Delta$.

**Remark 18.2.** Note that a set $\Delta = \{\varepsilon \mid \varepsilon \approx 0\}$ are #-open and #-closed simultaneously.

**Definition 18.12.** A subset $A$ of a non-Archimedean metric space $X$ is *admissible* if $A$ is exactly #-closed or exactly #-open but not #-open and #-closed simultaneously.

**Theorem 18.16.** (i) The union (of an arbitrary number) of #-open admissible sets is #-open. (ii) The intersection of finitely or hyper finitely many #-open admissible sets
is the diameter of S. If $d(S) < \infty^{\#}$, S is bounded or hyperbounded. If $d(S) = \infty^{\#}$, S is hyperunbounded.

**Theorem 18.18.** (Principle of Nested Sets) If $S_1, S_2, \dots$, are #-closed nonempty subsets of $\mathbb{R}_c^{\#n}$ such that

$$ \forall r (r \in \mathbb{N}^{\#}) [S_{r+1} \subset S_r] \qquad (18.49) $$

and

$$ \#\lim_{r \to \infty} \# d(S_r) = 0, \qquad (18.50) $$

then the intersection

$$ \Lambda = \bigcap_{r=1}^{\infty} S_r \qquad (18.51) $$

contains exactly one point:

**Proof.** Let $\{X_r\}$ be a hyper infinite sequence such that $X_r \in S_r, r \ge 1$. Because of (18.49), $X_r \in S_k$ if $r \ge k$, so $\|X_r - X_s\| < d(S_k)$ if $r, s \ge k$.

From (18.50) and Theorem 14.15., $X_r$ #-converges to a #-limit $\bar{X}$. Since $\bar{X}$ is a #-limit point of every $S_k$ and every $S_k$ is #-closed, $\bar{X}$ is in every $S_k$. Therefore, $\bar{X} \in \Lambda$, so $\Lambda \ne \emptyset$.

Moreover, $\bar{X}$ is the only point in $\Lambda$, since if $Y \in \Lambda$, then $\|\bar{X} - Y\| < d(S_k), k \ge 1$, and (18.50) implies that $Y = \bar{X}$.

**Definition 18.14.** If S is a nonempty admissible subset of $\mathbb{R}_c^{\#n}$ we say that S is a #-compact set in $\mathbb{R}_c^{\#n}$ if is a #-closed and bounded or hyperbounded set.

**Definition 18.15.** Collection H of admissible #-open sets is an #-open covering of a set S if $S \subset \cup \{H | H \in \mathbf{H}\}$.

**Theorem 18.19.** (Heine-Borel Theorem) If H is an #-open covering of a #-compact subset S, then S can be covered by hyper finitely many sets from H.

**Proof.** The proof is by contradiction. We first consider the case where $n = 2$. Suppose that there is a #-open covering H for S from which it is impossible to select a hyperfinite

subcovering. Since S is bounded or hyperbounded, S is contained in a #-closed square $T = \{\{x, y\} | a_1 \le x \le a_1 + L, a_2 \le ya_2 + L\}$ with sides of length L (Pic. 14.5.1).

<figure><img src="image_2.png" /><figcaption>Pic.18.5.1.</figcaption></figure>

Bisecting the sides of T as shown by the dashed lines in Figure 14.5.1 leads to four #-closed squares, $T^{(1)}, T^{(2)}, T^{(3)},$ and $T^{(4)}$, with sides of length $L/2$. Let $S^{(i)} = S \cap T, 1 \le i \le 4$. Each $S^{(i)}$, being the intersection of admissible #-closed sets, is
is #-open.

**Proof.** (i) Let $x \in \bigcup A_i = A$. Then $x \in A_i$ for some $i$. Since this is #-open, $x$ has an #-neighbourhood lying completely inside $A_i$ and this is also inside $A$.

(ii) It is enough to show this for just two #-open sets $A$ and $B$. So suppose $x \in A \cap B$. Then $x \in A$ and so has an #-neighbourhood $N_{\varepsilon_1}(p), \varepsilon_1 \approx 0$ lying in $A$. Similarly $x$ has an #-neighbourhood $N_{\varepsilon_2}(p), \varepsilon_2 \approx 0$ lying in $B$. So if $\varepsilon = \min\{\varepsilon_1, \varepsilon_2\}$ the #-neighbourhood $N_{\varepsilon}(p)$ lies in both $A$ and $B$ and hence in $A \cap B$. By hyper infinite induction statement (ii) holds in general.

**Theorem 18.17.** Any admissible subset $A$ of a metric space $X$ is #-closed if and only if its complement $X\setminus A$ is admissible and is #-open subset of a metric space $X$.

**Proof.** 1. Suppose $A$ is admissible and $A$ is #-closed. We need to show that $X\setminus A$ is #-open.

So suppose that $x$ belongs $X\setminus A$. Then some #-neighbourhood of $x$ does not meet $A$ (otherwise $x$ would be a #-limit point of $A$ and hence in $A$). Thus this #-neighbourhood of $x$ lies completely in $X\setminus A$ which is what we needed to prove.

2. Conversely, suppose that $X\setminus A$ is #-open. We need to show that $A$ contains all its #-limit points. So suppose $x$ is a #-limit point of $A$ and that $x \notin A$. Then $x \in X\setminus A$ and hence

has an #-neighbourhood subset $X\setminus A$. But this is an #-neighbourhood that does not meet $A$

and we have a contradiction.

