**Proof.** If a set $S$ in $*\mathbb{R}_c^{\#n}$ is hyper bounded, then it can be enclosed within an $n$-box
$T_0 = [-a, a]^n$ where $a > 0$. By the property above, it is enough to show that $T_0$ is
\#-compact.

Assume, by way of contradiction, that $T_0$ is not \#-compact. Then there exists a hyper
infinite open cover $C_{\infty}^{\#}$ of $T_0$ that does not admit any hyperfinite subcover. Through
bisection of each of the sides of $T_0$, the box $T_0$ can be broken up into $2n$ sub $n$-boxes,
each of which has diameter equal to half the diameter of $T_0$. Then at least one of the
$2n$ sections of $T_0$ must require an hyper infinite subcover of $C_{\infty}^{\#}$, otherwise $C_{\infty}^{\#}$ itself
would have a hyperfinite subcover, by uniting together the hyperfinite covers of the
sections. Call this section $T_1$. Likewise, the sides of $T_1$ can be bisected, yielding $2^n$
sections of $T_1$, at least one of which must require an hyper infinite subcover of $C_{\infty}^{\#}$.
Continuing in like manner yields a decreasing hyper infinite sequence of nested
$n$-boxes: $T_0 \supset T_1 \supset T_2 \supset \dots \supset T_k \supset \dots$, $k \in \mathbb{N}^{\#}$, where the side length of $T_k$ is $(2a)/2^k$,
which \#-converges to 0 as $k$ tends to hyper infinity, $k \to \infty^{\#}$. Let us define a hyper
infinite sequence $\{x_k\}_{k \in \mathbb{N}^{\#}}$ such that each $x_k : x_k \in T_k$. This hyper infinite sequence
is Cauchy, so it must \#-converge to some \#-limit $L$. Since each $T_k$ is \#-closed, and
for each $k$ the sequence $\{x_k\}_{k \in \mathbb{N}^{\#}}$ is eventually always inside $T_k$, we see that $L \in T_k$
for each $k \in \mathbb{N}^{\#}$. Since $C_{\infty}^{\#}$ covers $T_0$, then it has some member $U \in C_{\infty}^{\#}$ such that
$L \in U$. Since $U$ is open, there is an $n$-ball $B(L) \subseteq U$. For large enough $k$, one has
$T_k \subseteq B(L) \subseteq U$, but then the hyper infinite number of members of $C_{\infty}^{\#}$ needed to
cover $T_k$ can be replaced by just one: $U$, a contradiction. Thus, $T_0$ is \#-compact. Since
$S$ is \#-closed and a subset of the \#-compact set $T_0$, then $S$ is also \#-compact.
As an application of the Generalized Heine-Borel theorem, we give a short proof of
the Generalized Bolzano-Weierstrass Theorem.

**Theorem 13.21.** (Generalized Bolzano–Weierstrass Theorem) Every hyper bounded hyper infinite set $S \subset *\mathbb{R}_c^\#$ has at least one \#-limit point.

**Proof.** We will show that a hyper bounded nonempty set without a \#-limit point can
contain only finite or a hyper finite number of points. If $S$ has no \#-limit points, then $S$
is \#-closed and every point $x \in S$ has an \#-open neighborhood $N_x$ that
contains no point of $S$ other than $x$. The collection $H = \{N_x | x \in S\}$ is an \#-open
covering for $S$. Since $S$ is also hyper bounded, Theorem 13.20 implies that $S$ can be
covered by finite or a hyper finite collection of sets from $H$, say $N_{x_1}, \dots, N_{x_n}, n \in \mathbb{N}^\#$.
Since these sets contain only $x_1, \dots, x_n$ from $S$, it follows that $S = \{x_k\}_{1 \le k \le n}, n \in \mathbb{N}^\#$.

13. External Cauchy hyperreals $\mathbb{R}_c^\#$ and $*\mathbb{R}_c^\#$ axiomatically.

A model for the Cauchy hyperreal number system consists of a set $\mathbb{R}_c^\#$, two distinct
elements 0 and 1 of $\mathbb{R}_c^\#$, two binary operations $+$ and $\times$ on $\mathbb{R}^\#$ (called addition and
multiplication, respectively), and a binary relation $\leq$ on $\mathbb{R}^\#$, satisfying the following
properties.

Axioms:

I.($\mathbb{R}_c^\#$, $+$, $\times$) forms a field i.e.,

(i) For all $x, y$, and $z$ in $\mathbb{R}^\#$, $x + (y + z) = (x + y) + z$ and $x \times (y \times z) = (x \times y) \times z$.
(associativity of addition and multiplication)

(ii) For all $x$ and $y$ in $\mathbb{R}^\#$, $x + y = y + x$ and $x \times y = y \times x$.
(commutativity of addition and multiplication)

(iii) For all $x, y$, and $z$ in $\mathbb{R}^\#$, $x \times (y + z) = (x \times y) + (x \times z)$.
standard

part of $x$ and is denoted by $st(x)$.

**Proof.** Let $x \in \mathbb{R}^\#$ be finite. Let $D_1$, be the set of $r \in \mathbb{R}$ such that $r < x$ and $D_2$ the set of $r' \in \mathbb{R}$ such that $x < r'$. The pair $(D_1, D_2)$ forms a Dedekind cut in $\mathbb{R}$, hence determines a

unique $r_0 \in \mathbb{R}$. A simple argument shows that $|x - r_0|$ is infinitesimal, i.e., $st(x) = r_0$.

**Notation 13.1.** We usually write $x \approx 0$ iff $x \in \mathbb{I}_\approx^\#$.

**Definition 13.2.** A hypersequence of hyperreal numbers is any function $a : \mathbb{N}^\# \to \mathbb{R}^\#$. Often hypersequences such as these are called hyperreal hypersequences, hypersequences of hyperreal numbers or hypersequences in $\mathbb{R}^\#$ to make it clear that the

elements of the sequence are hyperreal numbers. Analogous definitions can be given for sequences of hypernatural numbers, hyperintegers, etc.

**Notation 13.2.** However, we usually write $a_n$ for the image of $n \in \mathbb{N}^\#$ under $a$, rather than

$a(n)$. The values $a_n$ are often called the elements of the hypersequence $(x_n)_{n \in \mathbb{N}^\#}$.

**Definition 13.3.** We call $x \in \mathbb{R}^\#$ the limit of the hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ if the following condition holds: for each hyperreal number $\varepsilon \in \mathbb{R}^\#$ such that $\varepsilon \approx 0, \varepsilon > 0$, there exists a hypernatural number $N \in \mathbb{N}^\#$ such that, for every hypernatural number $n \ge N$, we have $|x_n - x| < \varepsilon$.

**Definition 13.4.** The hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ is said to #-converge to the #-limit $x$, written $x_n \to x, n \to \infty^\#$ or $\lim_{n \to \infty^\#} (x_n) = x$. Symbolically, this reads:

$$ \forall \varepsilon [(\varepsilon \approx 0) \wedge (\varepsilon > 0)] [\exists N \in \mathbb{N}^\# (\forall n \in \mathbb{N}^\# (n \ge N \Rightarrow |x_n - x| < \varepsilon))]. \quad (13.1) $$

If a hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ converges to some limit, then it is convergent; otherwise it is #-divergent. A hypersequence that has zero as a #-limit is sometimes called a null hypersequence.

Limits of hypersequences behave well with respect to the usual arithmetic operations.

If $a_n \to a, n \to \infty^\#$ and $b_n \to b, n \to \infty^\#$, then $a_n + b_n \to a + b, n \to \infty^\#$ and

$a_n \times b_n \to a \times b, n \to \infty^\#$ if neither $b_n$ or any $b_n$ is zero, $a_n \times b_n \to a \times b, n \to \infty^\#$.

The following properties of limits of real hypersequences provided, in each equation below, that the limits on the right exist.

The limit of a hypersequence is unique.

1. \#-lim$_{n \to \infty^\#}$ $(a_n \pm b_n) = \#-\lim_{n \to \infty^\#} a_n \pm \#-\lim_{n \to \infty^\#} b_n

2. \#-lim$_{n \to \infty^\#}$ $(c \times a_n) = c \times \#-\lim_{n \to \infty^\#} a_n$

3. \#-lim$_{n \to \infty^\#}$ $(a_n \times b_n) = (\#-\lim_{n \to \infty^\#}) \times (\#-\lim_{n \to \infty^\#} b_n)$

4. \#-lim$_{n \to \infty^\#}$ $(a_n/b_n) = \#-\lim_{n \to \infty^\#} a_n/\#-\lim_{n \to \infty^\#} b_n \text{ provided } \#-\lim_{n \to \infty^\#} b_n \neq 0$

5. \#-lim$_{n \to \infty^\#}$ $a_n^p = [\#-\lim_{n \to \infty^\#} a_n]^p$

6. If $a_n \le b_n$ where $n$ greater than some $N$, then $\#-\lim_{n \to \infty^\#} a_n \le \#-\lim_{n \to \infty^\#} b_n$

7. (Squeeze theorem) If $a_n \le c_n \le b_n$, and $\#-\lim_{n \to \infty^\#} a_n = \#-\lim_{n \to \infty^\#} b_n = L$, then $\#-\lim_{n \to \infty^\#} c_n = L$.

**Definition 13.5.** A hyper infinite sequence $(x_n)$ is said to tend to hyperinfinity, written $x_n \to \infty^\#$ or $\#-\lim_{n \to \infty^\#} x_n = \infty^\#$, if for every $K \in \mathbb{R}^\#$, there is an $N \in \mathbb{N}^\#$ such that for every $n \ge N$; that is, the hypersequence terms are eventually larger than any fixed $K$.
(distributivity of multiplication over addition)

(iv) For all $x \in \mathbb{R}^\#$, $x + 0 = x$.

(existence of additive identity)

0 is not equal to 1, and for all $x \in \mathbb{R}^\#$, $x \times 1 = x$.

(existence of multiplicative identity)

(v) For every $x \in \mathbb{R}^\#$, there exists an element $-x$ in $\mathbb{R}^\#$, such that $x + (-x) = 0$.

(existence of additive inverses)

(vi) For every $x \neq 0$ in $\mathbb{R}^\#$, there exists an element $x - 1$ in $\mathbb{R}^\#$, such that $x \times (x - 1) = 1$.

(existence of multiplicative inverses)

II. ($\mathbb{R}^\#$, $\le$) forms a totally ordered set. In other words,

(i) For all $x \in \mathbb{R}^\#$, $x \leq x$. (reflexivity)

(ii) For all $x$ and $y$ in $\mathbb{R}^\#$, if $x \leq y$ and $y \leq x$, then $x = y$. (antisymmetry)

(iii) For all $x, y$, and $z$ in $\mathbb{R}^\#$, if $x \leq y$ and $y \leq z$, then $x \leq z$. (transitivity)

(iv) For all $x$ and $y$ in $\mathbb{R}^\#$, $x \leq y$ or $y \leq x$. (totality)

The field operations $+$ and $\times$ on $\mathbb{R}^\#$ are compatible with the order $\leq$. In other words,

(v) For all $x, y$ and $z$ in $\mathbb{R}^\#$, if $x \leq y$, then $x + z \leq y + z$. (preservation of order under addition)

(vi) For all $x$ and $y$ in $\mathbb{R}^\#$, if $0 \leq x$ and $0 \leq y$, then $0 \leq x \times y$ (preservation of order under multiplication)

III. Non-Archimedean property

$\mathbb{Q}^\#$ $\subsetneq \mathbb{R}^\#$ i.e., $\mathbb{R}^\#$ is non-Archimedean ordered field.

**Remark 13.1.** Here a hyperrational is by definition a ratio of two hyperintegers.
Consider

the ring $\mathbb{Q}_{\text{fin}}^\#$ of all limited (i.e. finite) elements in $\mathbb{Q}^\#$. Then $\mathbb{Q}_{\text{fin}}^\#$ has a unique maximal ideal $\mathbb{I}_{\approx}^\#$, the infinitesimals or infinitesimal numbers are quantities that are closer to zero

than any real number from the field $\mathbb{R}$, but are not zero. The quotient ring $\mathbb{Q}_{\text{fin}}^\#/\mathbb{I}_{\approx}^\#$ gives the field $\mathbb{R}$ of real numbers.

**Definition 13.1.** An element $x \in \mathbb{R}^\#$ is called finite if $|x| < r$ for some $r \in \mathbb{Q}$, $r > 0$. As we shall see in a moment in bivalent case,

**Theorem 13.1.** Every finite $x \in \mathbb{R}^\#$ is infinitely close to some (unique) $r \in \mathbb{R}$ in the sense that $|x - r|$ is either 0 or positively infinitesimal in $\mathbb{R}^\#$. This unique $r$ is called the standard

part of $x$ and is denoted by $st(x)$.

**Proof.** Let $x \in \mathbb{R}^\#$ be finite. Let $D_1$, be the set of $r \in \mathbb{R}$ such that $r < x$ and $D_2$ the set of $r' \in \mathbb{R}$ such that $x < r'$. The pair $(D_1, D_2)$ forms a Dedekind cut in $\mathbb{R}$, hence determines a unique $r_0 \in \mathbb{R}$. A simple argument shows that $|x - r_0|$ is infinitesimal, i.e., $st(x) = r_0$.

**Notation 13.1.** We usually write $x \approx 0$ iff $x \in \mathbb{I}_{\approx}^\#$.

**Definition 13.2.** A hypersequence of hyperreal numbers is any function $a : \mathbb{N}^\# \to \mathbb{R}^\#$. Often hypersequences such as these are called hyperreal hypersequences, hypersequences of hyperreal numbers or hypersequences in $\mathbb{R}^\#$ to make it clear that the

elements of the sequence are hyperreal numbers. Analogous definitions can be given
Similarly, $x_n \to -\infty^\#$ if for every $K \in \mathbb{R}^\#$, there is an $N \in \mathbb{N}^\#$ such that for every $n \ge N$, $x_n < K$. If a hypersequence tends to infinity or minus infinity, then it is divergent.

However, a divergent hypersequence need not tend to plus or minus hyperinfinity

**Definition 13.6.** A hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ of hyperreal numbers is called a *Cauchy hypersequence* if for every positive hyperreal number $\epsilon$, there is a positive

hyperinteger

$N \in \mathbb{N}^\#$ such that for all hypernatural numbers $m, n > N$ : $|x_m - x_n| < \epsilon$, where the vertical bars denote the absolute value. In a similar way one can define

Cauchy hypersequences

of hyperrational numbers, etc. Cauchy formulated such a condition by requiring $|x_m - x_n| \approx 0$ i.e., to be infinite small for every pair of infinite large $m, n \in \mathbb{N}^\#$.

**Definition 13.7.** Let $\mathbb{R}_c^\#$ be the set of Cauchy hypersequences of hyperrational numbers.

That is, hypersequences $(x_n)_{n \in \mathbb{N}^\#}$ of hyperrational numbers such that for every hyperrational $\epsilon > 0$, there exists an hyperinteger $N \in \mathbb{N}^\# \mathbb{N}$ such that for all hypernatural numbers $m, n > N$, $|x_m - x_n| < \epsilon$. Here the vertical bars as usual denote the absolute value.

**Definition 13.8.** A standard procedure to force all Cauchy hypersequences in a metric space to converge is adding new points to the metric space in a process called completion. $\mathbb{R}_c^\#$ is defined as the completion of $\mathbb{Q}^\#$ with respect to the metric $|x - y|$, as will be detailed below.

**Definition 13.9.** Cauchy hypersequences $(x_n)_{n \in \mathbb{N}^\#}$ and $(y_n)_{n \in \mathbb{N}^\#}$ can be added and multiplied as follows:

$$ (x_n)_{n \in \mathbb{N}^\#} + (y_n)_{n \in \mathbb{N}^\#} = (x_n + y_n)_{n \in \mathbb{N}^\#}, \qquad (13.2) $$

and

$$ (x_n)_{n \in \mathbb{N}^\#} \times (y_n)_{n \in \mathbb{N}^\#} = (x_n \times y_n)_{n \in \mathbb{N}^\#}. \qquad (13.3) $$

**Definition 13.10.** Two Cauchy hypersequences are called equivalent if and only if the difference between them tends to zero. This defines an equivalence relation that is compatible with the operations (16.2)-(16.3) defined above, and the set $\mathbb{R}_c^\#$ of all equivalence classes $cl[(x_n)_{n \in \mathbb{N}^\#}]$ can be shown to satisfy all axioms of the hyperreal numbers.

We can embed $\mathbb{Q}^\#$ into $\mathbb{R}_c^\#$ by identifying the rational number $r \in \mathbb{Q}^\#$ with the equivalence

class of the hypersequence $(r_n)_{n \in \mathbb{N}^\#}$ with $r_n = r$ for all $n \in \mathbb{N}^\#$.

**Remark 13.2.** Comparison between hyperreal numbers is obtained by defining the following comparison between Cauchy hypersequences:

$$ (x_n)_{n \in \mathbb{N}^\#} \ge (y_n)_{n \in \mathbb{N}^\#} \qquad (13.4) $$

if and only if $x$ is equivalent to $y$ or there exists an hyperinteger $N \in \mathbb{N}^\#$ such that

$$ x_n \ge y_n $$

for all $n > N$.

**Remark 13.3.** By construction, every hyperreal number $x \in \mathbb{R}_c^\#$ is represented by a Cauchy hyper infinite sequence of hyperrational numbers. This representation is far from unique; every hyperrational hypersequence that converges to $x$ is a representation of $x$. This reflects the observation that one can often use different
for

sequences of hypernatural numbers, hyperintegers, etc.

**Notation 13.2.** However, we usually write $a_n$ for the image of $n \in \mathbb{N}^\#$ under $a$, rather than

$a(n)$. The values $a_n$ are often called the elements of the hypersequence $(x_n)_{n \in \mathbb{N}^\#}$.

**Definition 13.3.** We call $x \in \mathbb{R}^\#$ the limit of the hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ if the following condition holds: for each hyperreal number $\varepsilon \in \mathbb{R}^\#$ such that $\varepsilon \approx 0, \varepsilon > 0$, there exists a hypernatural number $N \in \mathbb{N}^\#$ such that, for every hypernatural number $n \ge N$, we have $|x_n - x| < \varepsilon$.

**Definition 13.4.** The hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ is said to #-converge to the #-limit $x$, written $x_n \to x, n \to \infty^\#$ or $\lim_{n \to \infty^\#} (x_n) = x$. Symbolically, this reads:

$$ \forall \varepsilon [(\varepsilon \approx 0) \land (\varepsilon > 0)][\exists N \in \mathbb{N}^\# (\forall n \in \mathbb{N}^\# (n \ge N \Rightarrow |x_n - x| < \varepsilon))]. \quad (13.1) $$

If a hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ converges to some limit, then it is convergent; otherwise it is #-divergent. A hypersequence that has zero as a #-limit is sometimes called a null hypersequence.

Limits of hypersequences behave well with respect to the usual arithmetic operations.

If $a_n \to a, n \to \infty^\#$ and $b_n \to b, n \to \infty^\#$, then $a_n + b_n \to a + b, n \to \infty^\#$ and

$a_n \times b_n \to a \times b, n \to \infty^\#$ if neither $b_n$ or any $b_n$ is zero, $a_n \times b_n \to a \times b, n \to \infty^\#$.

The following properties of limits of real hypersequences provided, in each equation below, that the limits on the right exist.

The limit of a hypersequence is unique.

1. $\#\lim_{n \to \infty^\#} (a_n \pm b_n) = \#\lim_{n \to \infty^\#} a_n \pm \#\lim_{n \to \infty^\#} b_n$

2. $\#\lim_{n \to \infty^\#} (c \times a_n) = c \times \#\lim_{n \to \infty^\#} a_n$

3. $\#\lim_{n \to \infty^\#} (a_n \times b_n) = (\#\lim_{n \to \infty^\#}) \times (\#\lim_{n \to \infty^\#} b_n)$

4. $\#\lim_{n \to \infty^\#} (a_n/b_n) = \#\lim_{n \to \infty^\#} a_n/\#\lim_{n \to \infty^\#} b_n$ provided $\#\lim_{n \to \infty^\#} b_n \neq 0$

5. $\#\lim_{n \to \infty^\#} a_n^p = [\#\lim_{n \to \infty^\#} a_n]^p$

6. If $a_n \le b_n$ where $n$ greater than some $N$, then $\#\lim_{n \to \infty^\#} a_n \le \#\lim_{n \to \infty^\#} b_n$

7. (Squeeze theorem) If $a_n \le c_n \le b_n$, and $\#\lim_{n \to \infty^\#} a_n = \#\lim_{n \to \infty^\#} b_n = L$, then $\#\lim_{n \to \infty^\#} c_n = L$.

**Definition 13.5.** A hyper infinite sequence $(x_n)$ is said to tend to hyperinfinity, written $x_n \to \infty^\#$ or $\#\lim_{n \to \infty^\#} x_n = \infty^\#$, if for every $K \in \mathbb{R}^\#$, there is an $N \in \mathbb{N}^\#$ such that for every $n \ge N$; that is, the hypersequence terms are eventually larger than any fixed $K$.

Similarly, $x_n \to -\infty^\#$ if for every $K \in \mathbb{R}^\#$, there is an $N \in \mathbb{N}^\#$ such that for every $n \ge N$, $x_n < K$. If a hypersequence tends to infinity or minus infinity, then it is divergent.