**Definition 18.13.** If $S$ is a nonempty subset of $\mathbb{R}_c^{\#n}$, then

$$d(S) = \sup\{\|\mathbf{X} - \mathbf{Y}\|\mid\mathbf{X}, \mathbf{Y} \in S\} \qquad (18.48)$$

is the diameter of $S$. If $d(S) < \infty^{\#}$, $S$ is bounded or hyperbounded. If $d(S) = \infty^{\#}$, $S$ is hyperunbounded.

**Theorem 18.18.** (Principle of Nested Sets) If $S_1, S_2, \dots$, are #-closed nonempty subsets of $\mathbb{R}_c^{\#n}$ such that

$$\forall r(r \in \mathbb{N}^{\#})[S_{r+1} \subset S_r] \qquad (18.49)$$

and

$$\#\lim_{r \to \infty} d(S_r) = 0, \qquad (18.50)$$

then the intersection

$$\Lambda = \bigcap_{r=1}^{\infty} S_r \qquad (18.51)$$

contains exactly one point:

**Proof.** Let $\{X_r\}$ be a hyper infinite sequence such that $X_r \in S_r, r \ge 1$. Because of (18.49), $X_r \in S_k$ if $r \ge k$, so $\|X_r - X_s\| < d(S_k)$ if $r, s \ge k$.

From (18.50) and Theorem 14.15., $X_r$ #-converges to a #-limit $\bar{X}$. Since $\bar{X}$ is a #-limit point of every $S_k$ and every $S_k$ is #-closed, $\bar{X}$ is in every $S_k$. Therefore, $\bar{X} \in \Lambda$, so $\Lambda \ne \emptyset$.

Moreover, $\bar{X}$ is the only point in $\Lambda$, since if $Y \in \Lambda$, then $\|\bar{X} - Y\| < d(S_k), k \ge 1$, and (18.50) implies that $Y = \bar{X}$.

**Definition 18.14.** If $S$ is a nonempty admissible subset of $\mathbb{R}_c^{\#n}$ we say that $S$ is a #-compact set in $\mathbb{R}_c^{\#n}$ if is a #-closed and bounded or hyperbounded set.
#-closed, and $S = \bigcup_{i=1}^{4} S^{(i)}$. Moreover, $\mathbf{H}$ covers each $S^{(i)}$, but at least one $S^{(i)}$ cannot be covered by any finite or hyperfinite subcollection of $\mathbf{H}$, since if all the $S^{(i)}$ could be, then so could $S$. Let $S_1$ be a set with this property, chosen from $S^{(1)}, S^{(2)}, S^{(3)}$, and $S^{(4)}$. We are now back to the situation we started from: a #-compact set $S_1$ covered by $\mathbf{H}$, but not by any hyperfinite subcollection of $\mathbf{H}$. However, $S_1$ is contained in a square $T_1$ with sides of length $L/2$ instead of $L$. Bisecting the sides of $T_1$ and repeating the argument, we obtain a subset $S_2$ of $S_1$ that has the same properties as $S$, except that it is contained in a square with sides of length $L/4$. Continuing in this way produces a hyper infinite sequence of nonempty #-closed sets $S_0 = S, S_1, S_2, \dots$, such that $S_k \supset S_{k+1}$ and $d(S_k) \le L/2^{k-1/2}, k \ge 0$. From Theorem 14.18, there is a point $\bar{X}$ in $\cap_{k=1}^{\infty} S_k$. Since $\bar{X} \in S$, there is an open set $H$ in $\mathbf{H}$ that contains $\bar{X}$, and this $H$ must also contain some #-neighborhood of $\bar{X}$. Since every $\mathbf{X}$ in $S_k$ satisfies the inequality $\|\mathbf{X} - \bar{X}\| < 2^{-k+1/2}L$, it follows that $S_k \subset H$ for $k \in \mathbb{N}^\# / \mathbb{N}$ sufficiently large. This contradicts our assumption on $\mathbf{H}$, which led us to believe that no $S_k$ could be covered by a hyperfinite

number of sets from $\mathbf{H}$. Consequently, this assumption must be false: $\mathbf{H}$ must have a finite or hyperfinite subcollection that covers $S$. This completes the proof for $n = 2$. The idea of the proof is the same for $n > 2$. The counterpart of the square $T$ is the hypercube with sides of length $L$:

$$T = \{(x_1, x_2, \dots, x_n) | a_i \le x_i \le a_i + L, 1 \le i \le n\}.$$

Halving the intervals of variation of the $n$ coordinates $x_1, x_2, \dots, x_n$ divides $T$ into $2^n$ closed hypercubes with sides of length $L/2$:

$$T^{(i)} = \{(x_1, x_2, \dots, x_n) | b_i \le x_i \le b_i + L/2, 1 \le i \le n\},$$

where $b_i = a_i$ or $b_i = a_i + L/2$. If no hyperfinite subcollection of $\mathbf{H}$ covers $S$, then at least one of these smaller hypercubes must contain a subset of $S$ that is not covered by any hyperfinite subcollection of $S$. Now the proof proceeds as for $n = 2$.

## 18.6. #-Neighborhoods and #-open sets in $\mathbb{R}_c^{\#n}$, $n \in \mathbb{N}^\#$. Connected Sets and Regions in $\mathbb{R}_c^{\#n}$.

**Definition 18.16.** Assum that $A$ is admissible subset of $\mathbb{R}_c^{\#n}$.

(i) The #-interior $\mathrm{-int}(A)$ of a set $A$ is the largest open subset $A$,

(ii) The #-closure $\mathrm{-cl}(A)$ of a set $A$ is the smallest #-closed set containing $A$.