However, a divergent hypersequence need not tend to plus or minus hyperinfinity

**Definition 13.6.** A hypersequence $(x_n)_{n \in \mathbb{N}^\#}$ of hyperreal numbers is called a *Cauchy hypersequence* if for every positive hyperreal number $\varepsilon$, there is a positive

hyperinteger

$N \in \mathbb{N}^\#$ such that for all hypernatural numbers $m, n > N : |x_m - x_n| < \varepsilon$, where the vertical bars denote the absolute value. In a similar way one can define

Cauchy hypersequences

of hyperrational numbers, etc. Cauchy formulated such a condition by requiring $|x_m - x_n| \approx 0$ i.e., to be infinite small for every pair of infinite large $m, n \in \mathbb{N}^\#$.

**Definition 13.7.** Let $\mathbb{R}^\#_c$ be the set of Cauchy hypersequences of hyperrational numbers.
hypersequences to approximate the same hyperreal number. The equation
0.999... = 1 states that the hyper infinite sequences
(0, 0.9, 0.99, 0.999, ... ) and (1, 1, 1, 1, ...) are equivalent, i.e., their difference
#-converges to 0.

#### IV. The field $\mathbb{R}^\#$ is complete in the following sense:

**Definition 13.11.** Let $S \subsetneq \mathbb{R}_c^\#$ be a non-empty set of hyperreal numbers.

A hyperreal number $x \in \mathbb{R}_c^\#$ is called an upper bound for $S$ if $x \ge s$ for all $s \in S$.

A hyperreal number $x$ is the least upper bound (or supremum $\sup S$) for $S$ if $x$ is an upper bound for $S$ and $x \le y$ for every upper bound $y$ of $S$.

**Remark 13.4.** The order $\le$ given by Eq. (14.4) obviously is $\le$-incomplete.

**Definition 13.12.** Let $S \subsetneq \mathbb{R}_c^\#$ be a nonempty subset of $\mathbb{R}_c^\#$. We will say that:

(1) $S$ is $\le$-admissible above if the following conditions are satisfied:

(i) $S$ bounded above;

(ii) let $A(S)$ be a set $\forall x[x \in A(S) \Leftrightarrow x \ge S]$ then for any $\varepsilon > 0, \varepsilon \approx 0$ there exist $\alpha \in S$ and $\beta \in A(S)$ such that $\beta - \alpha \le \varepsilon \approx 0$.

(2) $S$ is $\le$-admissible below if the following condition are satisfied:

(i) $S$ bounded below;

(ii) let $L(S)$ be a set $\forall x[x \in L(S) \Leftrightarrow x \le S]$ then for any $\varepsilon > 0, \varepsilon \approx 0$ there exist $\alpha \in S$ and $\beta \in L(S)$ such that $\alpha - \beta \le \varepsilon \approx 0$.

**Theorem 13.2.** (i) Every $\le$-admissible above subset $S \subsetneq \mathbb{R}_c^\#$ has a supremum $\sup S$.

(ii) Every $\le$-admissible below subset $S \subsetneq \mathbb{R}_c^\#$ has in finitum $inf S$.

**Proof.** Let $S \subsetneq \mathbb{R}_c^\#$ be a nonempty subset of $\mathbb{R}_c^\#$, and let $M \in \mathbb{Q}^\#$ be an hyperrational upper bound for $S$. We are going to construct two hypersequences of hyperrational numbers, $(u_n)_{n \in \mathbb{N}^\#}$ and $(l_n)_{n \in \mathbb{N}^\#}$. First, since $S$ is nonempty, there is some element $s_0 \in S$.

We can choose a hyperrational number $L \in \mathbb{Q}^\#$ such that $L < s_0$. Now, we go through the following hyperinductive procedure to produce hyperrational numbers $u_0, u_1, u_2, \dots$ and $l_0, l_1, l_2, l_3, \dots$.

(i) Set $u_0 = M$ and $l_0 = L$.

(ii) Suppose that we have already defined $u_n$ and $l_n, n \in \mathbb{N}^\#$.

Consider the number $m_n = (u_n + l_n)/2$, i.e., the average between $u_n$ and $l_n$.

(1) If $m_n$ is an upper bound for $S$, define $u_{n+1} = m_n$ and $l_{n+1} = l_n$.

(2) If $m_n$ is not an upper bound for $S$, define $u_{n+1} = u_n$ and $l_{n+1} = m_n$.

Since $l_0 < M$, it is easy to prove by hyperinfinite induction that $(u_n)_{n \in \mathbb{N}^\#}$ is a non-increasing hypersequence, i.e. $u_{n+1} \le u_n$ and $(l_n)_{n \in \mathbb{N}^\#}$ is a non-decreasing hypersequence, i.e. $l_{n+1} \ge l_n$.

**Remark 13.5.** Note that in the first case above, we have that

$$u_{n+1} - l_{n+1} = m_n - l_n = \frac{u_n + l_n}{2} - l_n = \frac{u_n - l_n}{2} \qquad (13.5)$$

In the second case, we also have that

$$u_{n+1} - l_{n+1} = u_n - m_n = u_n - \frac{u_n + l_n}{2} = \frac{u_n - l_n}{2} \qquad (13.6)$$

Now, this means that $u_1 - l_1 = \frac{1}{2}(M - L)$ and so $u_2 - l_2 = \frac{1}{2}(u_1 - l_1) = \frac{1}{2^2}(M - L)$,
and in general by hyperinfinite induction one obtains
That is, hypersequences $(x_n)_{n \in \mathbb{N}^\#}$ of hyperrational numbers such that for every hyperrational $\varepsilon > 0$, there exists an hyperinteger $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that for all hypernatural numbers $m, n > N, |x_m - x_n| < \varepsilon$. Here the vertical bars as usual denote the absolute value.

**Definition 13.8.** A standard procedure to force all Cauchy hypersequences in a metric space to converge is adding new points to the metric space in a process called completion. $\mathbb{R}_c^\#$ is defined as the completion of $\mathbb{Q}^\#$ with respect to the metric $|x - y|$, as will be detailed below.

**Definition 13.9.** Cauchy hypersequences $(x_n)_{n \in \mathbb{N}^\#}$ and $(y_n)_{n \in \mathbb{N}^\#}$ can be added and multiplied as follows:

$$ (x_n)_{n \in \mathbb{N}^\#} + (y_n)_{n \in \mathbb{N}^\#} = (x_n + y_n)_{n \in \mathbb{N}^\#}, \qquad (13.2) $$

and

$$ (x_n)_{n \in \mathbb{N}^\#} \times (y_n)_{n \in \mathbb{N}^\#} = (x_n \times y_n)_{n \in \mathbb{N}^\#}. \qquad (13.3) $$

**Definition 13.10.** Two Cauchy hypersequences are called equivalent if and only if the difference between them tends to zero. This defines an equivalence relation that is compatible with the operations (16.2)-(16.3) defined above, and the set $\mathbb{R}_c^\#$ of all equivalence classes $cl[(x_n)_{n \in \mathbb{N}^\#}]$ can be shown to satisfy all axioms of the hyperreal numbers.

We can embed $\mathbb{Q}^\#$ into $\mathbb{R}_c^\#$ by identifying the rational number $r \in \mathbb{Q}^\#$ with the equivalence class of the hypersequence $(r_n)_{n \in \mathbb{N}^\#}$ with $r_n = r$ for all $n \in \mathbb{N}^\#$.

**Remark 13.2.** Comparison between hyperreal numbers is obtained by defining the following comparison between Cauchy hypersequences:

$$ (x_n)_{n \in \mathbb{N}^\#} \ge (y_n)_{n \in \mathbb{N}^\#} \qquad (13.4) $$

if and only if $x$ is equivalent to $y$ or there exists an hyperinteger $N \in \mathbb{N}^\#$ such that $x_n \ge y_n$ for all $n > N$.

**Remark 13.3.** By construction, every hyperreal number $x \in \mathbb{R}_c^\#$ is represented by a Cauchy hyper infinite sequence of hyperrational numbers. This representation is far from unique; every hyperrational hypersequence that converges to $x$ is a representation of $x$. This reflects the observation that one can often use different hypersequences to approximate the same hyperreal number. The equation $0.999... = 1$ states that the hyper infinite sequences $(0, 0.9, 0.99, 0.999, ...)$ and $(1, 1, 1, 1, ...)$ are equivalent, i.e., their difference $\#-\text{converges to } 0$.

**IV.** The field $\mathbb{R}^\#$ is complete in the following sense:

**Definition 13.11.** Let $S \subsetneq \mathbb{R}_c^\#$ be a non-empty set of hyperreal numbers.

A hyperreal number $x \in \mathbb{R}_c^\#$ is called an upper bound for $S$ if $x \ge s$ for all $s \in S$.

A hyperreal number $x$ is the least upper bound (or supremum $\sup S$) for $S$ if $x$ is an upper bound for $S$ and $x \le y$ for every upper bound $y$ of $S$.

**Remark 13.4.** The order $\le$ given by Eq. (14.4) obviously is $\le$-incomplete.

**Definition 13.12.** Let $S \subsetneq \mathbb{R}_c^\#$ be a nonempty subset of $\mathbb{R}_c^\#$. We will say that:

(1) $S$ is $\le$-admissible above if the following conditions are satisfied:

(i) $S$ bounded above;
$$u_n - l_n = 2^{-n}(M - L). \qquad (13.7)$$

Since $M > L$ so $M - L > 0$, and since $2^{-n} < n^{-1}$ we have for any $\varepsilon > 0$, $\varepsilon \approx 0$ that $2^{-n}(M - L) < \varepsilon$ for all sufficiently large $n \in \mathbb{N}^*\mathbb{N}$. Thus, $u_n - l_n < \varepsilon$ as well, and so

$$\# - \lim_{n \to \infty} (u_n - l_n) = 0. \qquad (13.8)$$

This defines two hypersequences of hyperrationals, and so we have hyperreal numbers

$l = (l_n)_{n \in \mathbb{N}^*}$ and $u = (u_n)_{n \in \mathbb{N}^*}$. It is easy to prove, by induction on $n \in \mathbb{N}^*$ that:

(i) $u_n$ is an upper bound for $S$ for all $n \in \mathbb{N}^*$ and

(ii) $l_n$ is never an upper bound for $S$ for any $n \in \mathbb{N}^*$.

Thus $u$ is an upper bound for $S$. To see that it is a least upper bound, notice that the \#-limit of $(u_n - l_n)_{n \in \mathbb{N}^*}$ is 0, and so $l = u$. Now suppose $b < u = l$ is a smaller upper bound

for $S$. Since $(l_n)_{n \in \mathbb{N}^*}$ is monotonic increasing it is easy to see that $b < l_n$ for some $n \in \mathbb{N}^*$.

But $l_n$ is not an upper bound for $S$ and so neither is $b$. Hence $u$ is a least upper bound for $S$.

## 14. §14.1. External non-Archimedean field $\tilde{\mathbb{R}}_c^\#$ via special extension of external non-Archimedean field $\mathbb{R}_c^\#$.

**Notation 14.1.3.** Let $\Delta \subset \mathbb{R}_c^\#$ and $\Delta \neq \{0\}$. Then we write $\Delta > 0$ iff $a \in \Delta \Rightarrow a > 0$.

**Definition 14.1.13.** Let $\Delta \subset \mathbb{R}_c^\#$ and $\Delta > 0$. Assume that: $a, b \in \Delta \Rightarrow a + b \in \Delta$. Then we say that $\Delta$ is a positive idempotent in $\mathbb{R}_c^\#$.

**Notation 14.1.4.** We will denote by $\mathbb{R}_{c+,\text{fin}}^\#$ a set of the all positive finite number in $\mathbb{R}_c^\#$ except infinitesimals in $\mathbb{R}_c^\#$.

**Remark 14.1.6.** Note that a set $\mathbb{R}_{c+,\text{fin}}^\# \setminus \{0_{\mathbb{R}_c^\#}\} \subset \mathbb{R}_c^\#$ is a positive idempotent in $\mathbb{R}_c^\#$.

**Proposition 14.1.1.** Let $\Delta \subset \mathbb{R}_c^\#$ is a positive idempotent in $\mathbb{R}_c^\#$. Then the following are equivalent. [In what follows assume $a, b > 0_{\mathbb{R}_c^\#}$].

(i) $a \in \Delta \Rightarrow 2a \in \Delta$,

(ii) $a \in \Delta \Rightarrow na \in \Delta$ for all standard integers $n \in \mathbb{N}$,

(iii) $a \in \Delta \Rightarrow ra \in \Delta$ for all finite $r \in \mathbb{R}_c^\#$.

**Proof.** All parts are immediate from the Definition 14.1.13.

**Notation 14.1.4.** $\Delta_{\approx}^{\#+} \triangleq \{\delta \in \mathbb{R}_c^\# | \delta > 0, \delta \approx 0\}$, i.e. $\Delta_{\approx}^{\#+}$ is a set of the all positive infinitesimals in $\mathbb{R}_c^\#$; $\Delta_{\approx}^{\#-} \triangleq \{\delta \in \mathbb{R}_c^\# | \delta < 0, \delta \approx 0_{\mathbb{R}_c^\#}\}$, i.e. $\Delta_{\approx}^{\#+}$ is a set of the all negative infinitesimals in $\mathbb{R}_c^\#$. Note that $\Delta_{\approx}^{\#-} = -\Delta_{\approx}^{\#+}$.

**Remark 14.1.7.** Note that a set $\Delta_{\approx}^{\#+} \subset \mathbb{R}_c^\#$ is a positive idempotent in $\mathbb{R}_c^\#$ and $\Delta_{\approx}^{\#-}$ is a negative idempotent in $\mathbb{R}_c^\#$.

**Definition 14.1.14.** Let $\{a_n\}_{n=0}^\infty$ be $\mathbb{R}_{c+,\text{fin}}^\#$-valued countable sequence $a : \mathbb{N} \to \mathbb{R}_{c+,\text{fin}}^\#$ such that:

(i) there is $M \in \mathbb{N}$ such that $\{a_n\}_{n=M}^\infty$ is monotonically decreasing $\mathbb{R}_{c+,\text{fin}}^\#$-valued countable sequence $a : \mathbb{N} \to \mathbb{R}_{c+,\text{fin}}^\# \setminus \{0_{\mathbb{R}_c^\#}\}$

(ii) there is $M \in \mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{\mathbb{R}_c^\#}$ [it follows from (ii)]

(iii) for all $n \in \mathbb{N}$, $a_n \not\approx 0_{\mathbb{R}_c^\#}$ and for any $\epsilon > 0$, $\epsilon \not\approx 0_{\mathbb{R}_c^\#}$, $\epsilon \in \mathbb{R}_{c+,\text{fin}}^\#$ there is $N \in \mathbb{N}$ such
(ii) let $A(S)$ be a set $\forall x[x \in A(S) \Leftrightarrow x \geq S]$ then for any $\varepsilon > 0, \varepsilon \approx 0$ there exist $\alpha \in S$ and $\beta \in A(S)$ such that $\beta - \alpha \leq \varepsilon \approx 0$.

(2) S is ≤-admissible below if the following condition are satisfied:

(i) S bounded below;

(ii) let $L(S)$ be a set $\forall x[x \in L(S) \Leftrightarrow x \leq S]$ then for any $\varepsilon > 0, \varepsilon \approx 0$ there exist $\alpha \in S$ and $\beta \in L(S)$ such that $\alpha - \beta \leq \varepsilon \approx 0$.

**Theorem 13.2.** (i) Every ≤-admissible above subset $S \subsetneq \mathbb{R}_c^\#$ has a supremum $\sup S$.

(ii) Every ≤-admissible below subset $S \subsetneq \mathbb{R}_c^\#$ has infinum $\inf S$.

**Proof.** Let $S \subsetneq \mathbb{R}_c^\#$ be a nonempty subset of $\mathbb{R}_c^\#$, and let $M \in \mathbb{Q}^\#$ be an hyperrational upper bound for $S$. We are going to construct two hypersequences of hyperrational numbers, $(u_n)_{n \in \mathbb{N}^\#}$ and $(l_n)_{n \in \mathbb{N}^\#}$. First, since $S$ is nonempty, there is some element $s_0 \in S$.

We can choose a hyperrational number $L \in \mathbb{Q}^\#$ such that $L < s_0$. Now, we go through the following hyperinductive procedure to produce hyperrational numbers $u_0, u_1, u_2, \dots$ and $l_0, l_1, l_2, l_3, \dots$.

(i) Set $u_0 = M$ and $l_0 = L$.

(ii) Suppose that we have already defined $u_n$ and $l_n, n \in \mathbb{N}^\#$.

Consider the number $m_n = (u_n + l_n)/2$, i.e., the average between $u_n$ and $l_n$.

(1) If $m_n$ is an upper bound for $S$, define $u_{n+1} = m_n$ and $l_{n+1} = l_n$.

(2) If $m_n$ is not an upper bound for $S$, define $u_{n+1} = u_n$ and $l_{n+1} = m_n$.

Since $l_0 < M$, it is easy to prove by hyperinfinite induction that $(u_n)_{n \in \mathbb{N}^\#}$ is a non-increasing hypersequence, i.e. $u_{n+1} \leq u_n$ and $(l_n)_{n \in \mathbb{N}^\#}$ is a non-decreasing hypersequence, i.e. $l_{n+1} \geq l_n$.

**Remark 13.5.** Note that in the first case above, we have that

$$u_{n+1} - l_{n+1} = m_n - l_n = \frac{u_n + l_n}{2} - l_n = \frac{u_n - l_n}{2}. \quad (13.5)$$

In the second case, we also have that

$$u_{n+1} - l_{n+1} = u_n - m_n = u_n - \frac{u_n + l_n}{2} = \frac{u_n - l_n}{2}. \quad (13.6)$$

Now, this means that $u_1 - l_1 = \frac{1}{2}(M - L)$ and so $u_2 - l_2 = \frac{1}{2}(u_1 - l_1) = \frac{1}{2^2}(M - L)$, and in general by hyperinfinite induction one obtains

$$u_n - l_n = 2^{-n}(M - L). \quad (13.7)$$

Since $M > L$ so $M - L > 0$, and since $2^{-n} < n^{-1}$ we have for any $\varepsilon > 0, \varepsilon \approx 0$ that $2^{-n}(M - L) < \varepsilon$ for all sufficiently large $n \in \mathbb{N}^\#\mathbb{N}$. Thus, $u_n - l_n < \varepsilon$ as well, and so

$$\#\lim_{n \to \infty} (u_n - l_n) = 0. \quad (13.8)$$

This defines two hypersequences of hyperrationals, and so we have hyperreal numbers

$l = (l_n)_{n \in \mathbb{N}^\#}$ and $u = (u_n)_{n \in \mathbb{N}^\#}$. It is easy to prove, by induction on $n \in \mathbb{N}^\#$ that:

(i) $u_n$ is an upper bound for $S$ for all $n \in \mathbb{N}^\#$ and

(ii) $l_n$ is never an upper bound for $S$ for any $n \in \mathbb{N}^\#$.

Thus $u$ is an upper bound for $S$. To see that it is a least upper bound, notice that the $\#\text{-limit of } (u_n - l_n)_{n \in \mathbb{N}^\#}$ is 0, and so $l = u$. Now suppose $b < u = l$ is a smaller upper bound

for $S$. Since $(l_n)_{n \in \mathbb{N}^\#}$ is monotonic increasing it is easy to see that $b < l_n$ for some $n \in \mathbb{N}^\#$.
that for all $n > N: a_n < \epsilon$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\downarrow 0}$.

We define a set $\Delta_{\omega}^{-\downarrow 0}$ by $c_n \in \Delta_{\omega}^{-\downarrow 0} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$. Note that $\Delta_{\omega}^{-\downarrow 0} = -\Delta_{\omega}^{+\downarrow 0}$.

**Remark 14.1.8.** Note that a set $\Delta_{\omega}^{+\downarrow 0}$ is a positive idempotent in $\mathbb{R}_c^\#$ and a set $\Delta_{\omega}^{-\downarrow 0}$ is a negative idempotent in $\mathbb{R}_c^\#$.

**Proposition 14.1.2.** (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ then:

(i) $\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$

(ii) $\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0} \cup \Delta_{\approx}^{+\#} \cup \Delta_{\approx}^{-\#} \cup \{0_{\mathbb{R}_c^\#}\}_{n=0}^{\infty}$
where $\{0_{\mathbb{R}_c^\#}\}_{n=0}^{\infty}$ is a countable $0_{\mathbb{R}_c^\#}$-valued sequence.

(iii) $\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$.

(2) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ then we define

(i) $\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$

(ii) $\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0}$

(iii) $\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$

(3) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0}$ and $x, y \in \mathbb{R}_c^\#$ then we define

(iv) $x + y\{a_n\}_{n=0}^{\infty} \triangleq \{x + ya_n\}_{n=0}^{\infty}$

**Proof.** Immediately by definitions and by Definition 14.1.14.

**Definition 14.1.15.** We define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{+\downarrow 0}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N: a_n \le b_n$ and similarly we define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{-\downarrow 0}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N: a_n \le b_n$

**Definition 14.1.16.** (1) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega}^{+\downarrow} \times \mathbb{R}_{c+,\text{fin}}^\#$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $x \in \mathbb{R}_{c+,\text{fin}}^\#$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N: a_n < x$.