**Theorem 18.20.** 1. $\mathrm{-cl}(\emptyset) = \emptyset$

2. $A \subset \mathrm{-cl}(A)$ for any set $A$.

3. $\mathrm{-cl}(A \cup B) = \mathrm{-cl}(A) \cup \mathrm{-cl}(B)$ for any sets $A$ and $B$.

4. $\mathrm{cl}(\mathrm{-cl}(A)) = \mathrm{-cl}(A)$ for any set $A$.

**Proof.** 1. and 2. follow from the definition.

To prove 3 note that $\mathrm{-cl}(A) \cup \mathrm{-cl}(B)$ is a #-closed set which contains $A \cup B$ and so $\mathrm{-cl}(A) \subset \mathrm{-cl}(A \cup B)$. Similarly, $\mathrm{-cl}(B) \subset \mathrm{-cl}(A \cup B)$ and so $\mathrm{-cl}(A) \cup \mathrm{-cl}(B) \subset \mathrm{-cl}(A \cup B)$ and the result follows.

To prove 4 we have $\mathrm{-cl}(A) \subset \mathrm{-cl}(\mathrm{-cl}(A))$ from 2. Also $\mathrm{-cl}(A)$ is a #-closed set which contains $\mathrm{-cl}(A)$ and hence it contains $\mathrm{-cl}(\mathrm{-cl}(A))$.

**Example 18.3.** For $\mathbb{R}_c^\#$ with its usual topology induced by its canonical metric $d(x, y) = |x - y|$, $\mathrm{-cl}((a, b)) = [a, b]$ and $\mathrm{-int}([a, b]) = (a, b)$.

**Definition 18.17.** If $\varepsilon > 0, \varepsilon \approx 0$, the $\varepsilon$-neighborhood of a point $X_0$ in $\mathbb{R}_c^{\#n}$ is the set
**Definition 18.15.** Collection $\mathbf{H}$ of admissible #-open sets is an #-open covering of a set $S$ if $S \subset \bigcup \{\mathbf{H} | \mathbf{H} \in \mathbf{H}\}$.

**Theorem 18.19.** (Heine-Borel Theorem) If $\mathbf{H}$ is an #-open covering of a #-compact subset $S$, then $S$ can be covered by hyper finitely many sets from $\mathbf{H}$.

**Proof.** The proof is by contradiction. We first consider the case where $n = 2$. Suppose that there is a #-open covering $\mathbf{H}$ for $S$ from which it is impossible to select a hyperfinite

subcovering. Since $S$ is bounded or hyperbounded, $S$ is contained in a #-closed square $T = \{(x,y) | a_1 \le x \le a_1 + L, a_2 \le xa_2 + L\}$ with sides of length $L$ (Pic. 14.5.1).

<figure><img src="image_2.png" /><figcaption>Pic.18.5.1.</figcaption></figure>

Bisecting the sides of $T$ as shown by the dashed lines in Figure 14.5.1 leads to four #-closed squares, $T^{(1)}, T^{(2)}, T^{(3)},$ and $T^{(4)}$, with sides of length $L/2$. Let $S^{(i)} = S \cap T, 1 \le i \le 4$. Each $S^{(i)}$, being the intersection of admissible #-closed sets, is #-closed, and $S = \cup_{i=1}^4 S^{(i)}$. Moreover, $\mathbf{H}$ covers each $S^{(i)}$, but at least one $S^{(i)}$ cannot be covered by any finite or hyperfinite subcollection of $\mathbf{H}$, since if all the $S^{(i)}$ could be, then so could $S$. Let $S_1$ be a set with this property, chosen from $S^{(1)}, S^{(2)}, S^{(3)},$ and $S^{(4)}$.

We are now back to the situation we started from: a #-compact set $S_1$ covered by $\mathbf{H}$, but not by any hyperfinite subcollection of $\mathbf{H}$. However, $S_1$ is contained in a square $T_1$ with sides of length $L/2$ instead of $L$. Bisecting the sides of $T_1$ and repeating the argument, we obtain a subset $S_2$ of $S_1$ that has the same properties as $S$, except that it is contained in a square with sides of length $L/4$. Continuing in this way produces a hyper infinite sequence of nonempty #-closed sets $S_0 = S, S_1, S_2, \dots$, such that $S_k \supset S_{k+1}$ and $d(S_k) \le L/2^{k-1/2}, k \ge 0$. From Theorem 14.18, there is a point $\bar{X}$ in $\cap_{k=1}^\infty S_k$. Since $\bar{X} \in S$, there is an open set $H$ in $\mathbf{H}$ that contains $\bar{X}$, and this $H$ must also contain some #-neighborhood of $\bar{X}$. Since every $X$ in $S_k$ satisfies the inequality $||X - \bar{X}|| < 2^{-k+1/2}L$, it follows that $S_k \subset H$ for $k \in \mathbb{N}^\# \cap \mathbb{N}$ sufficiently large. This contradicts our assumption on $\mathbf{H}$, which led us to believe that no $S_k$ could be covered by a hyperfinite

number of sets from $\mathbf{H}$. Consequently, this assumption must be false: $\mathbf{H}$ must have a finite or hyperfinite subcollection that covers $S$. This completes the proof for $n = 2$. The idea of the proof is the same for $n > 2$. The counterpart of the square $T$ is the hypercube with sides of length $L$:

$$T = \{(x_1, x_2, \dots, x_n) | a_i \le x_i \le a_i + L, 1 \le i \le n\}.$$

Halving the intervals of variation of the $n$ coordinates $x_1, x_2, \dots, x_n$ divides $T$ into $2^n$ closed hypercubes with sides of length $L/2$:
$$N_{\varepsilon}(\mathbf{X}_0) = \{\mathbf{X} \mid \|\mathbf{X} - \mathbf{X}_0\| < \varepsilon\}. \qquad (18.52)$$