(2) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\approx}^{+\#} \times \Delta_{\omega}^{+\downarrow}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $x \in \Delta_{\approx}^{+\#}$ then $x < \{a_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N: x < a_n$.

(3) Let $\{a_n\}_{n=0}^{\infty}$ be $\Delta_{\approx}^{+\#}$-valued countable sequence $a: \mathbb{N} \to \Delta_{\approx}^{+\#}$, and we denote a set of the all these sequences by $\Delta_{\approx,\omega}^{+\#}$.

We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\approx,\omega}^{+\#} \times \Delta_{\omega}^{+\downarrow}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\approx,\omega}^{+\#}$ and $x \in \Delta_{\approx}^{+\#}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N: a_n < x$.

**Proposition 14.1.2.** Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $\{a_n\}_{n=0}^{\infty} \neq 0_{\mathbb{R}_c^\#}$ then there is $N \in \mathbb{N}$ such that $0_{\mathbb{R}_c^\#} < \Delta_{\approx}^{+\#} < \{a_n\}_{n=0}^{\infty} < \mathbb{R}_{c+,\text{fin}}^\# \setminus \{0_{\mathbb{R}_c^\#}\}$.

**Proof.** Immediately by definitions and by Definition 14.1.15.

**Remark 14.1.9.** Note that it follows from Proposition 14.1.2 that

$$0_{\mathbb{R}_c^\#} < \Delta_{\approx}^{+\#} < \Delta_{\omega}^{+\downarrow 0} < \mathbb{R}_{c+,\text{fin}}^\# \setminus \{0_{\mathbb{R}_c^\#}\}. \quad (14.1.9)$$

**Definition 14.1.17.** Let $\{a_n\}_{n=0}^{\infty}$ be monotonically increasing $\mathbb{R}_{c+,\text{fin}}^\#$-valued countable sequence $a: \mathbb{N} \to \mathbb{R}_{c+,\text{fin}}^\# \Delta_{\approx}^{+\#}$ such that:

(i) there is $M \in \mathbb{N}$ such that for all $n > M, a_n \neq 0_{\mathbb{R}_c^\#}$

(ii) there is $N \in \mathbb{N}$ such that for all $n > N$ and for any $\xi > 0_{\mathbb{R}_c^\#}, \xi \in \mathbb{R}_{c+,\text{fin}}^\#$, $a_n > \xi$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\downarrow\infty}$. We define a set $\Delta_{\omega}^{-\downarrow\infty}$ by $c_n \in \Delta_{\omega}^{-\downarrow\infty} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}$. Note that $\Delta_{\omega}^{-\downarrow\infty} = -\Delta_{\omega}^{+\downarrow\infty}$.

**Proposition 14.1.3.** (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty}$ then:
But $l_n$ is not an upper bound for $S$ and so neither is $b$. Hence $u$ is a least upper bound for $S$.

## 14. §14.1. External non-Archimedean field $\mathbb{R}_c^\#$ via special extension of external non-Archimedean field $\mathbb{R}_c^\#$.

**Notation 14.1.3.** Let $\Delta \subset \mathbb{R}_c^\#$ and $\Delta \neq \{0\}$. Then we write $\Delta > 0$ iff $a \in \Delta \Rightarrow a > 0$.

**Definition 14.1.13.** Let $\Delta \subset \mathbb{R}_c^\#$ and $\Delta > 0$. Assume that: $a, b \in \Delta \Rightarrow a + b \in \Delta$. Then we say that $\Delta$ is a positive idempotent in $\mathbb{R}_c^\#$.

**Notation 14.1.4.** We will denote by $\mathbb{R}_{c+,\text{fin}}^\#$ a set of the all positive finite number in $\mathbb{R}_c^\#$ except infinitesimals in $\mathbb{R}_c^\#$.

**Remark 14.1.6.** Note that a set $\mathbb{R}_{c+,\text{fin}}^\# \setminus \{0_{\mathbb{R}_c^\#}\} \subset \mathbb{R}_c^\#$ is a positive idempotent in $\mathbb{R}_c^\#$.

**Proposition 14.1.1.** Let $\Delta \subset \mathbb{R}_c^\#$ is a positive idempotent in $\mathbb{R}_c^\#$. Then the following are equivalent. [In what follows assume $a, b > 0_{\mathbb{R}_c^\#}$].

(i) $a \in \Delta \Rightarrow 2a \in \Delta$,

(ii) $a \in \Delta \Rightarrow na \in \Delta$ for all standard integers $n \in \mathbb{N}$,

(iii) $a \in \Delta \Rightarrow ra \in \Delta$ for all finite $r \in \mathbb{R}_c^\#$.

*Proof.* All parts are immediate from the Definition 14.1.13.

**Notation 14.1.4.** $\Delta_{\approx}^{+\#} \triangleq \{\delta \in \mathbb{R}_c^\# | \delta > 0, \delta \approx 0\}$, i.e. $\Delta_{\approx}^{+\#}$ is a set of the all positive infinitesimals in $\mathbb{R}_c^\#$; $\Delta_{\approx}^{-\#} \triangleq \{\delta \in \mathbb{R}_c^\# | \delta < 0, \delta \approx 0_{\mathbb{R}_c^\#}\}$, i.e. $\Delta_{\approx}^{-\#}$ is a set of the all negative infinitesimals in $\mathbb{R}_c^\#$. Note that $\Delta_{\approx}^{-\#} = -\Delta_{\approx}^{+\#}$.

**Remark 14.1.7.** Note that a set $\Delta_{\approx}^{+\#} \subset \mathbb{R}_c^\#$ is a positive idempotent in $\mathbb{R}_c^\#$ and $\Delta_{\approx}^{-\#}$ is a negative idempotent in $\mathbb{R}_c^\#$.

**Definition 14.1.14.** Let $\{a_n\}_{n=0}^\infty$ be $\mathbb{R}_{c+,\text{fin}}^\#$-valued countable sequence $a : \mathbb{N} \to \mathbb{R}_{c+,\text{fin}}^\#$ such that:

(i) there is $M \in \mathbb{N}$ such that $\{a_n\}_{n=M}^\infty$ is monotonically decreasing $\mathbb{R}_{c+,\text{fin}}^\#$-valued countable sequence $a : \mathbb{N} \to \mathbb{R}_{c+,\text{fin}}^\# \setminus \{0_{\mathbb{R}_c^\#}\}$

(ii) there is $M \in \mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{\mathbb{R}_c^\#}$ [it follows from (ii)]

(iii) for all $n \in \mathbb{N}$, $a_n \not\approx 0_{\mathbb{R}_c^\#}$ and for any $\epsilon$ such that $\epsilon > 0$, $\epsilon \not\approx 0_{\mathbb{R}_c^\#}$, $\epsilon \in \mathbb{R}_{c+,\text{fin}}^\#$ there is $N \in \mathbb{N}$ such that for all $n > N$: $a_n < \epsilon$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\downarrow 0}$.

We define a set $\Delta_{\omega}^{-\downarrow 0}$ by $a_n \in \Delta_{\omega}^{-\downarrow 0} \Leftrightarrow \{-a_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0}$. Note that $\Delta_{\omega}^{-\downarrow 0} = -\Delta_{\omega}^{+\downarrow 0}$.

**Remark 14.1.8.** Note that a set $\Delta_{\omega}^{+\downarrow 0}$ is a positive idempotent in $\mathbb{R}_c^\#$ and a set $\Delta_{\omega}^{-\downarrow 0}$ is a negative idempotent in $\mathbb{R}_c^\#$.

**Proposition 14.1.2.** (1) Let $\{a_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0}$ and $\{b_n\}_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow 0}$ then:

(i) $\{a_n\}_{n=0}^\infty + \{b_n\}_{n=0}^\infty \triangleq \{a_n + b_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0}$

(ii) $\{a_n\}_{n=0}^\infty - \{b_n\}_{n=0}^\infty \triangleq \{a_n - b_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0} \cup \Delta_{\approx}^{+\#} \cup \Delta_{\approx}^{-\#} \cup \{0_{\mathbb{R}_c^\#}\}_{n=0}^\infty$
where $\{0_{\mathbb{R}_c^\#}\}_{n=0}^\infty$ is a set of countable $0_{\mathbb{R}_c^\#}$-valued sequence.

(iii) $\{a_n\}_{n=0}^\infty \times \{b_n\}_{n=0}^\infty \triangleq \{a_n \times b_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0}$.

(2) Let $\{a_n\}_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow 0}$ and $\{b_n\}_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow 0}$ then we define

(i) $\{a_n\}_{n=0}^\infty + \{b_n\}_{n=0}^\infty \triangleq \{a_n + b_n\}_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow 0}$

(ii) $\{a_n\}_{n=0}^\infty - \{b_n\}_{n=0}^\infty \triangleq \{a_n - b_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0}$
$$
(i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}
$$

$$
\begin{array}{l}
(\text{ii}) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \cup \Delta_{\omega}^{-\downarrow\infty} \cup \Delta_{\approx}^{+\#} \cup \Delta_{\approx}^{-\#} \cup \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0} \{0_{\mathbb{R}_c^{\#}}\}_{n=0}^{\infty} \\
\text{where } \{0_{\mathbb{R}_c^{\#}}\}_{n=0}^{\infty} \text{ is a countable } 0_{\mathbb{R}_c^{\#}}\text{-valued sequence.}
\end{array}
$$

$$
(iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}.
$$

$$
(2) \text{ Let } \{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty} \text{ and } \{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty} \text{ then we define}
$$

$$
(i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty}
$$

$$
(\text{ii}) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \cup \Delta_{\omega}^{-\downarrow\infty}
$$

$$
(iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}
$$

$$
(3) \text{ Let } \{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \text{ and } x, y \in \mathbb{R}_c^{\#} \text{ then we define}
$$

$$
\begin{array}{l}
(\text{iv}) x_n + y_n \{a_n\}_{n=0}^{\infty} \triangleq \{x_n + y_n a_n\}_{n=0}^{\infty} \quad \text{and we denote a set of the all these sequences} \\
\text{by } \{\Delta_{\omega}^{+\downarrow\infty}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}.
\end{array}
$$

**Proof.** Immediately by definitions and by Definition 14.1.16.

$$
\textbf{Remark 14.1.10.} \text{Note that } \{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \Leftrightarrow \{a_n^{-1}\}_{n=N}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}.
$$

$$
\textbf{Definition 14.1.18.} (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and let $\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ be a hyper infinite sequence
$$

$$
\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots), \quad (14.1.10)
$$

i.e. for any infinite $m \in \mathbb{N}^{\#} \setminus \mathbb{N}$, $A_m \equiv \{a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\overline{\Delta_{\omega}^{+\downarrow 0}}$ and a set of the all hyper infinite sequences $\overline{\{-a_n\}_{n=0}^{\infty}}$ by $\overline{\Delta_{\omega}^{-\downarrow 0}}$. (2) Let $\{x_n + y_n a_n\}_{n=0}^{\infty} \in \{\overline{\Delta_{\omega}^{+\downarrow 0}}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}$ and let

$$
\{x_n + y_n A_n\}_{n=0}^{\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} \qquad (14.1.11)
$$

$$
(x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, \{x_k + y_k a_k\}, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots),
$$

i.e. for any infinite $m \in \mathbb{N}^{\#} \setminus \mathbb{N}$, $A_m \equiv \{x_n + y_n a_n\}_{n=0}^{\infty}$. We will denote a set of the all these
hyper infinite sequences by $\overline{\{\Delta_{\omega}^{+\downarrow 0}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}}$.

**Definition 14.1.19.** Let $\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ and $\{B_n\}_{n=0}^{\infty} = \overline{\{b_n\}_{n=0}^{\infty}}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$.
Then we define:

$$
(i) \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n + b_n\}_{n=0}^{\infty}} = \{A_n + B_n\}_{n=0}^{\infty} \in \overline{\Delta_{\omega}^{+\downarrow 0}}
$$

$$
\begin{align*}
(\text{ii}) \{A_n\}_{n=0}^{\infty} - \{B_n\}_{n=0}^{\infty} &= \overline{\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n - b_n\}_{n=0}^{\infty}} = \\
&= \{A_n - B_n\}_{n=0}^{\infty} \in \overline{\Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0} \cup \{0_{\mathbb{R}_c^{\#}}\}_{n=0}^{\infty}}
\end{align*}
$$

$$
\begin{align*}
(\text{iii}) \{A_n\}_{n=0}^{\infty} \times \{B_n\}_{n=0}^{\infty} &= \overline{\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n \times b_n\}_{n=0}^{\infty}} = \{A_n \times B_n\}_{n=0}^{\infty} \in \overline{\Delta_{\omega}^{+\downarrow 0}} \\
\text{Let } \{A_n\}_{n=0}^{\infty} &= \overline{\{a_n\}_{n=0}^{\infty}} \text{ and } \{B_n\}_{n=0}^{\infty} = \overline{\{b_n\}_{n=0}^{\infty}} \text{ be in } \{\overline{\Delta_{\omega}^{+\downarrow 0}}, \{x_{1,n}\}_{n=0}^{\infty}, \{y_{1n}\}_{n=0}^{\infty}, \} \text{ and} \\
\{B_n\}_{n=0}^{\infty} &= \overline{\{b_n\}_{n=0}^{\infty}} \text{ be in } \{\overline{\Delta_{\omega}^{+\downarrow 0}}, \{x_{2,n}\}_{n=0}^{\infty}, \{y_{2,n}\}_{n=0}^{\infty}, \}. \text{ Then we define:}
\end{align*}
$$

$$
\begin{align*}
(\text{iv}) \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} &= \overline{\{x_{1,n} + y_{1,n} a_n\}_{n=0}^{\infty} + \{x_{2,n} + y_{2,n} b_n\}_{n=0}^{\infty}} \triangleq \\
&\triangleq \overline{\{x_{1,n} + x_{2,n} + y_{1,n} a_n + y_{2,n} b_n\}_{n=0}^{\infty}} = \{x_{1,n} + \dots + x_{2,n} + y_{1,n} A_n + y_{2,n} B_n\}_{n=0}^{\infty}
\end{align*}
$$

**Definition 14.1.20.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$, i.e. for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \overline{\Delta_{\omega}^{+\downarrow 0}}$. Say $\{\Psi_n\}_{n=0}^{\infty}$ \#-tends to $0_{\mathbb{R}_c^{\#}}$ as $n \to \infty^{\#}$ iff for any given $\varepsilon > 0$ $0_{\mathbb{R}_c^{\#}}$, $\varepsilon \approx 0_{\mathbb{R}_c^{\#}}$ there is a hypernatural number $N \in \mathbb{N}^{\#} \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.1.21.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be a hyper infinite sequence such that for all
$$ (iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0} $$

(3) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0}$ and $x, y \in \mathbb{R}_c^{\#}$ then we define

$$ (iv) x + y \{a_n\}_{n=0}^{\infty} \triangleq \{x + ya_n\}_{n=0}^{\infty} $$

**Proof.** Immediately by definitions and by Definition 14.1.14.

**Definition 14.1.15.** We define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{+\downarrow 0} \times \Delta_{\omega}^{+\downarrow 0}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such

that for all $n > N : a_n \le b_n$ and similarly we define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{-\downarrow 0}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow 0}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such

that for all $n > N : a_n \le b_n$

**Definition 14.1.16.** (1) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega}^{+\downarrow 0} \times \mathbb{R}_{c+,fin}^{\#}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $x \in \mathbb{R}_{c+,fin}^{\#}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all

$n > N : a_n < x$.

(2) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega}^{\#+} \times \Delta_{\omega}^{+\downarrow 0}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0}$ and $x \in \Delta_{\omega}^{\#+}$ then $x < \{a_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N : x < a_n$.

(3) Let $\{a_n\}_{n=0}^{\infty}$ be $\Delta_{\omega}^{\#+}$- valued countable sequence $a : \mathbb{N} \to \Delta_{\omega}^{\#+}$, and we denote a set of the all these sequences by $\Delta_{\omega,\omega}^{\#+}$.

We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega,\omega}^{\#+} \times \Delta_{\omega}^{+\downarrow 0}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega,\omega}^{\#+}$ and $x \in \Delta_{\omega}^{\#+}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N : a_n < x$.

**Proposition 14.1.2.** Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow 0} \{a_n\}_{n=0}^{\infty} \neq 0_{\mathbb{R}_c^{\#}}$ then

$$ 0_{\mathbb{R}_c^{\#}} < \Delta_{\omega}^{\#+} < \{a_n\}_{n=0}^{\infty} < \mathbb{R}_{c+,fin}^{\#} \setminus \{0_{\mathbb{R}_c^{\#}}\}. $$

**Proof.** Immediately by definitions and by Definition 14.1.15.

**Remark 14.1.9.** Note that it follows from Proposition 14.1.2 that

$$ 0_{\mathbb{R}_c^{\#}} < \Delta_{\omega}^{\#+} < \Delta_{\omega}^{+\downarrow 0} < \mathbb{R}_{c+,fin}^{\#} \setminus \{0_{\mathbb{R}_c^{\#}}\}. \quad (14.1.9) $$

**Definition 14.1.17.** Let $\{a_n\}_{n=0}^{\infty}$ be monotonically increasing $\mathbb{R}_{c+,fin}^{\#}$- valued countable sequence $a : \mathbb{N} \to \mathbb{R}_{c+,fin}^{\#} \setminus \Delta_{\omega}^{+\downarrow 0}$ such that:

(i) there is $M \in \mathbb{N}$ such that for all $n > M, a_n \neq 0_{\mathbb{R}_c^{\#}}$

(ii) there is $N \in \mathbb{N}$ such that for all $n > N$ and for any $\xi$ such that $\xi > 0_{\mathbb{R}_c^{\#}}, \xi \in \mathbb{R}_{c+,fin}^{\#}$: $a_n > \xi$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\downarrow \infty}$. We define a set $\Delta_{\omega}^{-\downarrow \infty}$ by $c_n \in \Delta_{\omega}^{-\downarrow \infty} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$. Note that $\Delta_{\omega}^{-\downarrow \infty} = -\Delta_{\omega}^{+\downarrow \infty}$.

**Proposition 14.1.3.** (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$ then:

(i) $\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$

(ii) $\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty} \cup \Delta_{\omega}^{-\downarrow \infty} \cup \Delta_{\omega}^{\#+} \cup \Delta_{\omega}^{\#-} \cup \Delta_{\omega}^{+\downarrow 0} \cup \Delta_{\omega}^{-\downarrow 0} \{0_{\mathbb{R}_c^{\#}}\}_{n=0}^{\infty}$ where $\{0_{\mathbb{R}_c^{\#}}\}_{n=0}^{\infty}$ is a countable $0_{\mathbb{R}_c^{\#}}$- valued sequence.

(iii) $\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$.

(2) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow \infty}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow \infty}$ then we define

(i) $\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow \infty}$

(ii) $\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty} \cup \Delta_{\omega}^{-\downarrow \infty}$

(iii) $\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$

(3) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow \infty}$ and $x_n, y_n \in \mathbb{R}_c^{\#}$ then we define

(iv) $x_n + y_n \{a_n\}_{n=0}^{\infty} \triangleq \{x_n + y_n a_n\}_{n=0}^{\infty}$ and we denote a set of the all these sequences by $\{\Delta_{\omega}^{+\downarrow \infty}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}$.

**Proof.** Immediately by definitions and by Definition 14.1.16.
$n \in \mathbb{N}^{\#}, \Psi_n \in \widetilde{\Delta_{\omega}^{+10}}$. We call $\{\Psi_n\}_{n=0}^{\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{\mathbb{R}_c^{\#}}$. To be precise: given any $\varepsilon > 0_{\mathbb{R}_c^{\#}}, \varepsilon \approx 0_{\mathbb{R}_c^{\#}}$ there is a hypernatural number $N \in \mathbb{N}^{\#}\setminus N = N(\varepsilon)$ such that for any $m, n > N, |\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.1.3.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\widetilde{\Delta_{\omega}^{+10}}$. If $\{\Psi_n\}_{n=0}^{\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_{\#} \Phi$ as $n \to \infty^{\#}$ for some $\Phi \in \widetilde{\Delta_{\omega}^{+10}}$), then $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Proof.** We know that $\Psi_n \to_{\#} \Phi$. Here is a ubiquitous trick: instead of using $\varepsilon$ in the definition, start with an arbitrary infinitesimal $\varepsilon > 0, \varepsilon \approx 0_{\mathbb{R}_c^{\#}}$ and then choose $N$ so that $|\Psi_n - \Phi| < \varepsilon/2$ when $n > N$. Then if $m, n > N$, we have
$$|\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \varepsilon/2 + \varepsilon/2 = \varepsilon.$$
This shows that $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Theorem 14.1.4.** If $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence, then it is bounded in $\mathbb{R}_c^{\#}$; that is, there is some number $M \in \mathbb{R}_c^{\#}$ such that $|\{\Psi_n\}_{n=0}^{\infty}| \le M$ for all $n \in \mathbb{N}^{\#}$.

**Proof.** Since $\{\Psi_n\}_{n=0}^{\infty}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N$ such that $|\Psi_m - \Psi_n| < 1$ whenever $m, n > N$. Thus, $|\Psi_{N+1} - \Psi_n| < 1$ for $n > N$. We can rewrite this as $\Psi_{N+1} - 1 < \Psi_n < \Psi_{N+1} + 1$. This means that $|\Psi_n|$ is less than the maximum of $|\Psi_{N+1} - 1|$ and $|\Psi_{N+1} + 1|$. So, set $M \in \mathbb{R}_c^{\#}$ larger than any number in the following list: $\{|\Psi_0|, |\Psi_1|, \dots, |\Psi_N|, |\Psi_{N+1} - 1|, |\Psi_{N+1} + 1|\}$.
Then for any term $\Psi_n$, if $n \le N$, then $|\Psi_n|$ appears in the list and so $|\Psi_n| \le M$; if $n > N$, then (as shown above) $|\Psi_n|$ is less than at least one of the last two entries in the list, and so $|\Psi_n| \le M$. Hence, $M$ is a bound for the sequence.
Let $\Xi$ denote the set of all Cauchy hyper infinite sequences We must define an equivalence relation on $\Xi$.

**Definition 14.1.22.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

**Reflexivity:** for any $s \in S$, $s$ is related to $s$.

**Symmetry:** for any $s, t \in S$, if $s$ is related to $t$ then $t$ is related to $s$.

**Transitivity:** for any $s, t, r \in S$, if $s$ is related to $t$ and $t$ is related to $r$, then $s$ is related to $r$.

The following well known proposition goes most of the way to showing that an
equivalence relation divides a set into bins.

**Theorem 14.1.5.** Let $S$ be a set, with an equivalence relation on pairs of elements.
For $s \in S$, denote by $[s]$ the set of all elements in $S$ that are related to $s$. Then for
any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.
The sets $[s]$ for $s \in S$ are called the equivalence classes, and they are the bins.

**Corollary 14.1.1.** If $S$ is a set with an equivalence relation on pairs of elements, then the equivalence classes are non-empty disjoint sets whose union is all of $S$.

**Definition 14.1.23.** Let $\{\Psi_{1,n}\}_{n=0}^{\infty}$ and $\{\Psi_{2,n}\}_{n=0}^{\infty}$ be in $\widetilde{\Delta_{\omega}^{+10}}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_{\#} 0_{\mathbb{R}_c^{\#}}$ as $n \to \infty^{\#}$, i.e. if the sequence $\{|\Psi_{1,n} - \Psi_{2,n}|\}_{n=0}^{\infty}$ #-tends to $0_{\mathbb{R}_c^{\#}}$.

**Proposition 14.1.4.** Definition 4.1.23 yields an equivalence relation on $\Xi$.
**Remark 14.1.10.** Note that $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \Leftrightarrow \{a_n^{-1}\}_{n=N}^{\infty} \in \Delta_{\omega}^{+\uparrow 0}$.

**Definition 14.1.18.** (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\uparrow 0}$ and let $\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ be a hyper infinite sequence

$$ \{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, \{a_1\}_{n=0}^{n=1}, \dots, \{a_k\}_{n=0}^{n=k}, \dots, \{a_n\}_{n=0}^{\infty}, \dots), \quad (14.1.10) $$

i.e. for any infinite $m \in \mathbb{N}^{\#}\mathbb{N}$, $A_m \equiv \{a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\overline{\Delta_{\omega}^{+\downarrow 0}}$ and a set of the all hyper infinite sequences $\overline{\{-a_n\}_{n=0}^{\infty}}$ by $\overline{\Delta_{\omega}^{-\uparrow 0}}$. (2) Let $\{x_n + y_n a_n\}_{n=0}^{\infty} \in \{\overline{\Delta_{\omega}^{+\downarrow 0}}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}$ and let

$$ \{x_n + y_n A_n\}_{n=0}^{\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} \quad (14.1.11) $$

$$ (x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, \{x_k + y_k a_k\}, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots), $$

i.e. for any infinite $m \in \mathbb{N}^{\#}\mathbb{N}$, $A_m \equiv \{x_n + y_n a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\overline{\{\Delta_{\omega}^{+\downarrow 0}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}}$.

**Definition 14.1.19.** Let $\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ and $\{B_n\}_{n=0}^{\infty} = \overline{\{b_n\}_{n=0}^{\infty}}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$. Then we define:

$$ (i) \quad \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n + b_n\}_{n=0}^{\infty}} = \{A_n + B_n\}_{n=0}^{\infty} \in \overline{\Delta_{\omega}^{+\downarrow 0}} $$

$$ (ii) \quad \{A_n\}_{n=0}^{\infty} - \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n - b_n\}_{n=0}^{\infty}} = \\ = \{A_n - B_n\}_{n=0}^{\infty} \in \overline{\Delta_{\omega}^{+\downarrow 0}} \cup \overline{\Delta_{\omega}^{-\uparrow 0}} \cup \{0_{\mathbb{R}^c}\}_{n=0}^{\infty} $$

$$ (iii) \quad \{A_n\}_{n=0}^{\infty} \times \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n \times b_n\}_{n=0}^{\infty}} = \{A_n \times B_n\}_{n=0}^{\infty} \in \overline{\Delta_{\omega}^{+\downarrow 0}} \\ \text{Let } \{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}} \text{ and } \{B_n\}_{n=0}^{\infty} = \overline{\{b_n\}_{n=0}^{\infty}} \text{ be in } \{\overline{\Delta_{\omega}^{+\downarrow 0}}, \{x_{1,n}\}_{n=0}^{\infty}, \{y_{1n}\}_{n=0}^{\infty}\} \text{ and } \{B_n\}_{n=0}^{\infty} = \overline{\{b_n\}_{n=0}^{\infty}} \text{ be in } \{\overline{\Delta_{\omega}^{+\downarrow 0}}, \{x_{2,n}\}_{n=0}^{\infty}, \{y_{2,n}\}_{n=0}^{\infty}\}. \text{ Then we define:} $$

$$ (iv) \quad \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} = \overline{\{x_{1,n} + y_{1,n}a_n\}_{n=0}^{\infty} + \{x_{2,n} + y_{2,n}b_n\}_{n=0}^{\infty}} \triangleq \\ \triangleq \overline{\{x_{1,n} + x_{2,n} + y_{1,n}a_n + y_{2,n}b_n\}_{n=0}^{\infty}} = \{x_{1,n} + x_{2,n} + y_{1,n}A_n + y_{2,n}B_n\}_{n=0}^{\infty} $$

**Definition 14.1.20.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$, i.e. for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \overline{\Delta_{\omega}^{+\uparrow 0}}$. Say $\{\Psi_n\}_{n=0}^{\infty}$ #-tends to $0_{\mathbb{R}^c}$ as $n \to \infty$ iff for any given $\varepsilon > 0_{\mathbb{R}^c}$, $\varepsilon \approx 0_{\mathbb{R}^c}$ there is a hypernatural number $N \in \mathbb{N}^{\#}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.1.21.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be a hyper infinite sequence such that for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \overline{\Delta_{\omega}^{+\downarrow 0}}$. We call $\{\Psi_n\}_{n=0}^{\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{\mathbb{R}^c}$. To be precise: given any $\varepsilon > 0_{\mathbb{R}^c}$, $\varepsilon \approx 0_{\mathbb{R}^c}$ there is a hypernatural number $N \in \mathbb{N}^{\#}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.1.3.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$. If $\{\Psi_n\}_{n=0}^{\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_{\#} \Phi$ as $n \to \infty$ for some $\Phi \in \overline{\Delta_{\omega}^{+\uparrow 0}}$), then $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Proof.** We know that $\Psi_n \to_{\#} \Phi$. Here is a ubiquitous trick: instead of using $\varepsilon$ in the definition, start with an arbitrary infinitesimal $\varepsilon > 0$, $\varepsilon \approx 0_{\mathbb{R}^c}$ and then choose $N$ so that $|\Psi_n - \Phi| < \varepsilon/2$ when $n > N$. Then if $m, n > N$, we have $|\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.
Proof. we need to show that this relation is reflexive, symmetric, and transitive.

• Reflexive: $\Psi_n - \Psi_n = 0_{\mathbb{R}_c^#}$, and the sequence all of whose terms are $0_{\mathbb{R}_c^#}$ clearly converges to $0_{\mathbb{R}_c^#}$. So $\{\Psi_n\}_{n=0}^{\infty}$ is related to $\{\Psi_n\}_{n=0}^{\infty}$.

• Symmetric: Suppose $\{\Psi_{1,n}\}_{n=0}^{\infty}$ is related to $\{\Psi_{2,n}\}_{n=0}^{\infty}$, so $\Psi_{1,n} - \Psi_{2,n} \to_# 0_{\mathbb{R}_c^#}$.
But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}|$ comes into play in Definition 14.1.23, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_# 0_{\mathbb{R}_c^#}$ as well. Hence, $\{\Psi_{2,n}\}_{n=0}^{\infty}$ is related to $\{\Psi_{1,n}\}_{n=0}^{\infty}$.

• Transitive: Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 14.1.4.
Suppose $\{\Psi_{1,n}\}_{n=0}^{\infty}$ is related to $\{\Psi_{2,n}\}_{n=0}^{\infty}$, and $\{\Psi_{2,n}\}_{n=0}^{\infty}$ is related to $\{\Psi_{3,n}\}_{n=0}^{\infty}$. This means that $\Psi_{1,n} - \Psi_{2,n} \to_# 0_{\mathbb{R}_c^#}$ and $\Psi_{2,n} - \Psi_{3,n} \to_# 0_{\mathbb{R}_c^#}$.

To be fully precise, let us fix infinite small $\varepsilon > 0_{\mathbb{R}_c^#}$; then there exists an $N \in \mathbb{N}^*\mathbb{N}$ such that for all $n > N$, $|\Psi_{1,n} - \Psi_{2,n}| < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$, $|\Psi_{2,n} - \Psi_{3,n}| < \varepsilon/2$. Well, then, as long as $n > \max(N,M)$, we have that $|\Psi_{2,n} - \Psi_{3,n}| = |(\Psi_{1,n} - \Psi_{2,n}) + (\Psi_{2,n} - \Psi_{3,n})| \le |\Psi_{1,n} - \Psi_{2,n}| + |\Psi_{2,n} - \Psi_{3,n}| < \varepsilon2 + \varepsilon2 = \varepsilon$.
So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|\Psi_{1,n} - \Psi_{3,n}| < \varepsilon$. This means that $\Psi_{1,n} - \Psi_{3,n} \to_# 0_{\mathbb{R}_c^#}$ i.e. $\{\Psi_{1,n}\}_{n=0}^{\infty}$ is related to $\{\Psi_{3,n}\}_{n=0}^{\infty}$.

So, we really have equivalence relation, and so by Corollary 14.1.1, the set $\Xi$ is partitioned into disjoint subsets (equivalence classes).

**Definition 14.1.24.** The hyperreal numbers $\tilde{\mathbb{R}}_c^\#$ are the equivalence classes $[\{\Psi_{1,n}\}_{n=0}^{\infty}]$ of Cauchy hyper infinite sequences of, as per Definition 14.1.23.
That is, each such equivalence class is a hyperreal number in $\tilde{\mathbb{R}}_c^\#$.

**Definition 14.1.25.** Let $s, t \in \tilde{\mathbb{R}}_c^\#$, so there are Cauchy hyper infinite sequences $\{\Psi_n\}_{n=0}^{\infty}$ and $\{\Phi_n\}_{n=0}^{\infty}$ with $s = [\{\Psi_n\}_{n=0}^{\infty}]$ and $t = [\{\Phi_n\}_{n=0}^{\infty}]$.

(a) Define $s + t$ to be the equivalence class of the hyper infinite sequence $\{\Psi_n + \Phi_n\}_{n=0}^{\infty}$.

(b) Define $s \times t$ to be the equivalence class of the hyper infinite sequence $\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}$.

**Proposition 14.1.5.** The operations $+$, $\cdot$ in Definition 14.1.25 (a), (b) are well-defined.

**Proof.** Suppose that $[\{\Psi_n\}_{n=0}^{\infty}] = [\{\Psi_{1,n}\}_{n=0}^{\infty}]$ and $[\{\Phi_n\}_{n=0}^{\infty}] = [\{\Phi_{1,n}\}_{n=0}^{\infty}]$.
Thus means that $\Psi_n - \Psi_{1,n} \to_# 0_{\mathbb{R}_c^#}$ and $\Phi_n - \Phi_{1,n} \to_# 0_{\mathbb{R}_c^#}$. Then $(\Psi_n + \Phi_n) - (\Psi_{1,n} + \Phi_{1,n}) = (\Psi_n - \Psi_{1,n}) + (\Phi_n - \Phi_{1,n})$. Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to $0_{\mathbb{R}_c^#}$, and so $[(\Psi_n + \Phi_n)] = [(\Psi_{1,n} + \Phi_{1,n})]$.

Multiplication is a little trickier; this is where we will use Theorem 14.1.4. We will also use another ubiquitous technique: adding $0_{\mathbb{R}_c^#}$ in the form of $s - s$. Again, suppose that

$[\{\Psi_n\}_{n=0}^{\infty}] = [\{\Psi_{1,n}\}_{n=0}^{\infty}]$ and $[\{\Phi_n\}_{n=0}^{\infty}] = [\{\Phi_{1,n}\}_{n=0}^{\infty}]$; we wish to show that $[\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}] = [\{\Psi_{1,n} \times \Phi_{1,n}\}_{n=0}^{\infty}]$, or, in other words, that $\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_# 0_{\mathbb{R}_c^#}$. Well, we add and subtract one of the other cross terms, say $\Phi_n \times \Psi_{1,n}$:

$$ \Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} = \Psi_n \times \Phi_n + (\Phi_n \times \Psi_{1,n} - \Phi_n \times \Psi_{1,n}) - \Psi_{1,n} \times \Phi_{1,n} = $$
This shows that ${\Psi_n}^\infty_{n=0}$ is a Cauchy hyper infinite sequence.

**Theorem 14.1.4.** If ${\Psi_n}^\infty_{n=0}$ is a Cauchy hyper infinite sequence, then it is bounded in $\mathbb{R}_c^\#$; that is, there is some number $M \in \mathbb{R}_c^\#$ such that $|\{\Psi_n\}^\infty_{n=0}| \le M$ for all $n \in \mathbb{N}^\#$.

**Proof.** Since $\{\Psi_n\}^\infty_{n=0}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N$ such that $|\Psi_m - \Psi_n| < 1$ whenever $m, n > N$. Thus, $|\Psi_{N+1} - \Psi_n| < 1$ for $n > N$. We can rewrite this as $\Psi_{N+1} - 1 < \Psi_n < \Psi_{N+1} + 1$. This means that $|\Psi_n|$ is less than the maximum of $|\Psi_{N+1} - 1|$ and $|\Psi_{N+1} + 1|$. So, set $M \in \mathbb{R}_c^\#$ larger than any number in the following list: $\{|\Psi_0|, |\Psi_1|, \dots, |\Psi_N|, |\Psi_{N+1} - 1|, |\Psi_{N+1} + 1|\}$.

Then for any term $\Psi_n$, if $n \le N$, then $|\Psi_n|$ appears in the list and so $|\Psi_n| \le M$; if $n > N$, then (as shown above) $|\Psi_n|$ is less than at least one of the last two entries in the list, and so $|\Psi_n| \le M$. Hence, $M$ is a bound for the sequence.

Let $\Xi$ denote the set of all Cauchy hyper infinite sequences We must define an equivalence relation on $\Xi$.

**Definition 14.1.22.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

**Reflexivity:** for any $s \in S$, $s$ is related to $s$.

**Symmetry:** for any $s, t \in S$, if $s$ is related to $t$ then $t$ is related to $s$.

**Transitivity:** for any $s, t, r \in S$, if $s$ is related to $t$ and $t$ is related to $r$, then $s$ is related to $r$.

The following well known proposition goes most of the way to showing that an equivalence relation divides a set into bins.

**Theorem 14.1.5.** Let $S$ be a set, with an equivalence relation on pairs of elements. For $s \in S$, denote by $[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.

The sets $[s]$ for $s \in S$ are called the equivalence classes, and they are the bins.

**Corollary 14.1.1.** If $S$ is a set with an equivalence relation on pairs of elements, then the equivalence classes are non-empty disjoint sets whose union is all of $S$.

**Definition 14.1.23.** Let $\{\Psi_{1,n}\}^\infty_{n=0}$ and $\{\Psi_{2,n}\}^\infty_{n=0}$ be in $\Delta_\omega^{+|0|}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \xrightarrow{\#} 0_{\mathbb{R}_c^\#}$ as $n \to \infty^\#$, i.e. if the sequence $\{|\Psi_{1,n} - \Psi_{2,n}|\}^\infty_{n=0}$ \#-tends to $0_{\mathbb{R}_c^\#}$.

**Proposition 14.1.4.** Definition 4.1.23 yields an equivalence relation on $\Xi$.

Proof. we need to show that this relation is reflexive, symmetric, and transitive.

• **Reflexive:** $\Psi_n - \Psi_n = 0_{\mathbb{R}_c^\#}$, and the sequence all of whose terms are $0_{\mathbb{R}_c^\#}$ clearly converges to $0_{\mathbb{R}_c^\#}$. So $\{\Psi_n\}^\infty_{n=0}$ is related to $\{\Psi_n\}^\infty_{n=0}$.

• **Symmetric:** Suppose $\{\Psi_{1,n}\}^\infty_{n=0}$ is related to $\{\Psi_{2,n}\}^\infty_{n=0}$, so $\Psi_{1,n} - \Psi_{2,n} \xrightarrow{\#} 0_{\mathbb{R}_c^\#}$.

But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.1.23, it follows that $\Psi_{2,n} - \Psi_{1,n} \xrightarrow{\#} 0_{\mathbb{R}_c^\#}$ as well. Hence, $\{\Psi_{2,n}\}^\infty_{n=0}$ is related to $\{\Psi_{1,n}\}^\infty_{n=0}$.

• **Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 14.1.4. Suppose $\{\Psi_{1,n}\}^\infty_{n=0}$ is related to $\{\Psi_{2,n}\}^\infty_{n=0}$, and $\{\Psi_{2,n}\}^\infty_{n=0}$ is related to $\{\Psi_{3,n}\}^\infty_{n=0}$. This means that $\Psi_{1,n} - \Psi_{2,n} \xrightarrow{\#} 0_{\mathbb{R}_c^\#}$ and $\Psi_{2,n} - \Psi_{3,n} \xrightarrow{\#} 0_{\mathbb{R}_c^\#}$.

To be fully precise, let us fix infinite small $\varepsilon > 0_{\mathbb{R}_c^\#}$; then there exists an $N \in \mathbb{N}^\# \setminus \mathbb{N}$ such that for all $n > N$, $|\Psi_{1,n} - \Psi_{2,n}| < \varepsilon/2$; also, there exists an $M$ such that for all
$$ = (\Psi_n \times \Phi_n - \Phi_n \times \Psi_{1,n}) + (\Phi_n \times \Psi_{1,n} - \Psi_{1,n} \times \Phi_{1,n}) = \\ = \Phi_n \times (\Psi_n - \Psi_{1,n}) + \Psi_{1,n} \times (\Phi_n - \Phi_{1,n}). $$

Hence, we have $|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}|$. Now, from Theorem 14.1.4, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in \mathbb{N}^\#$. Taking some number $R$ (for example $R = M + L$) which is bigger than both, we have

$$ |\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}| \le R(|\Psi_n - \Psi_{1,n}| + |\Phi_n - \Phi_{1,n}|). $$

Now, noting that both an $-cn$ and $\Phi_n - \Phi_{1,n}$ $\#$-tend to $0_{\mathbb{R}^\#}$ and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2R$, we see that

$$ \Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_{\#} 0_{\mathbb{R}^\#} $$

**Theorem 14.1.6.** Given any hyperreal number

$$ s \in \tilde{\mathbb{R}}^\#, s \neq 0_{\tilde{\mathbb{R}}^\#} = [0_{\tilde{\mathbb{R}}^\#}] = [(0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, \dots)], $$

there is a hyperreal number $t \in \tilde{\mathbb{R}}^\#$ such that

$$ s \times t = 1_{\tilde{\mathbb{R}}^\#} = [1_{\tilde{\mathbb{R}}^\#}] = [(1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, \dots)]. $$

**Proof.** First we must properly understand what the theorem says. The premise is that $s$ is nonzero, which means that $s$ is not in the equivalence class of $0_{\tilde{\mathbb{R}}^\#} = (0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, \dots)$. In other words, $s = [\{\Psi_n\}_{n=0}^\infty]$ where $\Psi_n - 0_{\tilde{\mathbb{R}}^\#}$ does not $\#$-converge to $0_{\tilde{\mathbb{R}}^\#}$ as $n \to \infty^\#$. From this, we are to deduce the existence of a hyperreal number

$$ t = [\{\Phi_n\}_{n=0}^\infty] \text{ such that } s \times t = [\{\Psi_n \times \Phi_n\}_{n=0}^\infty] \text{ is the same equivalence class as } 1_{\tilde{\mathbb{R}}^\#} = [(1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, \dots)]. \text{ Doing so is actually an easy consequence of the fact} $$

that nonzero hyperreal numbers from $\tilde{\mathbb{R}}^\#$ have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{\Psi_n\}_{n=0}^\infty$ does not $\#$-tend to $0_{\tilde{\mathbb{R}}^\#}$), there's no reason any number of the terms in $\{\Psi_n\}_{n=0}^\infty$ can't equal $0_{\tilde{\mathbb{R}}^\#}$. However, it turns out that eventually, $\Psi_n \neq 0_{\tilde{\mathbb{R}}^\#}$.