**Definition 18.18.** If $\mathbf{X}_0$ is a point in $\mathbb{R}_c^{\#n}$ and $r > 0$, the sphere of radius $r$ about $\mathbf{X}_0$ is the set $S_r(\mathbf{X}_0) = \{\mathbf{X} \mid \|\mathbf{X} - \mathbf{X}_0\| = r\}$

**Definition 18.19.** If $\mathbf{X}_0$ is a point in $\mathbb{R}_c^{\#n}$ and $r > 0$, the \#-open $n$-ball of radius $r$ about $\mathbf{X}_0$ is the set $B_r(\mathbf{X}_0) = \{\mathbf{X} \mid \|\mathbf{X} - \mathbf{X}_0\| < r\}$. Thus, $\varepsilon$-neighborhoods are \#-open $n$-balls. If $\mathbf{X}_1$ is in $B_r(\mathbf{X}_0)$ and $\|\mathbf{X} - \mathbf{X}_1\| < \varepsilon = r - \|\mathbf{X} - \mathbf{X}_0\|$, then $\mathbf{X}$ is in $B_r(\mathbf{X}_0)$.

Thus, $B_r(\mathbf{X}_0)$ contains an $\varepsilon$-neighborhood of each of its points, and is therefore \#-open.

The \#-closure of $B_r(\mathbf{X}_0)$ is the \#-closed $n$-ball of radius $r$ about $\mathbf{X}_0$, defined by $\#-cl(B_r(\mathbf{X}_0)) = \{\mathbf{X} \mid \|\mathbf{X} - \mathbf{X}_0\| < r\}, r = \|\mathbf{X}_1 - \mathbf{X}_0\|$.

**Proposition 18.1.** If $\mathbf{X}_1$ and $\mathbf{X}_2$ are in $S_r(\mathbf{X}_0)$ for some $r > 0$, then so is every point on the line segment from $\mathbf{X}_1$ to $\mathbf{X}_2$.

**Definition 18.20.** A subset $S \subset \mathbb{R}_c^{\#n}$ is \#-connected if it is impossible to represent $S$ as the union of two disjoint nonempty sets such that neither contains a \#-limit point of the other; that is, if $S$ cannot be expressed as $S = A \cup B$, where

$$A \neq \emptyset, B \neq \emptyset, \#-cl(A) \cap B = \emptyset, \#-cl(B) \cap A = \emptyset. \qquad (18.53)$$

If $S$ can be expressed in this way, then $S$ is \#-disconnected.

**Definition 18.21.** A region $S$ in $\mathbb{R}_c^{\#n}$ is the union of an \#-open \#-connected set with some, all, or none of its \#-boundary; thus, $\#-int(S)$ is \#-connected, and every point of $S$ is a \#-limit point of $\#-int(S)$.

## 18.7. The \#-limits and \#-continuity $\mathbb{R}_c^\#$-valued functions of $n \in \mathbb{N}^\#$ variables.

We denote the domain of a function $f$ by $D_f$ and the value of $f$ at a point $\mathbf{X} = \{x_i\}_{i=1}^n$ by $f(\mathbf{X})$ or $f(\{x_i\}_{i=1}^n)$.

**Definition 18.22.** We say that $f(\mathbf{X})$ \#-approaches the \#-limit $L$ as $\mathbf{X}$ \#-approaches $\mathbf{X}_0$ and write

$$\#-\lim_{\mathbf{X} \to \#} x_0 f(\mathbf{X}) = L \qquad (18.54)$$

if $\mathbf{X}_0$ is a \#-limit point of $D_f$ and, for every $\varepsilon > 0$, $\varepsilon \approx 0$, there is a $\delta > 0$, $\delta \approx 0$, such that $|f(\mathbf{X}) - L| < \varepsilon$ for all $\mathbf{X} \in D_f$ such that $0 < \|\mathbf{X} - \mathbf{X}_0\| < \delta$.

**Theorem 18.21.** If $\#-\lim_{\mathbf{X} \to \#} x_0 f(\mathbf{X})$ exists, then it is unique.

**Theorem 18.22.** Suppose that $f$ and $g$ are defined on a set $D \subset \mathbb{R}_c^{\#n}$, $\mathbf{X}_0$ is a \#-limit point of $D$, and $\#-\lim_{\mathbf{X} \to \#} x_0 f(\mathbf{X}) = L_1$, $\#-\lim_{\mathbf{X} \to \#} x_0 g(\mathbf{X}) = L_2$. Then

$$\begin{align}
\#-\lim_{\mathbf{X} \to \#} x_0 (f \pm g)(\mathbf{X}) &= L_1 \pm L_2, \nonumber \\
\#-\lim_{\mathbf{X} \to \#} x_0 (f \times g)(\mathbf{X}) &= L_1 L_2, \tag{18.55} \\
\text{and} \nonumber
\end{align}$$

$$\#-\lim_{\mathbf{X} \to \#} x_0 (f/g)(\mathbf{X}) = L_1 L_2,$$

if $L_2 \neq 0$.