That is,

**Lemma 14.1.1.** If $\{\Psi_n\}_{n=0}^\infty$ is a Cauchy hyper infinite sequence which does not $\#$-tend to $0_{\tilde{\mathbb{R}}^\#}$, then there is an $N \in \mathbb{N}^\#$ such that, for $n > N$, $\Psi_n \neq 0_{\tilde{\mathbb{R}}^\#}$.

We will now use it to complete the proof of Theorem 14.1.6.

Let $N \in \mathbb{N}^\#$ be such that $\Psi_n \neq 0_{\tilde{\mathbb{R}}^\#}$ for $n > N$. Define hyper infinite sequence $\Phi_n$ of hyperreal numbers from $\tilde{\mathbb{R}}^\#$ as follows:

for $n \le N$, $\Phi_n = 0_{\tilde{\mathbb{R}}^\#}$, and for $n > N$, $\Phi_n = 1/\Psi_n$:

$$ \{\Phi_n\}_{n=0}^\infty = (0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, \dots, 0_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}/\Psi_{N+1}, 1/\Psi_{N+2}, \dots). $$

This makes sense since, for $n > N$, an is a nonzero hyperreal number, so $1_{\tilde{\mathbb{R}}^\#}/\Psi_n$ exists.

Then $\Psi_n \times \Phi_n$ is equal to $\Psi_n \times 0_{\tilde{\mathbb{R}}^\#} = 0_{\tilde{\mathbb{R}}^\#}$ for $n \le N$, and equals

$$ \Psi_n \times \Phi_n = \Psi_n \times 1_{\tilde{\mathbb{R}}^\#} / \Psi_n = 1_{\tilde{\mathbb{R}}^\#} \text{ for } n > N $$

Well, then, if we look at the hyper infinite sequence $1_{\tilde{\mathbb{R}}^\#} = (1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, \dots)$, we have $(1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, \dots) - (\Psi_n \times \Phi_n)$ is the sequence which is $1_{\tilde{\mathbb{R}}^\#} - 0_{\tilde{\mathbb{R}}^\#} = 1_{\tilde{\mathbb{R}}^\#}$.
$n > M, |\Psi_{2,n} - \Psi_{3,n}| < \varepsilon/2$. Well, then, as long as $n > \max(N,M)$, we have that
$|\Psi_{2,n} - \Psi_{3,n}| = |(\Psi_{1,n} - \Psi_{2,n}) + (\Psi_{2,n} - \Psi_{3,n})| \le |\Psi_{1,n} - \Psi_{2,n}| + |\Psi_{2,n} - \Psi_{3,n}| < \varepsilon 2 + \varepsilon 2 = \varepsilon$.
So, choosing $L$ equal to the max of $N,M$, we see that given $\varepsilon > 0$ we can always
choose $L$ so that for $n > L, |\Psi_{1,n} - \Psi_{3,n}| < \varepsilon$. This means that $\Psi_{1,n} - \Psi_{3,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$ i.e.
$\{\Psi_{1,n}\}_{n=0}^{\infty}$ is related to $\{\Psi_{3,n}\}_{n=0}^{\infty}$.

So, we really have equivalence relation, and so by Corollary 14.1.1, the set $\Xi$ is
partitioned into disjoint subsets (equivalence classes).

**Definition 14.1.24.** The hyperreal numbers $\tilde{\mathbb{R}}_{c,\omega}^{\#}$ are the equivalence classes
$[\{\Psi_{1,n}\}_{n=0}^{\infty}]$ of Cauchy hyper infinite sequences of, as per Definition 14.1.23.
That is, each such equivalence class is a hyperreal number in $\tilde{\mathbb{R}}_{c}^{\#}$.

**Definition 14.1.25.** Let $s, t \in \tilde{\mathbb{R}}_{c}^{\#}$, so there are Cauchy hyper infinite sequences $\{\Psi_n\}_{n=0}^{\infty}$ and $\{\Phi_n\}_{n=0}^{\infty}$ with $s = [\{\Psi_n\}_{n=0}^{\infty}]$ and $t = [\{\Phi_n\}_{n=0}^{\infty}]$.

(a) Define $s + t$ to be the equivalence class of the hyper infinite sequence
$\{\Psi_n + \Phi_n\}_{n=0}^{\infty}$.

(b) Define $s \times t$ to be the equivalence class of the hyper infinite sequence
$\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}$.

**Proposition 14.1.5.** The operations $+, \times$ in Definition 14.1.25 (a),(b) are well-defined.

**Proof.** Suppose that $[\{\Psi_n\}_{n=0}^{\infty}] = [\{\Psi_{1,n}\}_{n=0}^{\infty}]$ and $[\{\Phi_n\}_{n=0}^{\infty}] = [\{\Phi_{1,n}\}_{n=0}^{\infty}]$.
Thus means that $\Psi_n - \Psi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$ and $\Phi_n - \Phi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$. Then
$(\Psi_n + \Phi_n) - (\Psi_{1,n} + \Phi_{1,n}) = (\Psi_n - \Psi_{1,n}) + (\Phi_n - \Phi_{1,n})$. Now, using the familiar $\varepsilon/2$
trick, you can construct a proof that this tends to $0_{\mathbb{R}_c^{\#}}$, and so
$$[(\Psi_n + \Phi_n)] = [(\Psi_{1,n} + \Phi_{1,n})].$$

Multiplication is a little trickier; this is where we will use Theorem 14.1.4. We will
also use another ubiquitous technique: adding $0_{\mathbb{R}_c^{\#}}$ in the form of $s - s$. Again,
suppose that

$[\{\Psi_n\}_{n=0}^{\infty}] = [\{\Psi_{1,n}\}_{n=0}^{\infty}]$ and $[\{\Phi_n\}_{n=0}^{\infty}] = [\{\Phi_{1,n}\}_{n=0}^{\infty}]$; we wish to show that
$[\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}] = [\{\Psi_{1,n} \times \Phi_{1,n}\}_{n=0}^{\infty}]$, or, in other words, that
$\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$. Well, we add and subtract one of the other cross
terms, say $\Phi_n \times \Psi_{1,n}$:

$$
\begin{align*}
\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} &= \Psi_n \times \Phi_n + (\Phi_n \times \Psi_{1,n} - \Phi_n \times \Psi_{1,n}) - \Psi_{1,n} \times \Phi_{1,n} \\
&= (\Psi_n \times \Phi_n - \Phi_n \times \Psi_{1,n}) + (\Phi_n \times \Psi_{1,n} - \Psi_{1,n} \times \Phi_{1,n}) \\
&= \Phi_n \times (\Psi_n - \Psi_{1,n}) + \Psi_{1,n} \times (\Phi_n - \Phi_{1,n}).
\end{align*}
$$

Hence, we have $|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \cdot |\Phi_n - \Phi_{1,n}|$.

Now, from Theorem 14.1.4, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and
$|\Psi_{1,n}| \le L$ for all $n \in \mathbb{N}^\#$. Taking some number $R$ (for example $R = M + L$) which is
bigger than both, we have

$$
|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}| \le \\
\le R(|\Psi_n - \Psi_{1,n}| + |\Phi_n - \Phi_{1,n}|).
$$

Now, noting that both $\Psi_n - \Psi_{1,n}$ and $\Phi_n - \Phi_{1,n}$ \#-tend to $0_{\mathbb{R}_c^{\#}}$ and using the $\varepsilon/2$ trick
(actually, this time we'll want to use $\varepsilon/2R$, we see that

$$
\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}
$$

**Theorem 14.1.6.** Given any hyperreal number

$$
s \in \tilde{\mathbb{R}}_{c}^{\#}, s \neq 0_{\tilde{\mathbb{R}}_{c}^{\#}} = [\tilde{0}_{\tilde{\mathbb{R}}_{c}^{\#}}] = [(0_{\mathbb{R}_{c}^{\#}}, 0_{\mathbb{R}_{c}^{\#}}, 0_{\mathbb{R}_{c}^{\#}}, 0_{\mathbb{R}_{c}^{\#}}, \dots)],
$$
for $n \le N$ and equals $1_{\mathbb{R}_c^{\#}} - 1_{\mathbb{R}_c^{\prime}} = 0_{\mathbb{R}_c^{\#}}$ for $n > N$. Since this hyper infinite sequence is eventually equal to $0_{\mathbb{R}_c^{\#}}$, it #-converges to $0_{\mathbb{R}_c^{\#}}$ as $n \to \infty^{\#}$, and so

$$ [ \{ \Psi_n \times \Phi_n \}_{n=0}^{\infty} ] = [(1_{\mathbb{R}_c^{\#}}, 1_{\mathbb{R}_c^{\#}}, 1_{\mathbb{R}_c^{\#}}, 1_{\mathbb{R}_c^{\#}, \dots})] = 1_{\mathbb{R}_c^{\#}} \in \tilde{\mathbb{R}}_c^{\#}. \text{This shows that } t = [ \{ \Phi_n \}_{n=0}^{\infty} ] \text{ is a multiplicative inverse to } s = [ \{ \Psi_n \}_{n=0}^{\infty} ]. $$

**Definition 14.1.26.** Let $s \in \tilde{\mathbb{R}}_c^{\#}$. Say that $s$ is positive if $s \neq 0_{\tilde{\mathbb{R}}_c^{\#}}$, and if $s = [ \{ \Psi_n \}_{n=0}^{\infty} ]$ for some Cauchy hyper infinite sequence such that for some $N$, $\Psi_n > 0_{\mathbb{R}_c^{\#}}$ for all $n > N$. Given two hyperreal numbers $s, t$, say that $s > t$ if $s - t$ is positive.

**Theorem 14.1.7.** Let $s, t \in \tilde{\mathbb{R}}_c^{\#}$ be hyperreal numbers such that $s > t$, and let $r \in \tilde{\mathbb{R}}_c^{\#}$. Then $s + r > t + r$.

**Proof.** Let $s = [ \{ \Psi_n \}_{n=0}^{\infty} ]$, $t = [ \{ \Phi_n \}_{n=0}^{\infty} ]$, and $r = [ \{ \Theta_n \}_{n=0}^{\infty} ]$. Since $s > t$, i.e. $s - t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have

$\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\mathbb{R}_c^{\#}}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\mathbb{R}_c^{\#}}$ as $n \to \infty^{\#}$, by the assumption that $s - t > 0_{\tilde{\mathbb{R}}_c^{\#}}$. Thus, by Definition 14.1.26, this means that:

$$ s + r = [ \{ \Psi_n + \Theta_n \}_{n=0}^{\infty} ] > [ \{ \Phi_n + \Theta_n \}_{n=0}^{\infty} ] = t + r. $$

**Definition 14.1.27.** There is canonical imbeding

$$ \mathbb{R}_c^{\#} \hookrightarrow \tilde{\mathbb{R}}_c^{\#} \qquad (14.1.14) $$

defined by

$$ a \mapsto [\tilde{a}] \qquad (14.1.15) $$

where $\tilde{a}$ is hyper infinite sequence $\tilde{a} = (a, a, \dots), a \in \mathbb{R}_c^{\#}$.

**Notation 14.1.5.** $\hat{a} = (a, a, \dots) \in \tilde{\mathbb{R}}_c^{\#}, a \in \tilde{\mathbb{R}}_c^{\#}$.

**Remark 14.1.11.** If $a \in \tilde{\mathbb{R}}_c^{\#}$ we will identify hyperreal $a$ with hyper infinite sequence $\{a_n\}_{n=0}^{\infty} = a_0, a_1, \dots, a_{N-1}, \hat{a}_N, N \in \mathbb{N}^{\#}$ since $a = \#\text{lim}_{n \to \infty^{\#}} a_n$.

**Definition 14.1.28.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\tilde{\mathbb{R}}_c^{\#}$, $\{a_n\}_{n=0}^k \subset \tilde{\mathbb{R}}_c^{\#}$. We define external hyper infinite sequence $\{\overbrace{a_n}^k\}_{n=0} \subset \tilde{\mathbb{R}}_c^{\#}$ by

$$ \begin{aligned} \{A_n; k\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^k}^{\{a_n\}_{n=0}^{\infty}} = \\ &= (a_0, a_1, \dots, a_m, \dots, \overbrace{a_{k-1}}^{k}, \overbrace{a_k}^{k}, \hat{a}_k). \end{aligned} \qquad (14.1.16) $$

(ii) Let $\{a_n\}_{n=0}^{\infty}$ be countable sequence in $\tilde{\mathbb{R}}_c^{\#} : \{a_n\}_{n=0}^{\infty} \subset \tilde{\mathbb{R}}_c^{\#}$.

We define hyper infinite sequence $\{A_n\}_{n=0}^{\infty} = \overbrace{\{a_n\}_{n=0}^{\infty}}^{\{a_n\}_{n=0}^{\infty}} \subset \tilde{\mathbb{R}}_c^{\#}$ by

$$ \begin{aligned} \{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}}^{\{a_n\}_{n=0}^{\infty}} = \\ &= (a_0, a_1, \dots, a_k, \dots, \overbrace{a_n}^{\{a_n\}_{n=0}^{\infty}}, \overbrace{\{a_n\}_{n=0}^{\infty}}^{\{a_n\}_{n=0}^{\infty}}) \in [[\{a_n\}_{n=0}^{\infty}]]. \end{aligned} \qquad (14.1.17) $$

(iii) Let $\{a_n\}_{n=0}^N, N \in \mathbb{N}^{\#} \mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^{\#} : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^{\#}$.

We define hyper infinite sequence $\{\overbrace{a_n}^N\}_{n=0} \subset \tilde{\mathbb{R}}_c^{\#}$ by
there is a hyperreal number $t \in \tilde{\mathbb{R}}_c^\#$ such that

$$s \times t = 1_{\tilde{\mathbb{R}}_c^\#} = [\tilde{1}_{\tilde{\mathbb{R}}_c^\#}] = [(1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#}, 1_{\mathbb{R}^\#},...)].$$

**Proof.** First we must properly understand what the theorem says. The premise is that $s$ is nonzero, which means that $s$ is not in the equivalence class of $0_{\tilde{\mathbb{R}}_c^\#} = (0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#}, 0_{\mathbb{R}^\#},...)$.

In other words, $s = [\{\Psi_n\}_{n=0}^\infty]$ where $\Psi_n - 0_{\tilde{\mathbb{R}}_c^\#}$ does not #-converge to $0_{\tilde{\mathbb{R}}_c^\#}$ as $n \to \infty^\#$. From this, we are to deduce the existence of a hyperreal number

$t = [\{\Phi_n\}_{n=0}^\infty]$ such that $s \times t = [\{\Psi_n \times \Phi_n\}_{n=0}^\infty]$ is the same equivalence class as $1_{\tilde{\mathbb{R}}_c^\#} = [(1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#},...)]$. Doing so is actually an easy consequence of the fact that nonzero hyperreal numbers from $\mathbb{R}_c^\#$ have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{\Psi_n\}_{n=0}^\infty$ does not #-tend to $0_{\tilde{\mathbb{R}}_c^\#}$), there's no reason any number of the terms in $\{\Psi_n\}_{n=0}^\infty$ can't equal $0_{\tilde{\mathbb{R}}_c^\#}$. However, it turns out that eventually, $\Psi_n \neq 0_{\tilde{\mathbb{R}}_c^\#}$.

That is,

**Lemma 14.1.1.** If $\{\Psi_n\}_{n=0}^\infty$ is a Cauchy hyper infinite sequence which does not #-tend to $0_{\tilde{\mathbb{R}}_c^\#}$, then there is an $N \in \mathbb{N}^\#$ such that, for $n > N$, $\Psi_n \neq 0_{\tilde{\mathbb{R}}_c^\#}$.

We will now use it to complete the proof of Theorem 14.1.6.

Let $N \in \mathbb{N}^\#$ be such that $\Psi_n \neq 0_{\tilde{\mathbb{R}}_c^\#}$ for $n > N$. Define hyper infinite sequence $\Phi_n$ of hyperreal numbers from $\tilde{\mathbb{R}}_c^\#$ as follows:

for $n \le N$, $\Phi_n = 0_{\tilde{\mathbb{R}}_c^\#}$, and for $n > N$, $\Phi_n = 1/\Psi_n$ :

$$\{\Phi_n\}_{n=0}^\infty = (0_{\tilde{\mathbb{R}}_c^\#}, 0_{\tilde{\mathbb{R}}_c^\#},...,0_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}/\Psi_{N+1}, 1/\Psi_{N+2},...).$$

This makes sense since, for $n > N$, an is a nonzero hyperreal number, so $1_{\tilde{\mathbb{R}}_c^\#}/\Psi_n$ exists.

Then $\Psi_n \times \Phi_n$ is equal to $\Psi_n \times 0_{\tilde{\mathbb{R}}_c^\#} = 0_{\tilde{\mathbb{R}}_c^\#}$ for $n \le N$, and equals

$$\Psi_n \times \Phi_n = \Psi_n \times 1_{\tilde{\mathbb{R}}_c^\#}/\Psi_n = 1_{\tilde{\mathbb{R}}_c^\#} \text{ for } n > N$$

Well, then, if we look at the hyper infinite sequence $1_{\tilde{\mathbb{R}}_c^\#} = (1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#},...)$, we have $(1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#},...)-(\Psi_n \times \Phi_n)$ is the sequence which is $1_{\tilde{\mathbb{R}}_c^\#} - 0_{\tilde{\mathbb{R}}_c^\#} = 1_{\tilde{\mathbb{R}}_c^\#}$ for $n \le N$ and equals $1_{\tilde{\mathbb{R}}_c^\#}-1_{\tilde{\mathbb{R}}_c^\#} = 0_{\tilde{\mathbb{R}}_c^\#}$ for $n > N$. Since this hyper infinite sequence is eventually equal to $0_{\tilde{\mathbb{R}}_c^\#}$, it #-converges to $0_{\tilde{\mathbb{R}}_c^\#}$ as $n \to \infty^\#$, and so

$$[\{\Psi_n \times \Phi_n\}_{n=0}^\infty] = [(1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#}, 1_{\tilde{\mathbb{R}}_c^\#},...)] = 1_{\tilde{\mathbb{R}}_c^\#} \in \tilde{\mathbb{R}}_c^\#. \text{ This shows that } t = [\{\Phi_n\}_{n=0}^\infty] \\ \text{is a multiplicative inverse to } s = [\{\Psi_n\}_{n=0}^\infty].$$

**Definition 14.1.26.** Let $s \in \tilde{\mathbb{R}}_c^\#$. Say that $s$ is positive if $s \neq 0_{\tilde{\mathbb{R}}_c^\#}$, and if $s = [\{\Psi_n\}_{n=0}^\infty]$ for some Cauchy hyper infinite sequence such that for some $N$, $\Psi_n > 0_{\tilde{\mathbb{R}}_c^\#}$ for all $n > N$. Given two hyperreal numbers $s,t$, say that $s > t$ if $s - t$ is positive.

**Theorem 14.1.7.** Let $s,t \in \tilde{\mathbb{R}}_c^\#$ be hyperreal numbers such that $s > t$, and let $r \in \tilde{\mathbb{R}}_c^\#$. Then $s+r > t+r$.

**Proof.** Let $s = [\{\Psi_n\}_{n=0}^\infty]$, $t = [\{\Phi_n\}_{n=0}^\infty]$, and $r = [\{\Theta_n\}_{n=0}^\infty]$. Since $s > t$, i.e. $s-t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have $\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\tilde{\mathbb{R}}_c^\#}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\tilde{\mathbb{R}}_c^\#}$ as $n \to \infty^\#$, by the
$$ \begin{aligned} \{A_n; N\}^{*\infty}_{n=0} &= \overbrace{\{a_n\}^N_{n=0}}^{\widetilde{a_n}} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{N-1}, a_N, \hat{a}_N). \end{aligned} \quad (14.1.18) $$

**Definition 14.1.29.** (i) Let $\{a_n\}^k_{n=0}, k \in \mathbb{N}$ be finite sequence in $\widetilde{\mathbb{R}}^\#_c, \{a_n\}^k_{n=0} \subset \widetilde{\mathbb{R}}^\#_c$.
We define external finite sum $\mathrm{Ext}-\sum_{n=0}^{n=k} a_n$ by

$$ \mathrm{Ext}-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}^k_{n=0}}^{\widetilde{c_n}} = (c_0, c_1, \dots, \{c_j\}^{n=j}_{n=0}, \dots, c_{k-1}, \dots, c_k, \hat{c}_k) \in [[c_k]] \quad (14.1.19) $$

where $c_0 = a_0, c_j = \mathrm{Ext}-\sum_{n=0}^{n=j} a_n, 0 \le n \le k$.

(ii) Let $\{a_n\}^\infty_{n=0}$ be countable sequence in $\widetilde{\mathbb{R}}^\#_c : \{a_n\}^\infty_{n=0} \subset \widetilde{\mathbb{R}}^\#_c$. We define external countable sum $\mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n$ by

$$ \begin{aligned} \mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}^\infty_{n=0}}^{\widetilde{c_n}} = \\ &= (c_0, c_1, \dots, c_k, \dots, \{c_n\}^\infty_{n=0}, \overbrace{\{c_n\}^\infty_{n=0}}^{\widetilde{c_n}}) \in [[\{c_n\}^\infty_{n=0}]] \end{aligned} \quad (14.1.20) $$

where $c_0 = a_0, c_k = \mathrm{Ext}-\sum_{n=0}^{n=k} a_n, k \in \mathbb{N}$.

(iii) Let $\{a_n\}^{n=N}_{n=0}, N \in \mathbb{N}^\# \setminus \mathbb{N}$ be external hyperfinite sequence in $\widetilde{\mathbb{R}}^\#_c : \{a_n\}^N_{n=0} \subset \widetilde{\mathbb{R}}^\#_c$.
We define external hyperfinite sum $\mathrm{Ext}-\sum_{n=0}^{n=N} a_n$ by

$$ \mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}^{n=N}_{n=0}}^{\widetilde{c_n}} = (c_0, c_1, \dots, c_k, \dots, c_{N-1}, c_N, \hat{c}_N) \quad (14.1.21) $$

where $c_0 = a_0, c_k = \mathrm{Ext}-\sum_{n=0}^{n=k} a_n, 0 \le k \le N, c_N = \mathrm{Ext}-\sum_{n=0}^{n=N} a_n$.

(iv) Let $\{a_n\}^{n=N}_{n=0}, N \in \mathbb{N}^\# \setminus \mathbb{N}$ be external hyperfinite sequence in $\widetilde{\mathbb{R}}^\#_c : \{a_n\}^N_{n=0} \subset \widetilde{\mathbb{R}}^\#_c$ such that $a_n \equiv 0$ for all $n \in \mathbb{N}^\# \setminus \mathbb{N}$. We assume that

$$ \mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n. \quad (14.1.22) $$

**Example 14.1.1.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#, \alpha \in \widetilde{\mathbb{R}}^\#_c$,

$r \in \widetilde{\mathbb{R}}^\#_c$ be the first term and the ratio of the G.P respectively. Then for any $N \in \mathbb{N}^\#$ by Proposition 14.1.6 and Definition 14.1.29 one obtains that

$$ \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{\overbrace{1-\widetilde{r}^N}^{\widetilde{\mathbb{R}}^\#_c} - r^N}{1-\widetilde{r}^N} = \alpha \frac{\overbrace{1-\widetilde{r}^N}^{\widetilde{\mathbb{R}}^\#_c}}{1-\widetilde{r}^N} - \alpha \frac{\overbrace{r^N}^{\widetilde{r}^N}}{1-\widetilde{r}^N}. \quad (14.1.23) $$

and

$$ \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{\overbrace{1-\widetilde{r}^N}^{\widetilde{\mathbb{R}}^\#_c}}{1-\widetilde{r}^N} - \alpha \left\{ \frac{\overbrace{r^n}^{\widetilde{r}^n}}{1-\widetilde{r}^N} \right\}_{n=1}^{\infty}. \quad (14.1.24) $$

**Example 14.1.2.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#, \alpha \in \widetilde{\mathbb{R}}^\#_c, r \in \widetilde{\mathbb{R}}^\#_c$,
$r > 0, r \neq 1$. Note that
assumption that $s-t > 0_{\mathbb{R}_c^\#}$. Thus, by Definition 14.1.26, this means that:

$$s+r = [\{\Psi_n + \Theta_n\}_{n=0}^\infty] > [\{\Phi_n + \Theta_n\}_{n=0}^\infty] = t+r.$$

**Definition 14.1.27.** There is canonical imbeding

$$\mathbb{R}_c^\# \hookrightarrow \tilde{\mathbb{R}}_c^\# \qquad (14.1.14)$$

defined by

$$a \mapsto [\tilde{a}] \qquad (14.1.15)$$

where $\tilde{a}$ is hyper infinite sequence $\tilde{a} = (a, a, \dots), a \in \tilde{\mathbb{R}}_c^\#$.

**Notation 14.1.5.** $\hat{a} = (a, a, \dots) \in \tilde{\mathbb{R}}_c^\#$, $a \in \tilde{\mathbb{R}}_c^\#$.

**Remark 14.1.11.** If $a \in \tilde{\mathbb{R}}_c^\#$ we will identify hyperreal $a$ with hyper infinite sequence $\{a_n\}_{n=0}^\infty = a_0, a_1, \dots, a_{N-1}, \hat{a}_N, N \in \mathbb{N}^\#$.

**Definition 14.1.28.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\tilde{\mathbb{R}}_c^\#$, $\{a_n\}_{n=0}^k \subset \tilde{\mathbb{R}}_c^\#$.

We define external hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^k} \subset \tilde{\mathbb{R}}_c^\#$ by

$$\begin{aligned} \{A_n; k\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^k} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, a_k, \hat{a}_k). \end{aligned} \qquad (14.1.16)$$

(ii) Let $\{a_n\}_{n=0}^\infty$ be countable sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^\infty \subset \tilde{\mathbb{R}}_c^\#$.

We define hyper infinite sequence $\{A_n\}_{n=0}^\infty = \overbrace{\{a_n\}_{n=0}^\infty} \subset \tilde{\mathbb{R}}_c^\#$ by

$$\begin{aligned} \{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}} = \\ &= (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}_{n=0}^{\infty}}) \in [[\{a_n\}_{n=0}^{\infty}]]. \end{aligned} \qquad (14.1.17)$$

(iii) Let $\{a_n\}_{n=0}^N, N \in \mathbb{N}^\#\mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$.

We define hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^N} \subset \tilde{\mathbb{R}}_c^\#$ by

$$\begin{aligned} \{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^N} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{N-1}, a_N, \hat{a}_N). \end{aligned} \qquad (14.1.18)$$

**Definition 14.1.29.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\tilde{\mathbb{R}}_c^\#$, $\{a_n\}_{n=0}^k \subset \tilde{\mathbb{R}}_c^\#$.

We define external finite sum $\text{Ext}-\sum_{n=0}^{n=k} a_n$ by

$$\text{Ext}-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k} = (c_0, c_1, \dots, \{c_j\}_{n=0}^{n=j}, \dots, c_{k-1}, \dots, c_k, \hat{c}_k) \in [[c_k]] \qquad (14.1.19)$$

where $c_0 = a_0, c_j = \text{Ext}-\sum_{n=0}^{n=j} a_n, 0 \le n \le k$.

(ii) Let $\{a_n\}_{n=0}^\infty$ be countable sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^\infty \subset \tilde{\mathbb{R}}_c^\#$.
$$
\begin{align}
\alpha \frac{\widehat{1_{\mathbb{R}_c^#} - r^N}}{1_{\mathbb{R}_c^#} - r} &= \mathrm{Ext} - \sum_{n=1}^{n=N-1} \alpha r^{n-1} = \nonumber \\
&= \mathrm{Ext} - \sum_{n=1}^{\infty} \alpha r^{n-1} + \mathrm{Ext} - \sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \tag{14.1.25} \\
&= \alpha \frac{\widehat{1_{\mathbb{R}_c^#}}}{1_{\mathbb{R}_c^#} - r} - \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext} - \sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \nonumber
\end{align}
$$

From (14.1.25) we obtain

$$
\begin{align}
\mathrm{Ext} - \sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{\widehat{1_{\mathbb{R}_c^#} - r^N}}{1_{\mathbb{R}_c^#} - r} - \alpha \frac{\widehat{1_{\mathbb{R}_c^#}}}{1_{\mathbb{R}_c^#} - r} + \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^#} - r} \right\}_{n=1}^{\infty} = \tag{14.1.26} \\
&\quad \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^#} - r} \right\}_{n=1}^{\infty} - \alpha \frac{r^N}{1_{\mathbb{R}_c^#} - r}. \nonumber
\end{align}
$$

Assume that: (i) $r < 1_{\mathbb{R}_c^#}$, then from (14.1.26) we obtain

$$
\mathrm{Ext} - \sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} > 0_{\mathbb{R}_c^{\#}}. \quad (14.1.27)
$$

(ii) $r > 1_{\mathbb{R}_c^\#}$, then from (14.1.26) we obtain

$$
\mathrm{Ext} - \sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty} + \alpha \frac{r^N}{r - 1_{\mathbb{R}_c^{\#}}} > 0_{\mathbb{R}_c^{\#}}. \quad (14.1.28)
$$

**Proposition 14.1.6.** (i) Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#$. Let $S_N$, $\alpha \in \mathbb{R}_c^\#$, $r \in \mathbb{R}_c^\#$ be the sum of $N$ terms, first term and the ratio of the G.P respectively. Then for any $N \in \mathbb{N}^\#$ the statement $\Phi_N$ holds

$$
\Phi_N \iff_s \mathrm{Ext} - \sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r}. \quad (14.1.29)
$$

**Proof.** (i) Directly by hyperinfinite induction. Note that $\Phi_N \Rightarrow_s \Phi_{N+1}$:

$$
\begin{align}
S_{N+1} &= \mathrm{Ext} - \sum_{n=1}^{n=N} \alpha r^{n-1} = \mathrm{Ext} - \sum_{n=1}^{n=N-1} \alpha r^{n-1} + \alpha r^N = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha r^N = \nonumber \\
&= \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha \frac{(1_{\mathbb{R}_c^\#} - r) r^N}{1_{\mathbb{R}_c^\#} - r} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N + r^N - r^{N+1}}{1_{\mathbb{R}_c^\#} - r} = \tag{14.1.30} \\
&= \alpha \frac{1_{\mathbb{R}_c^\#} - r^{N+1}}{1_{\mathbb{R}_c^\#} - r}. \nonumber
\end{align}
$$

Thus $S_{N+1} = \alpha \frac{1_{\mathbb{R}_c^\#} + r^{N+1}}{1_{\mathbb{R}_c^\#} - r}$ and therefore $\Phi_{N+1}$ holds.

(ii) Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#$. Let $S_N$, $\alpha \in \mathbb{R}_c^\#$, $r \in \mathbb{R}_c^\#$ be the sum of $N$ terms, first term and the ratio of the G.P respectively. Then for any $N \in *\mathbb{N}$ the statement $\tilde{\Phi}_N$ holds
$$ \operatorname{Ext}-\sum_{n=0}^{n=\infty} a_n = \overbrace{\{c_n\}_{n=0}^{\infty}}^{\overline{\{c_n\}_{n=0}^{\infty}}} = \\ = (c_0, c_1, \dots, c_k, \dots, \overbrace{\{c_n\}_{n=0}^{\infty}, \overline{\{c_n\}_{n=0}^{\infty}}}^{\overline{\{c_n\}_{n=0}^{\infty}}}) \in [[\{c_n\}_{n=0}^{\infty}]] \qquad (14.1.20) $$

where $c_0 = a_0, c_k = \operatorname{Ext}-\sum_{n=0}^{n=k} a_n, k \in \mathbb{N}$.

(iii) Let $\{a_n\}_{n=0}^{n=N}, N \in \mathbb{N}^\#\mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$.

We define external hyperfinite sum $\operatorname{Ext}-\sum_{n=0}^{n=N} a_n$ by

$$ \operatorname{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{n=N}}^{\overline{\{c_n\}_{n=0}^{n=N}}} = (c_0, c_1, \dots, c_k, \dots, c_{N-1}, c_N, \hat{c}_N) \qquad (14.1.21) $$

where $c_0 = a_0, c_k = \operatorname{Ext}-\sum_{n=0}^{n=k} a_n, 0 \le k \le N, c_N = \operatorname{Ext}-\sum_{n=0}^{n=N} a_n$.

(iv) Let $\{a_n\}_{n=0}^{n=N}, N \in \mathbb{N}^\#\mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$ such that $a_n \equiv 0$ for all $n \in \mathbb{N}^\#\mathbb{N}$. We assume that

$$ \operatorname{Ext}-\sum_{n=0}^{n=N} a_n = \operatorname{Ext}-\sum_{n=0}^{n=\infty} a_n. \qquad (14.1.22) $$

**Example 14.1.1.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#, \alpha \in \tilde{\mathbb{R}}_c^\#$,

$r \in \tilde{\mathbb{R}}_c^\#$ be the first term and the ratio of the G.P respectively. Then for any $N \in \mathbb{N}^\#$ by Proposition 14.1.6 and Definition 14.1.29 one obtains that

$$ \operatorname{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{\overline{1 - r^N}}{\overline{1 - r}} = \alpha \frac{\overline{1 - r}}{\overline{1 - r}} - \alpha \frac{\overline{r^N}}{\overline{1 - r}}. \qquad (14.1.23) $$

and

$$ \operatorname{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{\overline{1 - r^N}}{\overline{1 - r}} - \alpha \left\{ \frac{\overline{r^n}}{\overline{1 - r}} \right\}_{n=1}^{\infty}. \qquad (14.1.24) $$

**Example 14.1.2.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#, \alpha \in \tilde{\mathbb{R}}_c^\#, r \in \tilde{\mathbb{R}}_c^\#$,

$r > 0, r \neq 1$. Note that

$$ \begin{align} \alpha \frac{\overline{1 - r^N}}{\overline{1 - r}} &= \operatorname{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \nonumber \\ &= \operatorname{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} + \operatorname{Ext}-\sum_{(n \in \mathbb{N}^\#\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \tag{14.1.25} \\ &= \alpha \frac{\overline{1 - r}}{\overline{1 - r}} - \alpha \left\{ \frac{\overline{r^n}}{\overline{1 - r}} \right\}_{n=1}^{\infty} + \operatorname{Ext}-\sum_{(n \in \mathbb{N}^\#\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \nonumber \end{align} $$

From (14.1.25) we obtain
$$ \tilde{\Phi}_N \Leftrightarrow_s \mathrm{Ext} - \sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#} - r^N}{1_{\tilde{\mathbb{R}}_c^\#} - r}. \qquad (14.1.31) $$

Notice that (i)⇒(ii) by definitions.

**Definition 14.1.30.** Let {$a_n$}$_{n=0}^{\infty}$, $n \in \mathbb{N}^\#$ be external hyperinfinite sequence in $\tilde{\mathbb{R}}_c^\#$:
{$a_n$}$_{n=0}^{\infty} \subset \tilde{\mathbb{R}}_c^\#$. We define external hyperinfinite sum $\mathrm{Ext} - \sum_{n=0}^{\infty} a_n$ by

$$ \mathrm{Ext} - \sum_{n=0}^{\infty} a_n = \#\lim_{N \to \infty} \left( \mathrm{Ext} - \sum_{n=0}^{N} a_n \right) \qquad (14.1.32) $$

if $\#$-limit in (14.1.31) exists.

**Example 14.1.3.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{n-1}, n \in \mathbb{N}^\#$, $\alpha \in \tilde{\mathbb{R}}_c^\#$, $r \in \tilde{\mathbb{R}}_c^\#$. From (14.1.27) we obtain

$$ \begin{aligned} \mathrm{Ext} - \sum_{n=0}^{\infty} \alpha r^{n-1} &= \#\lim_{N \to \infty} \left( \mathrm{Ext} - \sum_{n=0}^{N} \alpha r^{n-1} \right) = \#\lim_{N \to \infty} \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#} - r^N}{1_{\tilde{\mathbb{R}}_c^\#} - r} = \\ &= \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} \end{aligned} \qquad (14.1.33) $$

since $\#\lim_{N \to \infty} r^N = 0_{\tilde{\mathbb{R}}_c^\#}$ if $|r| < 1$. From (14.1.33) and (14.1.25) we obtain

$$ \begin{aligned} \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} &= \mathrm{Ext} - \sum_{n=0}^{\infty} \alpha r^{n-1} = \mathrm{Ext} - \sum_{n=0}^{\infty} \alpha r^{n-1} + \mathrm{Ext} - \sum_{n \in \mathbb{N}^\#} \alpha r^{n-1} = \\ &\quad \alpha \frac{\widehat{1_{\tilde{\mathbb{R}}_c^\#}}}{1_{\tilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{\widehat{r^n}}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext} - \sum_{n \in \mathbb{N}^\# \mathbb{N}} \alpha r^{n-1}. \end{aligned} \qquad (14.1.34) $$

From (14.1.34) we obtain

$$ \begin{aligned} \mathrm{Ext} - \sum_{n \in \mathbb{N}^\# \mathbb{N}} \alpha r^{n-1} &= \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} \alpha - \left( \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^\infty \right) = \\ &= \alpha \left\{ \frac{r^n}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^\infty > 0_{\tilde{\mathbb{R}}_c^\#}. \end{aligned} \qquad (14.1.35) $$

**Definition 14.1.31.** Let {$a_n$}$_{n=0}^{\infty}$ be $\mathbb{R}_c^\#$-valued countable sequence

$a: \mathbb{N} \to \mathbb{R}_c^\#$ such that:

(i) there is $M \in \mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{\mathbb{R}_c^\#}$, we denote a set of the all these sequences by $\Xi_{\tilde{\omega}}^{\pm, 0}$.
We define a set $-\Xi_{\tilde{\omega}}^{\pm, 0}$ by $\{c_n\}_{n=0}^{\infty} \in -\Xi_{\tilde{\omega}}^{\pm, 0} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Xi_{\tilde{\omega}}^{\pm, 0}$. Note that $\Xi_{\tilde{\omega}}^{\pm, 0} = -\Xi_{\tilde{\omega}}^{\pm, 0}$.

(ii) there is countable subsequence {$a_{n_k}$}$_{k=m}^{\infty} \subset \{a_n\}_{n=0}^{\infty}$ such that $a_{n_k} = 0_{\mathbb{R}_c^\#}, k \ge m$ and $a_n \ne 0_{\mathbb{R}_c^\#}$ iff $a_n \notin \{a_{n_k}\}_{k=m}^{\infty}$,

we denote a set of the all these countable sequences by $\Xi_{\tilde{\omega}}^{\pm, 0 \lor 0}$.

We define a set $-\Xi_{\tilde{\omega}}^{\pm, 0 \lor 0} = -\Xi_{\tilde{\omega}}^{\pm, 0 \lor 0} = \{-c_n\}_{n=0}^{\infty} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Xi_{\tilde{\omega}}^{\pm, 0 \lor 0}$. Note that $\Xi_{\tilde{\omega}}^{\pm, 0 \lor 0} = -\Xi_{\tilde{\omega}}^{\pm, 0 \lor 0}$.
$$
Ext-\sum_{(n \in \mathbb{N}^{\#N}) \wedge (n \le N-1)} \alpha r^{n-1} = \alpha \frac{\widehat{1_{\mathbb{R}_c^#} - r^N}}{1_{\mathbb{R}_c^#} - r} - \alpha \frac{\widehat{1_{\mathbb{R}_c^#}^* - r}}{1_{\mathbb{R}_c^#} - r} + \alpha \left\{ \frac{r^n}{\widehat{1_{\mathbb{R}_c^#} - r}} \right\}_{n=1}^{\infty} = \\
\alpha \left\{ \frac{r^n}{\widehat{1_{\mathbb{R}_c^#} - r}} \right\}_{n=1}^{\infty} - \alpha \frac{\widehat{r^N}}{\widehat{1_{\mathbb{R}_c^#} - r}}. \tag{14.1.26}
$$

Assume that: (i) $r < 1_{\mathbb{R}_c^\#}$, then from (14.1.26) we obtain

$$
\mathrm{Ext}-\sum_{(n \in \mathbb{N}^{\#N}) \wedge (n \le N-1)} \alpha r^{n-1} > 0_{\mathbb{R}_c^{\#}}. \quad (14.1.27)
$$

(ii) $r > 1_{\mathbb{R}_c^\#}$, then from (14.1.26) we obtain

$$
\mathrm{Ext}-\sum_{(n \in \mathbb{N}^{\#N}) \wedge (n \le N-1)} \alpha r^{n-1} = \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty} + \alpha \frac{r^N}{r - 1_{\mathbb{R}_c^{\#}}} > 0_{\mathbb{R}_c^{\#}}. \quad (14.1.28)
$$

**Proposition 14.1.6.** (i) Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#$. Let $S_N$, $\alpha \in \mathbb{R}_c^\#$, $r \in \mathbb{R}_c^\#$ be the sum of $N$ terms, first term and the ratio of the G.P respectively. Then for any $N \in \mathbb{N}^\#$ the statement $\Phi_N$ holds

$$
\Phi_N \iff_s \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r}. \quad (14.1.29)
$$

**Proof.**(i) Directly by hyper infinite induction. Note that $\Phi_N \Rightarrow_s \Phi_{N+1}$ :

$$
\begin{align}
S_{N+1} &= \mathrm{Ext}-\sum_{n=1}^{n=N} \alpha r^{n-1} = \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} + \alpha r^N = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha r^N = \nonumber \\
&= \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha \frac{(1_{\mathbb{R}_c^\#} - r) r^N}{1_{\mathbb{R}_c^\#} - r} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N + r^N - r^{N+1}}{1_{\mathbb{R}_c^\#} - r} = \nonumber \\
&= \alpha \frac{1_{\mathbb{R}_c^\#} - r^{N+1}}{1_{\mathbb{R}_c^\#} - r}. \tag{14.1.30}
\end{align}
$$

Thus $S_{N+1} = \alpha \frac{1_{\mathbb{R}_c^\#} + r^{N+1}}{1_{\mathbb{R}_c^\#} - r}$ and therefore $\Phi_{N+1}$ holds.