**Definition 18.23.** We say that $f(\mathbf{X})$ \#-approaches $\infty^\#$ as $\mathbf{X}$ \#-approaches $\mathbf{X}_0$ and write

$$\#-\lim_{\mathbf{X} \to \#} x_0 f(\mathbf{X}) = \infty^\# \qquad (14.56)$$

if $\mathbf{X}_0$ is a \#-limit point of $D_f$ and, for every hyperreal number $M$, there is a $\delta > 0$,
$$T^{(i)} = \{(x_1, x_2, \dots, x_n) | b_i \le x_i \le b_i + L/2, 1 \le i \le n\},$$

where $b_i = a_i$ or $b_i = a_i + L/2$. If no hyperfinite subcollection of $\mathbf{H}$ covers $S$, then at least one of these smaller hypercubes must contain a subset of $S$ that is not covered by any hyperfinite subcollection of $S$. Now the proof proceeds as for $n = 2$.

## 18.6.\#-Neighborhoods and \#-open sets in $\mathbb{R}_c^{\#n}, n \in \mathbb{N}^\#$. Connected Sets and Regions in $\mathbb{R}_c^{\#n}$.

**Definition 18.16.** Assum that $A$ is admissible subset of $\mathbb{R}_c^{\#n}$.

(i) The \#-interior \#-int($A$) of a set $A$ is the largest open subset $A$,

(ii) The \#-closure \#-cl($A$) of a set $A$ is the smallest \#-closed set containing $A$.

**Theorem 18.20.** 1. \#-cl($\emptyset$) = $\emptyset$

2. $A \subset \#-cl(A)$ for any set $A$.

3. $\#-cl(A \cup B) = \#-cl(A) \cup \#-cl(B)$ for any sets $A$ and $B$.

4. $cl(\#-cl(A)) = \#-cl(A)$ for any set $A$.

**Proof.** 1. and 2. follow from the definition.

To prove 3 note that $\#-cl(A) \cup \#-cl(B)$ is a \#-closed set which contains $A \cup B$ and so $\#-cl(A) \subset \#-cl(A \cup B)$. Similarly, $\#-cl(B) \subset \#-cl(A \cup B)$ and so $\#-cl(A) \cup \#-cl(B) \subset \#-cl(A \cup B)$ and the result follows.

To prove 4 we have $\#-cl(A) \subset \#-cl(\#-cl(A))$ from 2. Also $\#-cl(A)$ is a \#-closed set which contains $\#-cl(A)$ and hence it contains $\#-cl(\#-cl(A))$.

**Example 18.3.** For $\mathbb{R}_c^\#$ with its usual topology induced by its canonical metric $d(x, y) = |x - y|$, $\#-cl((a, b)) = [a, b]$ and $\#-int([a, b]) = (a, b)$.

**Definition 18.17.** If $\varepsilon > 0, \varepsilon \approx 0$, the $\varepsilon$-neighborhood of a point $X_0$ in $\mathbb{R}_c^{\#n}$ is the set

$$N_{\varepsilon}(X_0) = \{\mathbf{X} \mid \|\mathbf{X} - X_0\| < \varepsilon\}. \qquad (18.52)$$

**Definition 18.18.** If $X_0$ is a point in $\mathbb{R}_c^{\#n}$ and $r > 0$, the sphere of radius $r$ about $X_0$ is the set $S_r(X_0) = \{\mathbf{X} \mid \|\mathbf{X} - X_0\| = r\}$

**Definition 18.19.** If $X_0$ is a point in $\mathbb{R}_c^{\#n}$ and $r > 0$, the \#-open $n$-ball of radius $r$ about $X_0$ is the set $B_r(X_0) = \{\mathbf{X} \mid \|\mathbf{X} - X_0\| < r\}$. Thus, $\varepsilon$-neighborhoods are \#-open $n$-balls. If $X_1$ is in $B_r(X_0)$ and $\|\mathbf{X} - X_1\| < \varepsilon = r - \|\mathbf{X} - X_0\|$, then $X$ is in $B_r(X_0)$.

Thus, $B_r(X_0)$ contains an $\varepsilon$-neighborhood of each of its points, and is therefore \#-open.

The \#-closure of $B_r(X_0)$ is the \#-closed $n$-ball of radius $r$ about $X_0$, defined by $\#-cl(B_r(X_0)) = \{\mathbf{X} \mid \|\mathbf{X} - X_0\| < r\}, r = \|X_1 - X_0\|$.

**Proposition 18.1.** If $X_1$ and $X_2$ are in $S_r(X_0)$ for some $r > 0$, then so is every point on the line segment from $X_1$ to $X_2$.

**Definition 18.20.** A subset $S \subset \mathbb{R}_c^{\#n}$ is \#-connected if it is impossible to represent $S$ as the union of two disjoint nonempty sets such that neither contains a \#-limit point of the other; that is, if $S$ cannot be expressed as $S = A \cup B$, where

$$A \neq \emptyset, B \neq \emptyset, \#-cl(A) \cap B = \emptyset, \#-cl(B) \cap A = \emptyset. \qquad (18.53)$$

If $S$ can be expressed in this way, then $S$ is \#-disconnected.

**Definition 18.21.** A region $S$ in $\mathbb{R}_c^{\#n}$ is the union of an \#-open \#-connected set with some, all, or none of its \#-boundary; thus, $\#-int(S)$ is \#-connected, and every point of $S$ is a \#-limit point of $\#-int(S)$.

## 18.7. The \#-limits and \#-continuity $\mathbb{R}_c^\#$-valued functions of
$\delta \approx 0$, such that $f(\mathbf{X}) > M$ whenever $0 < \|\mathbf{X} - \mathbf{X}_0\| < \delta$ and $\mathbf{X} \in \mathbf{D}_f$. We say that
$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} f(\mathbf{X}) = -\infty^{\#} \qquad (18.57) $$

if $\#\lim_{\mathbf{X} \to \mathbf{X}_0} f(\mathbf{X}) = \infty^{\#}$.

**Definition 18.24.** If $\mathbf{D}_f$ is hyperunbounded, we say that

$$ \#\lim_{\|\mathbf{X}\| \to \infty} f(\mathbf{X}) = L, \qquad (18.58) $$

where $L$ finite or hyperfinite if for every $\delta > 0, \delta \approx 0$, there is a number $R \in \mathbb{R}_c^\#$ such that $|f(\mathbf{X}) - L| < \varepsilon|$ whenever $\|\mathbf{X}\| > R$ and $\mathbf{X} \in \mathbf{D}_f$.

**Definition 18.25.** If $\mathbf{X}_0 \in \mathbf{D}_f$ and is a $\#$-limit point of $\mathbf{D}_f$, then we say that $f$ is $\#$-continuous at $\mathbf{X}_0$ if

$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} f(\mathbf{X}) = f(\mathbf{X}_0). \qquad (18.59) $$

**Theorem 18.23.** Suppose that $\mathbf{X}_0 \in \mathbf{D}_f$ and is a $\#$-limit point of $\mathbf{D}_f$. Then $f$ is $\#$-continuous at $\mathbf{X}_0$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(\mathbf{X}) - f(\mathbf{X}_0)| < \varepsilon$ whenever $\|\mathbf{X} - \mathbf{X}_0\| < \delta$ and $\mathbf{X} \in \mathbf{D}_f$.

**Definition 18.26.** We will say that $f$ is $\#$-continuous on $S$ if $f$ is $\#$-continuous at every point of $S$.

**Theorem 18.24.** If $f$ and $g$ are $\#$-continuous on a set $S \subset \mathbb{R}_c^{\#n}$, then so are $f \pm g$, and $fg$. Also, $f/g$ is $\#$-continuous at each $\mathbf{X}_0 \in S$ such that $g(\mathbf{X}_0) \neq 0$.

**Definition 18.27.** Suppose that $g_1, g_2, \dots, g_n, n \in \mathbb{N}^\#$ are $\mathbb{R}_c^\#$-valued functions defined on a subset $T \subset \mathbb{R}_c^{\#n}$, and define the vector-valued function $G$ on $T$ by

$$ G(\mathbf{U}) = (g_1(\mathbf{U}), g_2(\mathbf{U}), \dots, g_n(\mathbf{U})), \mathbf{U} \in T. \qquad (18.60) $$

Then $g_1, g_2, \dots, g_n$ are the component functions of $G = (g_1, g_2, \dots, g_n)$. We say that

$$ \#\lim_{\mathbf{U} \to \mathbf{U}_0} G(\mathbf{U}) = \mathbf{L} = (L_1, \dots, L_n) \qquad (18.61) $$

if $\#\lim_{\mathbf{U} \to \mathbf{U}_0} g_i(\mathbf{U}) = L_i, 1 \le i \le n$ and that $G$ is $\#$-continuous at $\mathbf{U}_0$ if $g_1, g_2, \dots, g_n$ are each $\#$-continuous at $\mathbf{U}_0$.

**Theorem 18.25.** For a vector-valued function $G, \#\lim_{\mathbf{U} \to \mathbf{U}_0} G(\mathbf{U}) = \mathbf{L}$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $\|G(\mathbf{U}) - \mathbf{L}\| < \varepsilon$ whenever $0 < \|\mathbf{U} - \mathbf{U}_0\| < \delta$ and $\mathbf{U} \in \mathbf{D}_G$. Similarly, $G$ is $\#$-continuous at $\mathbf{U}_0$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $\|G(\mathbf{U}) - G(\mathbf{U}_0)\| < \varepsilon$ whenever $\|\mathbf{U} - \mathbf{U}_0\| < \delta$ and $\mathbf{U} \in \mathbf{D}_G$.

**Theorem 18.26.** Let $f$ be a $\mathbb{R}_c^\#$-valued function defined on a subset of $\mathbb{R}_c^{\#n}$, and let the vector-valued function $G = (g_1, g_2, \dots, g_n)$ be defined on a domain $\mathbf{D}_G$ in $\mathbb{R}_c^{\#n}$. Let the set $T = \{\mathbf{U} | \mathbf{U} \in \mathbf{D}_G \text{ and } G(\mathbf{U}) \in \mathbf{D}_f\}$, be nonempty; and define the $\mathbb{R}_c^\#$-valued composite function $h = f \circ G$ on $T$ by $h(\mathbf{U}) = f(G(\mathbf{U}))$, $\mathbf{U} \in T$. Now suppose that $\mathbf{U}_0 \in T$ and is a $\#$-limit point of $T$, $G$ is $\#$-continuous at $\mathbf{U}_0$, and $f$ is $\#$-continuous at $\mathbf{X}_0 = G(\mathbf{U}_0)$. Then $h$ is $\#$-continuous at $\mathbf{U}_0$.

**Theorem 18.27.** If $f$ is $\#$-continuous on a $\#$-compact set $S \subset \mathbb{R}_c^{\#n}$, then $f$ is bounded or hyperbounded on $S$.