(ii) Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#$. Let $S_N$, $\alpha \in \mathbb{R}_c^\#$, $r \in \mathbb{R}_c^\#$ be the sum of $N$ terms, first term and the ratio of the G.P respectively. Then for any $N \in \mathbb{N}^\#$ the statement $\tilde{\Phi}_N$ holds

$$
\tilde{\Phi}_N \iff_s \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r}. \quad (14.1.31)
$$

Notice that (i)⇒(ii) by definitions.

**Definition 14.1.30.** Let {$a_n$}$_{n=0}^{\infty}$, $n \in \mathbb{N}^\#$ be external hyperinfinite sequence in $\tilde{\mathbb{R}}_c^\#$ :
{$a_n$}$_{n=0}^{\infty} \subset \tilde{\mathbb{R}}_c^\#$. We define external hyperinfinite sum $\mathrm{Ext}-\sum_{n=0}^{\infty} a_n$ by

$$
\mathrm{Ext}-\sum_{n=0}^{\infty} a_n = \#\lim_{N\to\infty\#} \left( \mathrm{Ext}-\sum_{n=0}^{N} a_n \right) \quad (14.1.32)
$$

if #-limit in (14.1.31) exists.
**Definition 14.1.31.**

(1) Let {$a_n$}$_{n=0}^\infty \in \Xi_\omega^{\pm, \neq 0}$ and {$b_n$}$_{n=0}^\infty \in \Xi_\omega^{\pm, \neq 0}$ then we define

$$
(i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0}
$$

$$
(ii) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0}
$$

$$
(iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0}
$$

$$
(iv) \left( \{a_n\}_{n=0}^{\infty} \right)^{-1} \triangleq \{a_n^{-1}\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0}
$$

(2) Let {$a_n$}$_{n=0}^\infty \in \Xi_\omega^{\pm, \neq 0 \lor 0}$ and {$b_n$}$_{n=0}^\infty \in \Xi_\omega^{\pm, \neq 0 \lor 0}$ then we define

$$
(i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0}
$$

$$
(ii) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0}
$$

$$
(iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0}
$$

$$
(iv) (\{a_n\}_{n=0}^{\infty})^{-1*} \triangleq \{a_n^{1*}\}_{n=0}^{\infty} \text{ where}
$$

$$
a_n^{1*} = \begin{cases} a_n^{-1} & \text{if } a_n \neq 0_{*\mathbb{R}_c^#} \\ 0_{*\mathbb{R}_c^#} & \text{if } a_n = 0_{*\mathbb{R}_c^#} \end{cases} \qquad (14.1.36)
$$

Note that

$$
(i) \left( \left( \{a_n\}_{n=0}^{\infty} \right)^{-1*} \right)^{-1*} = \{a_n\}_{n=0}^{\infty}
$$

$$
\begin{align*}
\text{(ii)} \quad \{a_n\}_{n=0}^{\infty} \times (\{a_n\}_{n=0}^{\infty})^{-1*} &= \mathring{1}_{*\mathbb{R}_c^#} \quad \text{where } \mathring{1}_{*\mathbb{R}_c^#} = \{\alpha_n\}_{n=0}^{\infty} \text{ is countable sequence} \\
\text{such that}
\end{align*}
$$

$$
\alpha_n = \begin{cases} 1_{\mathbb{R}_c^#} & \text{if } a_n \neq 0_{\mathbb{R}_c^#} \\ 0_{\mathbb{R}_c^#} & \text{if } \alpha_n = 0_{\mathbb{R}_c^#} \end{cases} \qquad (14.1.37)
$$

**Definition 14.1.32.** We say that

$$
(\{a_n\}_{n=0}^{\infty})^{-1*} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0} \text{ is a quasi inverse of } \{a_n\}_{n=0}^{\infty}.
$$

**Definition 14.1.33. (1)** Let {$a_n$}$_{n=0}^\infty \in \Xi_\omega^{\pm, \neq 0 \lor 0}$ and let {$A_n$}$_{n=0}^* \coloneqq \overline{\{a_n\}_{n=0}^\infty}$ be a hyper infinite sequence

$$
\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots) \quad (14.1.38)
$$

i.e. for any infinite $m \in \mathbb{N}^\# \setminus \mathbb{N}$, $A_m \equiv \{a_n\}_{n=0}^\infty$. We will denote a set of the all these
hyper infinite sequences by $\overline{\Xi_\omega^{\pm, \neq 0 \lor 0}}$

$$
(2) \text{ Let } \{x_n + y_n a_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \neq 0 \lor 0} \text{ and let}
$$

$$
\begin{equation} \tag{14.1.39}
\begin{split}
\{x_n + y_n A_n\}_{n=0}^{\infty} &= \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} \\
(x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, x_k + y_k a_k, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots),
\end{split}
\end{equation}
$$

i.e. for any infinite $m \in \mathbb{N}^\# \setminus \mathbb{N}$, $A_m \equiv \{x_n + y_n a_n\}_{n=0}^\infty$. We will denote a set of the all these hyper infinite sequences by $\overline{\{\Xi_\omega^{\pm, \neq 0 \lor 0}, \{x_n\}_{n=0}^\infty, \{y_n\}_{n=0}^\infty\}}$.

**Definition 14.1.34.** Let {$A_n$}$_{n=0}^\infty = \overline{\{a_n\}_{n=0}^\infty}$ and {$B_n$}$_{n=0}^* = \overline{\{b_n\}_{n=0}^\infty}$ be in $\overline{\Xi_\omega^{\pm, \neq 0 \lor 0}}$.
Then we define:

$$
(i) \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n + b_n\}_{n=0}^{\infty}} = \{A_n + B_n\}_{n=0}^{\infty} \in \overline{\Xi_{\omega}^{\pm, \neq 0 \lor 0}}
$$

$$
\begin{align*}
\text{(ii)} \quad \{A_n\}_{n=0}^{\infty} - \{B_n\}_{n=0}^{\infty} &= \overline{\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n - b_n\}_{n=0}^{\infty}} = \\
&= \{A_n - B_n\}_{n=0}^{\infty} \in \overline{\Xi_{\omega}^{\pm, \neq 0 \lor 0}}
\end{align*}
$$

$$
\text{(iii)} \quad \{A_n\}_{n=0}^{\infty} \times \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n \times b_n\}_{n=0}^{\infty}} = \{A_n \times B_n\}_{n=0}^{\infty} \in \overline{\Xi_{\omega}^{\pm, \neq 0 \lor 0}}
$$
**Example 14.1.3.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{n-1}, n \in \mathbb{N}^\#, \alpha \in \tilde{\mathbb{R}}_c^\#, r \in \tilde{\mathbb{R}}_c^\#$. From (14.1.27) we obtain

$$ 
\begin{aligned}
\mathrm{Ext}-\sum_{n=0}^{\infty} \alpha r^{n-1} &= \#\mathrm{lim}_{N \to \infty} \left( \mathrm{Ext}-\sum_{n=0}^{N} \alpha r^{n-1} \right) = \#\mathrm{lim}_{N \to \infty} \alpha \frac{1 \tilde{\mathbb{R}}_c^\# - r^N}{1 \tilde{\mathbb{R}}_c^\# - r} = \\
&= \alpha \frac{1 \tilde{\mathbb{R}}_c^\#}{1 \tilde{\mathbb{R}}_c^\# - r}
\end{aligned}
\qquad (14.1.33) $$

since $\#\mathrm{lim}_{N \to \infty} r^N = 0_{\tilde{\mathbb{R}}_c^\#}$ if $|r| < 1$. From (14.1.33) and (14.1.25) we obtain

$$ 
\begin{aligned}
\alpha \frac{1 \tilde{\mathbb{R}}_c^\#}{1 \tilde{\mathbb{R}}_c^\# - r} &= \mathrm{Ext}-\sum_{n=0}^{\infty} \alpha r^{n-1} = \mathrm{Ext}-\sum_{n=0}^{\infty} \alpha r^{n-1} + \mathrm{Ext}-\sum_{n \in \mathbb{N}^\#} \alpha r^{n-1} = \\
&= \alpha \frac{1 \tilde{\mathbb{R}}_c^\#}{1 \tilde{\mathbb{R}}_c^\# - r} - \alpha \left\{ \frac{r^n}{1 \tilde{\mathbb{R}}_c^\# - r} \right\}_{n=1}^\infty + \mathrm{Ext}-\sum_{n \in \mathbb{N}^\# \mathbb{N}} \alpha r^{n-1}.
\end{aligned}
\qquad (14.1.34) $$

From (14.1.34) we obtain

$$ 
\begin{aligned}
\mathrm{Ext}-\sum_{n \in \mathbb{N}^\# \mathbb{N}} \alpha r^{n-1} &= \alpha \frac{1 \tilde{\mathbb{R}}_c^\#}{1 \tilde{\mathbb{R}}_c^\# - r} \alpha - \left( \frac{1 \tilde{\mathbb{R}}_c^\#}{1 \tilde{\mathbb{R}}_c^\# - r} - \alpha \left\{ \frac{r^n}{1 \tilde{\mathbb{R}}_c^\# - r} \right\}_{n=1}^\infty \right) = \\
&= \alpha \left\{ \frac{r^n}{1 \tilde{\mathbb{R}}_c^\# - r} \right\}_{n=1}^\infty > 0_{\tilde{\mathbb{R}}_c^\#}.
\end{aligned}
\qquad (14.1.35) $$

**Definition 14.1.31.** Let $\{a_n\}_{n=0}^\infty$ be $\mathbb{R}_c^\#$-valued countable sequence $a : \mathbb{N} \to \mathbb{R}_c^\#$ such that:

(i) there is $M \in \mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{\mathbb{R}_c^\#}$, we denote a set of the all these sequences by $\Xi_{\tilde{\omega}}^{\pm, \neq 0}$. We define a set $-\Xi_{\tilde{\omega}}^{\pm, \neq 0}$ by $\{c_n\}_{n=0}^\infty \in -\Xi_{\tilde{\omega}}^{\pm, \neq 0} \Leftrightarrow \{-c_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0}$. Note that $\Xi_{\tilde{\omega}}^{\pm, \neq 0} = -\Xi_{\tilde{\omega}}^{\pm, \neq 0}$.

(ii) there is countable subsequence $\{a_{n_k}\}_{k=m}^\infty \subset \{a_n\}_{n=0}^\infty$ such that $a_{n_k} = 0_{\mathbb{R}_c^\#}, k \ge m$ and $a_n \ne 0_{\mathbb{R}_c^\#}$ iff $a_n \notin \{a_{n_k}\}_{k=m}^\infty$, we denote a set of the all these countable sequences by $\Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$. We define a set $-\Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$ by $\{c_n\}_{n=0}^\infty \in -\Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0} \Leftrightarrow \{-c_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$. Note that $\Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0} = -\Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$.

**Definition 14.1.31.**

(1) Let $\{a_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0}$ and $\{b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0}$ then we define

(i) $\{a_n\}_{n=0}^\infty + \{b_n\}_{n=0}^\infty \triangleq \{a_n + b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$

(ii) $\{a_n\}_{n=0}^\infty - \{b_n\}_{n=0}^\infty \triangleq \{a_n - b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$

(iii) $\{a_n\}_{n=0}^\infty \times \{b_n\}_{n=0}^\infty \triangleq \{a_n \times b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0}$

(iv) $(\{a_n\}_{n=0}^\infty)^{-1} \triangleq \{a_n^{-1}\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0}$

(2) Let $\{a_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$ and $\{b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$ then we define

(i) $\{a_n\}_{n=0}^\infty + \{b_n\}_{n=0}^\infty \triangleq \{a_n + b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$

(ii) $\{a_n\}_{n=0}^\infty - \{b_n\}_{n=0}^\infty \triangleq \{a_n - b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$

(iii) $\{a_n\}_{n=0}^\infty \times \{b_n\}_{n=0}^\infty \triangleq \{a_n \times b_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \neq 0 \vee 0}$
**Definition 14.1.35.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \pm 0 \lor 0}$, i.e. for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \Xi_{\omega}^{\pm, \pm 0 \lor 0}$. Say $\{\Psi_n\}_{n=0}^{\infty}$ #-tends to $0_{\mathbb{R}_c^\#}$ as $n \to \infty^\#$ iff for any given $\varepsilon > 0_{\mathbb{R}_c^\#}$, $\varepsilon \approx 0_{\mathbb{R}_c^\#}$ there is a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.1.36.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be a hyper infinite sequence such that for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \tilde{\Xi}_{\omega}^{\pm, \pm 0 \lor 0}$. We call $\{\Psi_n\}_{n=0}^{\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{\mathbb{R}_c^\#}$. To be precise: given any $\varepsilon > 0_{\mathbb{R}_c^\#}$, $\varepsilon \approx 0_{\mathbb{R}_c^\#}$ there is a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.1.8.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \pm 0 \lor 0}$. If $\{\Psi_n\}_{n=0}^{\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_{\#} \Phi$ as $n \to \infty^\#$ for some $\Phi \in \tilde{\Xi}_{\omega}^{\pm, \pm 0 \lor 0}$), then $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Proof.** We know that $\Psi_n \to_{\#} \Phi$. Here is a ubiquitous trick: instead of using $\varepsilon$ in the definition, start with an arbitrary infinitesimal $\varepsilon > 0_{\mathbb{R}_c^\#}$, $\varepsilon \approx 0_{\mathbb{R}_c^\#}$ and then choose $N$ so that $|\Psi_n - \Phi| < \varepsilon/2$ when $n > N$. Then if $m, n > N$, we have

$$|\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \varepsilon/2 + \varepsilon/2 = \varepsilon.$$

This shows that $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Theorem 14.1.9.** If $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence, then it is bounded in $\mathbb{R}_c^\#$; that is, there is some number $M \in \mathbb{R}_c^\#$ such that

$$|\{\Psi_n\}_{n=0}^{\infty}| \le M \text{ for all } n \in \mathbb{N}^{\#}.$$ 

**Proof.** Since $\{\Psi_n\}_{n=0}^{\infty}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N$ such that $|\Psi_m - \Psi_n| < 1$ whenever $m, n > N$. Thus, $|\Psi_{N+1} - \Psi_n| < 1$ for $n > N$. We can rewrite this as $\Psi_{N+1} - 1 < \Psi_n < \Psi_{N+1} + 1$. This means that $|\Psi_n|$ is less than the maximum of $|\Psi_{N+1} - 1|$ and $|\Psi_{N+1} + 1|$. So, set $M \in \mathbb{R}_c^\#$ larger than any number in the following list: $\{|\Psi_0|, |\Psi_1|, \dots, |\Psi_N|, |\Psi_{N+1} - 1|, |\Psi_{N+1} + 1|\}$.

Then for any term $\Psi_n$, if $n \le N$, then $|\Psi_n|$ appears in the list and so $|\Psi_n| \le M$; if $n > N$, then (as shown above) $|\Psi_n|$ is less than at least one of the last two entries in the list, and so $|\Psi_n| \le M$. Hence, $M$ is a bound for the sequence.

Let $\tilde{\Xi}$ denote the set of all Cauchy hyper infinite sequences. We must define an equivalence relation on $\tilde{\Xi}$.

**Definition 14.1.37.** Let $\mathcal{S}$ be a set of objects. A relation among pairs of elements of $\mathcal{S}$ is said to be an equivalence relation if the following three properties hold:

**Reflexivity:** for any $s \in \mathcal{S}$, $s$ is related to $s$.

**Symmetry:** for any $s, t \in \mathcal{S}$, if $s$ is related to $t$ then $t$ is related to $s$.

**Transitivity:** for any $s, t, r \in \mathcal{S}$, if $s$ is related to $t$ and $t$ is related to $r$, then $s$ is related to $r$.

The following well known proposition goes most of the way to showing that an equivalence relation divides a set into bins.

**Theorem 14.1.5.10.** Let $\mathcal{S}$ be a set, with an equivalence relation on pairs of elements. For $s \in \mathcal{S}$, denote by $[s]$ the set of all elements in $\mathcal{S}$ that are related to $s$. Then for any $s, t \in \mathcal{S}$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.

The sets $[s]$ for $s \in \mathcal{S}$ are called the equivalence classes, and they are the bins.

**Corollary 14.1.2.** If $\mathcal{S}$ is a set with an equivalence relation on pairs of elements, then
(iv) $(\{a_n\}_{n=0}^{\infty})^{-1*} \triangleq \{a_n^{1*}\}_{n=0}^{\infty}$ where

$$a_n^{-1*} = \begin{cases} a_n^{-1} & \text{if } a_n \neq 0_{\mathbb{R}_c^*} \\ 0_{\mathbb{R}_c^*} & \text{if } a_n = 0_{\mathbb{R}_c^*} \end{cases} \qquad (14.1.36)$$

Note that

$$(i) \left( \left( \{a_n\}_{n=0}^{\infty} \right)^{-1*} \right)^{-1*} = \{a_n\}_{n=0}^{\infty}$$

(ii) $\{a_n\}_{n=0}^{\infty} \times (\{a_n\}_{n=0}^{\infty})^{-1*} = \tilde{1}_{\mathbb{R}_c^*}$ where $\tilde{1}_{\mathbb{R}_c^*} = \{a_n\}_{n=0}^{\infty}$ is countable sequence such that

$$\alpha_n = \begin{cases} 1_{\mathbb{R}_c^*} & \text{if } a_n \neq 0_{\mathbb{R}_c^*} \\ 0_{\mathbb{R}_c^*} & \text{if } a_n = 0_{\mathbb{R}_c^*} \end{cases} \qquad (14.1.37)$$

**Definition 14.1.32.** We say that

$(\{a_n\}_{n=0}^{\infty})^{-1*} \in \Xi_{\omega}^{\pm,0\vee=0}$ is a quasi inverse of $\{a_n\}_{n=0}^{\infty}$.

**Definition 14.1.33. (1)** Let $\{a_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm,0\vee=0}$ and let $\{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ be a hyper infinite sequence

$$\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots) \qquad (14.1.38)$$

i.e. for any infinite $m \in \mathbb{N}^{\#}\mathbb{N}$, $A_m \equiv \{a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\tilde{\Xi}_{\omega}^{\pm,0\vee=0}$

(2) Let $\{x_n + y_n a_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm,0\vee=0}$ and let

$$\{x_n + y_n A_n\}_{n=0}^{\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} = (x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, x_k + y_k a_k, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots), \qquad (14.1.39)$$

i.e. for any infinite $m \in \mathbb{N}^{\#}\mathbb{N}$, $A_m \equiv \{x_n + y_n a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\overline{\{\Xi_{\omega}^{\pm,0\vee=0}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}}$.

**Definition 14.1.34.** Let $\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ and $\{B_n\}_{n=0}^{*\infty} = \overline{\{b_n\}_{n=0}^{\infty}}$ be in $\tilde{\Xi}_{\omega}^{\pm,0\vee=0}$. Then we define:

$$(i) \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n + b_n\}_{n=0}^{\infty}} = \{A_n + B_n\}_{n=0}^{\infty} \in \tilde{\Xi}_{\omega}^{\pm,0\vee=0}$$

$$(ii) \{A_n\}_{n=0}^{\infty} - \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n - b_n\}_{n=0}^{\infty}} = \{A_n - B_n\}_{n=0}^{\infty} \in \tilde{\Xi}_{\omega}^{\pm,0\vee=0}$$

$$(iii) \{A_n\}_{n=0}^{\infty} \times \{B_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n \times b_n\}_{n=0}^{\infty}} = \{A_n \times B_n\}_{n=0}^{\infty} \in \tilde{\Xi}_{\omega}^{\pm,0\vee=0}$$

**Definition 14.1.35.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm,0\vee=0}$, i.e. for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \Xi_{\omega}^{\pm,0\vee=0}$. Say $\{\Psi_n\}_{n=0}^{\infty}$ #-tends to $0_{\mathbb{R}_c^*}$ as $n \to \infty^\#$ iff for any given $\varepsilon > 0_{\mathbb{R}_c^*}$, $\varepsilon \approx 0_{\mathbb{R}_c^*}$ there is a hypernatural number $N \in \mathbb{N}^{\#}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.1.36.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be a hyper infinite sequence such that for all $n \in \mathbb{N}^{\#}$, $\Psi_n \in \tilde{\Xi}_{\omega}^{\pm,0\vee=0}$. We call $\{\Psi_n\}_{n=0}^{\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{\mathbb{R}_c^*}$. To be precise: given any $\varepsilon > 0_{\mathbb{R}_c^*}$, $\varepsilon \approx 0_{\mathbb{R}_c^*}$ there is a hypernatural number $N \in \mathbb{N}^{\#}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.1.8.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm,0\vee=0}$. If $\{\Psi_n\}_{n=0}^{\infty}$ is a #-convergent hyper
the equivalence classes are non-empty disjoint sets whose union is all of S.

**Definition 14.1.38.** Let {$\Psi_{1,n}$$^{\infty}_{n=0}$} and {$\Psi_{2,n}$$^{\infty}_{n=0}$} be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_{\#} 0_{\mathbb{R}_c^{\#}}$ as $n \to \infty^{\#}$, i.e. if the hyper infinite sequence {$|\Psi_{1,n} - \Psi_{2,n}|$}$_{n=0}^{\infty}$ #-tends to $0_{\mathbb{R}_c^{\#}}$.

**Proposition 14.1.4.** Definition 4.1.38 yields an equivalence relation on $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$.