**Theorem 18.28.** Let $f$ be $\#$-continuous on a compact set $S \subset \mathbb{R}_c^{\#n}$ and $\alpha = \inf_{\mathbf{X} \in S} f(\mathbf{X})$, $\beta = \sup_{\mathbf{X} \in S} f(\mathbf{X})$. Then $f(\mathbf{X}_1) = \alpha$ and $f(\mathbf{X}_2) = \beta$ for some $\mathbf{X}_1$ and $\mathbf{X}_2$ in $S$.

**Theorem 18.29.** (Intermediate Value Theorem) Let $f$ be $\#$-continuous on a region $S \subset \mathbb{R}_c^{\#n}$. Suppose that $\mathbf{A}$ and $\mathbf{B}$ are in $S$ and $f(\mathbf{A}) < u < f(\mathbf{B})$. Then $f(\mathbf{C}) = u$
$n \in \mathbb{N}^\#$ variables.

We denote the domain of a function $f$ by $D_f$ and the value of $f$ at a point $\mathbf{X} = \{x_i\}_{i=1}^{i=n}$ by $f(\mathbf{X})$ or $f(\{x_i\}_{i=1}^{i=n})$.

**Definition 18.22.** We say that $f(\mathbf{X})$ #-approaches the #-limit $L$ as $\mathbf{X}$ #-approaches $\mathbf{X}_0$ and write

$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 f(\mathbf{X}) = L \qquad (18.54) $$

if $\mathbf{X}_0$ is a #-limit point of $D_f$ and, for every $\varepsilon > 0$, $\varepsilon \approx 0$, there is a $\delta > 0$, $\delta \approx 0$, such that $|f(\mathbf{X}) - L| < \varepsilon$ for all $\mathbf{X} \in D_f$ such that $0 < \|\mathbf{X} - \mathbf{X}_0\| < \delta$.

**Theorem 18.21.** If $\#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 f(\mathbf{X})$ exists, then it is unique.

**Theorem 18.22.** Suppose that $f$ and $g$ are defined on a set $D \subset \mathbb{R}_c^{\#n}$, $\mathbf{X}_0$ is a #-limit point of $D$, and $\#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 f(\mathbf{X}) = L_1$, $\#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 g(\mathbf{X}) = L_2$. Then

$$ \begin{aligned} \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 (f \pm g)(\mathbf{X}) &= L_1 \pm L_2, \\ \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 (f \times g)(\mathbf{X}) &= L_1 L_2, \end{aligned} \qquad (18.55) $$

$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 (f/g)(\mathbf{X}) = L_1 L_2, $$

if $L_2 \neq 0$.

**Definition 18.23.** We say that $f(\mathbf{X})$ #-approaches $\infty^\#$ as $\mathbf{X}$ #-approaches $\mathbf{X}_0$ and write

$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 f(\mathbf{X}) = \infty^\# \qquad (18.56) $$

if $\mathbf{X}_0$ is a #-limit point of $D_f$ and, for every hyperreal number $M$, there is a $\delta > 0$, $\delta \approx 0$, such that $f(\mathbf{X}) > M$ whenever $0 < \|\mathbf{X} - \mathbf{X}_0\| < \delta$ and $\mathbf{X} \in D_f$. We say that

$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 f(\mathbf{X}) = -\infty^\# \qquad (18.57) $$

if $\#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 [-f(\mathbf{X})] = \infty^\#$.

**Definition 18.24.** If $D_f$ is hyperunbounded, we say that

$$ \#\lim_{\|\mathbf{X} \to \mathbf{X}_0\| \to \infty^\#} f(\mathbf{X}) = L, \qquad (18.58) $$

where $L$ finite or hyperfinite if for every $\delta > 0$, $\delta \approx 0$, there is a number $R \in \mathbb{R}_c^\#$ such that $|f(\mathbf{X}) - L| < \varepsilon$ whenever $\|\mathbf{X}\| > R$ and $\mathbf{X} \in D_f$.

**Definition 18.25.** If $\mathbf{X}_0 \in D_f$ and is a #-limit point of $D_f$, then we say that $f$ is #-continuous at $\mathbf{X}_0$ if

$$ \#\lim_{\mathbf{X} \to \mathbf{X}_0} x_0 f(\mathbf{X}) = f(\mathbf{X}_0). \qquad (18.59) $$

**Theorem 18.23.** Suppose that $\mathbf{X}_0 \in D_f$ and is a #-limit point of $D_f$. Then $f$ is #-continuous at $\mathbf{X}_0$ if and only if for each $\varepsilon > 0$, $\varepsilon \approx 0$ there is a $\delta > 0$, $\delta \approx 0$ such that $|f(\mathbf{X}) - f(\mathbf{X}_0)| < \varepsilon$ whenever $\|\mathbf{X} - \mathbf{X}_0\| < \delta$ and $\mathbf{X} \in D_f$.

**Definition 18.26.** We will say that $f$ is #-continuous on $S$ if $f$ is #-continuous at every point of $S$.

**Theorem 18.24.** If $f$ and $g$ are #-continuous on a set $S \subset \mathbb{R}_c^{\#n}$, then so are $f \pm g$, and $fg$. Also, $f/g$ is #-continuous at each $\mathbf{X}_0 \in S$ such that $g(\mathbf{X}_0) \neq 0$.

**Definition 18.27.** Suppose that $g_1, g_2, \dots, g_n, n \in \mathbb{N}^\#$ are $\mathbb{R}_c^\#$-valued functions defined on a subset $T \subset \mathbb{R}_c^{\#n}$, and define the vector-valued function $G$ on $T$ by

$$ G(\mathbf{U}) = (g_1(\mathbf{U}), g_2(\mathbf{U}), \dots, g_n(\mathbf{U})), \mathbf{U} \in T. \qquad (18.60) $$
for some $C \in S$.

**Definition 18.28.** $f$ is uniformly #-continuous on a subset $S$ of its domain in $\mathbb{R}_c^{\#n}$ if for every $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(X) - f(X_0)| < \varepsilon$ whenever $\|X - X_0\| < \delta$ and $X, X_0 \in S$.

**Theorem 18.30.** If $f$ is #-continuous on a #-compact set $S \subset \mathbb{R}_c^{\#n}$ then $f$ is uniformly #-continuous on $S$.

## 18.8. Partial #-Derivatives and the #-Differential

**Definition 18.29.** Let $\Phi$ be a unit vector and $X$ a point in $\mathbb{R}_c^{\#n}$. The directional #-derivative

of $f(X)$ at $X$ in the direction of $\Phi$ is defined by

$$ \frac{\partial'^{\#} f(\mathbf{X})}{\partial'^{\#} \Phi} = \#- \lim_{t \to \# 0} \frac{f(\mathbf{X} + t\Phi) - f(\mathbf{X})}{t} \quad (18.62) $$

if the #-limit exists. That is, $\partial'^{\#}f(\mathbf{X})/\partial'^{\#}\Phi$ is the ordinary derivative of the function $H(t) = f(\mathbf{X} + t\Phi)$ at $t = 0$, if $H^{\#}(t)$ exists. The directional #-derivatives that we are most interested in are those in the directions of the unit vectors $E_i, 1 \le i \le n$, where all components of $E_i$ are zero except for the $i$-th, which is 1.

**Definition 18.30.** Since $X$ and $X + tE_i$ differ only in the $i$-th coordinate, $\partial'^{\#}f(\mathbf{X})/\partial'^{\#}E_i$ is called the partial #-derivative of $f$ with respect to $x_i$ at $X$. It is also denoted by $\partial'^{\#}f(\mathbf{X})/\partial'^{\#}x_i$ or $f_{x_i}^{\#}(\mathbf{X})$, thus,

$$ \frac{\partial'^{\#} f(\mathbf{X})}{\partial'^{\#} x_i} = f_{x_i}^{\#}(\mathbf{X}) = \#- \lim_{t \to \# 0} \frac{f(\{x_i + t\}_{i \in n}) - f(\{x_i\}_{i \in n})}{t} \quad (18.63) $$

If $X = (x, y)$, then we denote the partial #-derivatives accordingly; thus,

$$ \frac{\partial'^{\#} f(x, y)}{\partial'^{\#} x} = f_x^{\#}(x, y) = \#- \lim_{h \to \# 0} \frac{f(x + h, y) - f(x, y)}{h} \quad (18.64) $$

and

$$ \frac{\partial'^{\#} f(x, y)}{\partial'^{\#} y} = f_y^{\#}(x, y) = \#- \lim_{h \to \# 0} \frac{f(x, y + h) - f(x, y)}{h} \quad (18.65) $$

**Theorem 18.31.** If $f_{x_i}^{\#}(\mathbf{X})$ and $g_{x_i}^{\#}(\mathbf{X})$ exist, then

$$ \frac{\partial'^{\#} (f + g)(\mathbf{X})}{\partial'^{\#} x_i} = f_{x_i}^{\#}(\mathbf{X}) + g_{x_i}^{\#}(\mathbf{X}), \quad \frac{\partial'^{\#} (f \times g)(\mathbf{X})}{\partial'^{\#} x_i} = f_{x_i}^{\#}(\mathbf{X})g(\mathbf{X}) + g_{x_i}^{\#}(\mathbf{X})f(\mathbf{X}), \quad (18.66) $$

and, if $g(\mathbf{X}) \neq 0$,

$$ \frac{\partial'^{\#} (f/g)(\mathbf{X})}{\partial'^{\#} x_i} = \frac{g(\mathbf{X}) f_{x_i}^{\#}(\mathbf{X}) - f(\mathbf{X}) g_{x_i}^{\#}(\mathbf{X})}{[g(\mathbf{X})]^2} \quad (18.67) $$

If $f_{x_i}^{\#}(\mathbf{X})$ exists at every point of a set $D \subset \mathbb{R}_c^{\#n}$, then it defines a function $f_{x_i}^{\#}(\mathbf{X})$ on $D$. If this function has a partial #-derivative with respect to $x_j$ on a subset of $D$, we denote the partial #-derivative by

$$ \frac{\partial'^{\#}}{\partial'^{\#} x_j} \left( \frac{\partial'^{\#} f(\mathbf{X})}{\partial'^{\#} x_i} \right) = \frac{\partial'^{2\#} f(\mathbf{X})}{\partial'^{\#} x_j \partial'^{\#} x_i} = f_{x_i x_j}^{\#}(\mathbf{X}). \quad (18.68) $$

The function obtained by differentiating $f(\mathbf{X})$ successively with respect to $x_{i_1}, x_{i_2}, \dots, x_{i_r}$ is denoted by

$$ \frac{\partial^{r'\#} f(\mathbf{X})}{\partial'^{\#} x_{i_r} \partial'^{\#} x_{i_{r-1}} \dots \partial'^{\#} x_{i_1}} = f_{x_{i_1} \dots x_{i_{r-1}} x_{i_r}}^{\#}(\mathbf{X}) \quad (18.69) $$