**Proof.** We need to show that this relation is reflexive, symmetric, and transitive.

* Reflexive: $\Psi_n - \Psi_n = 0_{\mathbb{R}_c^{\#}}$, and the sequence all of whose terms are $0_{\mathbb{R}_c^{\#}}$ clearly #-converges to $0_{\mathbb{R}_c^{\#}}$. So {$\Psi_n$$^{\infty}_{n=0}$} is related to {$\Psi_n$$^{\infty}_{n=0}$}.

* Symmetric: Suppose {$\Psi_{1,n}$$^{\infty}_{n=0}$} is related to {$\Psi_{2,n}$$^{\infty}_{n=0}$}, so $\Psi_{1,n} - \Psi_{2,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$. But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.1.35, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$ as well. Hence, {$\Psi_{2,n}$$^{\infty}_{n=0}$} is related to {$\Psi_{1,n}$$^{\infty}_{n=0}$}.

* Transitive: Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 14.1.4. Suppose {$\Psi_{1,n}$$^{\infty}_{n=0}$} is related to {$\Psi_{2,n}$$^{\infty}_{n=0}$}, and {$\Psi_{2,n}$$^{\infty}_{n=0}$} is related to {$\Psi_{3,n}$$^{\infty}_{n=0}$}. This means that $\Psi_{1,n} - \Psi_{2,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$ and $\Psi_{2,n} - \Psi_{3,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$.

To be fully precise, let us fix infinite small $\varepsilon > 0_{\mathbb{R}_c^{\#}}$; then there exists an $N \in \mathbb{N}^{\#} \mathbb{N}$ such that for all $n > N$, $|\Psi_{1,n} - \Psi_{2,n}| < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$, $|\Psi_{2,n} - \Psi_{3,n}| < \varepsilon/2$. Well, then, as long as $n > \max(N,M)$, we have that $|\Psi_{2,n} - \Psi_{3,n}| = |(\Psi_{1,n} - \Psi_{2,n}) + (\Psi_{2,n} - \Psi_{3,n})| \le |\Psi_{1,n} - \Psi_{2,n}| + |\Psi_{2,n} - \Psi_{3,n}| < \varepsilon/2 + \varepsilon/2 = \varepsilon$. So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|\Psi_{1,n} - \Psi_{3,n}| < \varepsilon$. This means that $\Psi_{1,n} - \Psi_{3,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$, i.e. {$\Psi_{1,n}$$^{\infty}_{n=0}$} is related to {$\Psi_{3,n}$$^{\infty}_{n=0}$}.

So, we really have equivalence relation, and so by Corollary 14.1.2, the set $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$ is partitioned into disjoint subsets (equivalence classes).

**Definition 14.1.39.** The hyperreal numbers $\tilde{\mathbb{R}}_c^{\#}$ contain: (1) all the equivalence classes $[\{\Psi_{1,n}$$^{\infty}_{n=0}\}]$ of Cauchy hyper infinite sequences of, as per

Definition 14.1.38 and (2) the all gyperreals $\mathbb{R}_c^{\#} \subset \tilde{\mathbb{R}}_c^{\#}$ by canonical imbedding $\mathbb{R}_c^{\#} \hookrightarrow \tilde{\mathbb{R}}_c^{\#}$ (14.1.42)-(14.1.43).

That is, each such equivalence class is a hyperreal number in $\tilde{\mathbb{R}}_c^{\#}$.

**Definition 14.1.40.** Let $s, t \in \tilde{\mathbb{R}}_c^{\#}$, so there are Cauchy hyper infinite sequences {$\Psi_n$$^{\infty}_{n=0}$} and {$\Phi_n$$^{\infty}_{n=0}$} with $s = [\{\Psi_n$$^{\infty}_{n=0}\}]$ and $t = [\{\Phi_n$$^{\infty}_{n=0}\}]$.

(a) Define $s + t$ to be the equivalence class of the hyper infinite sequence {$\Psi_n + \Phi_n$$^{\infty}_{n=0}$}.

(b) Define $s \times t$ to be the equivalence class of the hyper infinite sequence {$\Psi_n \times \Phi_n$$^{\infty}_{n=0}$}.

**Proposition 14.1.5.** The operations $+, \times$ in Definition 14.1.25 (a),(b) are well-defined.

**Proof.** Suppose that $[\{\Psi_n$$^{\infty}_{n=0}\}] = [\{\Psi_{1,n}$$^{\infty}_{n=0}\}]$ and $[\{\Phi_n$$^{\infty}_{n=0}\}] = [\{\Phi_{1,n}$$^{\infty}_{n=0}\}]$. Thus means that $\Psi_n - \Psi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$ and $\Phi_n - \Phi_{1,n} \to_{\#} 0_{\mathbb{R}_c^{\#}}$. Then $(\Psi_n + \Phi_n) - (\Psi_{1,n} + \Phi_{1,n}) = (\Psi_n - \Psi_{1,n}) + (\Phi_n - \Phi_{1,n})$. Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to $0_{\mathbb{R}_c^{\#}}$, and so $[(\Psi_n + \Phi_n)] = [(\Psi_{1,n} + \Phi_{1,n})]$.
infinite sequence (that is, $\Psi_n \to_{\#} \Phi$ as $n \to \infty^{\#}$ for some $\Phi \in \tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$), then
$\{\Psi_n\}_{n=0}^{\infty^{\#}}$ is a Cauchy hyper infinite sequence.

**Proof.** We know that $\Psi_n \to_{\#} \Phi$. Here is a ubiquitous trick: instead of using $\epsilon$ in the definition, start with an arbitrary infinitesimal $\epsilon > 0_{\mathbb{R}^{\#}}$, $\epsilon \approx 0_{\mathbb{R}^{\#}}$ and then choose $N$ so that $|\Psi_n - \Phi| < \epsilon/2$ when $n > N$. Then if $m, n > N$, we have
$$|\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \epsilon/2 + \epsilon/2 = \epsilon.$$ 
This shows that $\{\Psi_n\}_{n=0}^{\infty^{\#}}$ is a Cauchy hyper infinite sequence.

**Theorem 14.1.9.** If $\{\Psi_n\}_{n=0}^{\infty^{\#}}$ is a Cauchy hyper infinite sequence, then it is bounded in $\mathbb{R}_c^{\#}$; that is, there is some number $M \in \mathbb{R}_c^{\#}$ such that
$$|\{\Psi_n\}_{n=0}^{\infty^{\#}}| \le M \text{ for all } n \in \mathbb{N}^{\#}.$$

**Proof.** Since $\{\Psi_n\}_{n=0}^{\infty^{\#}}$ is Cauchy, setting $\epsilon = 1$ we know that there is some $N$ such that $|\Psi_m - \Psi_n| < 1$ whenever $m, n > N$. Thus, $|\Psi_{N+1} - \Psi_n| < 1$ for $n > N$. We can rewrite this as $\Psi_{N+1} - 1 < \Psi_n < \Psi_{N+1} + 1$. This means that $|\Psi_n|$ is less than the maximum of $|\Psi_{N+1} - 1|$ and $|\Psi_{N+1} + 1|$. So, set $M \in \mathbb{R}_c^{\#}$ larger than any number in the following list: $\{|\Psi_0|, |\Psi_1|, \dots, |\Psi_N|, |\Psi_{N+1} - 1|, |\Psi_{N+1} + 1|\}$.

Then for any term $\Psi_n$, if $n \le N$, then $|\Psi_n|$ appears in the list and so $|\Psi_n| \le M$; if $n > N$, then (as shown above) $|\Psi_n|$ is less than at least one of the last two entries in the list, and so $|\Psi_n| \le M$. Hence, $M$ is a bound for the sequence.

Let $\tilde{\Xi}$ denote the set of all Cauchy hyper infinite sequences. We must define an equivalence relation on $\tilde{\Xi}$.

**Definition 14.1.37.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

**Reflexivity:** for any $s \in S$, $s$ is related to $s$.

**Symmetry:** for any $s, t \in S$, if $s$ is related to $t$ then $t$ is related to $s$.

**Transitivity:** for any $s, t, r \in S$, if $s$ is related to $t$ and $t$ is related to $r$, then $s$ is related to $r$.

The following well known proposition goes most of the way to showing that an
equivalence relation divides a set into bins.

**Theorem 14.1.5.10.** Let $S$ be a set, with an equivalence relation on pairs of elements. For $s \in S$, denote by $[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint. The sets $[s]$ for $s \in S$ are called the equivalence classes, and they are the bins.

**Corollary 14.1.2.** If $S$ is a set with an equivalence relation on pairs of elements, then the equivalence classes are non-empty disjoint sets whose union is all of $S$.

**Definition 14.1.38.** Let $\{\Psi_{1,n}\}_{n=0}^{\infty^{\#}}$ and $\{\Psi_{2,n}\}_{n=0}^{\infty^{\#}}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_{\#} 0_{\mathbb{R}^{\#}}$ as $n \to \infty^{\#}$, i.e. if the hyper infinite sequence $\{|\Psi_{1,n} - \Psi_{2,n}|\}_{n=0}^{\infty^{\#}}$ #-tends to $0_{\mathbb{R}^{\#}}$.

**Proposition 14.1.4.** Definition 4.1.38 yields an equivalence relation on $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$.

**Proof.** We need to show that this relation is reflexive, symmetric, and transitive.

• **Reflexive:** $\Psi_n - \Psi_n = 0_{\mathbb{R}^{\#}}$, and the sequence all of whose terms are $0_{\mathbb{R}^{\#}}$ clearly #-converges to $0_{\mathbb{R}^{\#}}$. So $\{\Psi_n\}_{n=0}^{\infty^{\#}}$ is related to $\{\Psi_n\}_{n=0}^{\infty^{\#}}$.

• **Symmetric:** Suppose $\{\Psi_{1,n}\}_{n=0}^{\infty^{\#}}$ is related to $\{\Psi_{2,n}\}_{n=0}^{\infty^{\#}}$, so $\Psi_{1,n} - \Psi_{2,n} \to_{\#} 0_{\mathbb{R}^{\#}}$.
Multiplication is a little trickier; this is where we will use Theorem 14.1.10. We will also use another ubiquitous technique: adding $0_{\mathbb{R}_c^#}$ in the form of $s-s$. Again, suppose that

$$[\{\Psi_n\}_{n=0}^{\infty}^#] = [\{\Psi_{1,n}\}_{n=0}^{\infty}^#] \text{ and } [\{\Phi_n\}_{n=0}^{\infty}^#] = [\{\Phi_{1,n}\}_{n=0}^{\infty}^#]; \text{ we wish to show that}$$

$$[\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}^#] = [\{\Psi_{1,n} \times \Phi_{1,n}\}_{n=0}^{\infty}^#], \text{ or, in other words, that}$$

$\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_{\#} 0_{\mathbb{R}_c^#}$. Well, we add and subtract one of the other cross terms, say $\Phi_n \times \Psi_{1,n}$:

$$\begin{aligned} \Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} &= \Psi_n \times \Phi_n + (\Phi_n \times \Psi_{1,n} - \Phi_n \times \Psi_{1,n}) - \Psi_{1,n} \times \Phi_{1,n} \\ &= (\Psi_n \times \Phi_n - \Phi_n \times \Psi_{1,n}) + (\Phi_n \times \Psi_{1,n} - \Psi_{1,n} \times \Phi_{1,n}) \\ &= \Phi_n \times (\Psi_n - \Psi_{1,n}) + \Psi_{1,n} \times (\Phi_n - \Phi_{1,n}). \end{aligned}$$

Hence, we have $|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}|$.

Now, from Theorem 14.1.9, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in \mathbb{N}^#$. Taking some number $R$ (for example $R = M+L$) which is bigger than both, we have

$$|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}| \le R(|\Psi_n - \Psi_{1,n}| + |\Phi_n - \Phi_{1,n}|).$$

Now, noting that both $\Psi_n - \Psi_{1,n}$ and $\Phi_n - \Phi_{1,n}$ #-tend to $0_{\mathbb{R}_c^#}$ and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2R$, we see that

$$\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_{\#} 0_{\mathbb{R}_c^#}$$

**Theorem 14.2.11.** Given any hyperreal number $s \in \tilde{\mathbb{R}}_c^#$, $s \ne 0_{\tilde{\mathbb{R}}_c^#}$, there is a hyperreal number $t \in \tilde{\mathbb{R}}_c^#$ such that $s \times t = 1_{\tilde{\mathbb{R}}_c^#}$ or $s \times t = \tilde{1}_{\tilde{\mathbb{R}}_c^#}$.

**Proof.** First we must properly understand what the theorem says. The premise is that $s$ is nonzero, which means that $s$ is not in the equivalence class of

$$0_{\tilde{\mathbb{R}}_c^#} = (0_{\mathbb{R}_c^#}, 0_{\mathbb{R}_c^#}, 0_{\mathbb{R}_c^#}, 0_{\mathbb{R}_c^#}, \dots). \qquad (14.1.40)$$

In other words, $s = [\{\Psi_n\}_{n=0}^{\infty}^#]$ where $\Psi_n - 0_{\tilde{\mathbb{R}}_c^#}$ does not #-converge to $0_{\mathbb{R}_c^#}$.

From this, we are to deduce the existence of a hyperreal number $t = [\{\Phi_n\}_{n=0}^{\infty}^#]$ such that $s \times t = [\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}^#]$ is the same equivalence class as $1_{\tilde{\mathbb{R}}_c^#} = [(1_{\mathbb{R}_c^#}, 1_{\mathbb{R}_c^#}, 1_{\mathbb{R}_c^#}, 1_{\mathbb{R}_c^#}, \dots)]$ or as some $\tilde{1}_{\tilde{\mathbb{R}}_c^#}$. Doing so is actually an easy consequence of the fact that nonzero hyperreal numbers from $\mathbb{R}_c^#$ have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{\Psi_n\}_{n=0}^{\infty}^#$ does not #-tend to $0_{\mathbb{R}_c^#}$ as $n \to \infty^#$), there's no reason any number of the terms in $\{\Psi_n\}_{n=0}^{\infty}^#$ can't equal $0_{\tilde{\mathbb{R}}_c^#}$. However, it turns out that eventually, $\Psi_n \ne 0_{\mathbb{R}_c^#}$.

That is,

**Lemma 14.1.2.** If $\{\Psi_n\}_{n=0}^{\infty}^#$ is a Cauchy hyper infinite sequence which does not #-tends to $0_{\mathbb{R}_c^#}$, then there is an $N \in \mathbb{N}^#$ such that, for $n > N$, $\Psi_n \ne 0_{\mathbb{R}_c^#}$.

We will now use it to complete the proof of Theorem 14.2.11.

Let $N \in \mathbb{N}^#$ be such that $\Psi_n \ne 0_{\mathbb{R}_c^#}$ for $n > N$. Define hyper infinite sequence $\Phi_n$ of hyperreal numbers from $\tilde{\mathbb{R}}_c^#$ as follows:

for $n \le N$, $\Phi_n = 0_{\mathbb{R}_c^#}$, and for $n > N$, $\Phi_n = 1_{\mathbb{R}_c^#}/\Psi_n$:

$$\{\Phi_n\}_{n=0}^{\infty} = (0_{\mathbb{R}_c^#}, 0_{\mathbb{R}_c^#}, \dots, 0_{\mathbb{R}_c^#}, 1_{\mathbb{R}_c^#}/\Psi_{N+1}, 1_{\mathbb{R}_c^#}/\Psi_{N+2}, \dots).$$
But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.1.35, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_# 0_{\mathbb{R}_c^#}$ as well. Hence, $\{\Psi_{2,n}\}_{n=0}^\infty$ is related to $\{\Psi_{1,n}\}_{n=0}^\infty$.

**Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 14.1.4. Suppose $\{\Psi_{1,n}\}_{n=0}^\infty$ is related to $\{\Psi_{2,n}\}_{n=0}^\infty$, and $\{\Psi_{2,n}\}_{n=0}^\infty$ is related to $\{\Psi_{3,n}\}_{n=0}^\infty$. This means that $\Psi_{1,n} - \Psi_{2,n} \to_# 0_{\mathbb{R}_c^#}$ and $\Psi_{2,n} - \Psi_{3,n} \to_# 0_{\mathbb{R}_c^#}$.

To be fully precise, let us fix infinite small $\varepsilon > 0_{\mathbb{R}_c^#}$; then there exists an $N \in \mathbb{N}^*\mathbb{N}$ such that for all $n > N$, $|\Psi_{1,n} - \Psi_{2,n}| < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$, $|\Psi_{2,n} - \Psi_{3,n}| < \varepsilon/2$. Well, then, as long as $n > \max(N, M)$, we have that $|\Psi_{2,n} - \Psi_{3,n}| = |(\Psi_{1,n} - \Psi_{2,n}) + (\Psi_{2,n} - \Psi_{3,n})| \le |\Psi_{1,n} - \Psi_{2,n}| + |\Psi_{2,n} - \Psi_{3,n}| < \varepsilon/2 + \varepsilon/2 = \varepsilon$. So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|\Psi_{1,n} - \Psi_{3,n}| < \varepsilon$. This means that $\Psi_{1,n} - \Psi_{3,n} \to_# 0_{\mathbb{R}_c^#}$, i.e. $\{\Psi_{1,n}\}_{n=0}^\infty$ is related to $\{\Psi_{3,n}\}_{n=0}^\infty$.

So, we really have equivalence relation, and so by Corollary 14.1.2, the set $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$ is partitioned into disjoint subsets (equivalence classes).

**Definition 14.1.39.** The hyperreal numbers $\tilde{\mathbb{R}}_c^\#$ contain: (1) all the equivalence classes $[\{\Psi_{1,n}\}_{n=0}^\infty]$ of Cauchy hyper infinite sequences of, as per

Definition 14.1.38 and (2) the all hyperreals $\tilde{\mathbb{R}}_c^\# \subset \tilde{\mathbb{R}}_c^\#$ by canonical imbedding $\tilde{\mathbb{R}}_c^\# \hookrightarrow \tilde{\mathbb{R}}_c^\# (14.1.42)-(14.1.43)$.

That is, each such equivalence class is a hyperreal number in $\tilde{\mathbb{R}}_c^\#$.

**Definition 14.1.40.** Let $s, t \in \tilde{\mathbb{R}}_c^\#$, so there are Cauchy hyper infinite sequences $\{\Psi_n\}_{n=0}^\infty$ and $\{\Phi_n\}_{n=0}^\infty$ with $s = [\{\Psi_n\}_{n=0}^\infty]$ and $t = [\{\Phi_n\}_{n=0}^\infty]$.

(a) Define $s+t$ to be the equivalence class of the hyper infinite sequence $\{\Psi_n + \Phi_n\}_{n=0}^\infty$.

(b) Define $s \times t$ to be the equivalence class of the hyper infinite sequence $\{\Psi_n \times \Phi_n\}_{n=0}^\infty$.

**Proposition 14.1.5.** The operations $+, \times$ in Definition 14.1.25 (a), (b) are well-defined.

**Proof.** Suppose that $[\{\Psi_n\}_{n=0}^\infty] = [\{\Psi_{1,n}\}_{n=0}^\infty]$ and $[\{\Phi_n\}_{n=0}^\infty] = [\{\Phi_{1,n}\}_{n=0}^\infty]$.

Thus means that $\Psi_n - \Psi_{1,n} \to_# 0_{\mathbb{R}_c^#}$ and $\Phi_n - \Phi_{1,n} \to_# 0_{\mathbb{R}_c^#}$. Then $(\Psi_n + \Phi_n) - (\Psi_{1,n} + \Phi_{1,n}) = (\Psi_n - \Psi_{1,n}) + (\Phi_n - \Phi_{1,n})$. Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to $0_{\mathbb{R}_c^#}$, and so $[(\Psi_n + \Phi_n)] = [(\Psi_{1,n} + \Phi_{1,n})]$.

Multiplication is a little trickier; this is where we will use Theorem 14.1.10. We will also use another ubiquitous technique: adding $0_{\mathbb{R}_c^#}$ in the form of $s-s$. Again, suppose that

$[\{\Psi_n\}_{n=0}^\infty] = [\{\Psi_{1,n}\}_{n=0}^\infty]$ and $[\{\Phi_n\}_{n=0}^\infty] = [\{\Phi_{1,n}\}_{n=0}^\infty]$; we wish to show that

$[\{\Psi_n \times \Phi_n\}_{n=0}^\infty] = [\{\Psi_{1,n} \times \Phi_{1,n}\}_{n=0}^\infty]$, or, in other words, that

$\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_# 0_{\mathbb{R}_c^#}$. Well, we add and subtract one of the other cross terms, say $\Phi_n \times \Psi_{1,n}$:

$$ \begin{align*} \Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} &= \Psi_n \times \Phi_n + (\Phi_n \times \Psi_{1,n} - \Phi_n \times \Psi_{1,n}) - \Psi_{1,n} \times \Phi_{1,n} = \\ &= (\Psi_n \times \Phi_n - \Phi_n \times \Psi_{1,n}) + (\Phi_n \times \Psi_{1,n} - \Psi_{1,n} \times \Phi_{1,n}) = \\ &= \Phi_n \times (\Psi_n - \Psi_{1,n}) + \Psi_{1,n} \times (\Phi_n - \Phi_{1,n}). \end{align*} $$
This makes sense since, for $n > N$, an is a nonzero hyperreal number, so $1_{\mathbb{R}^n}/\Psi_n$ exists.

Then $\Psi_n \times \Phi_n$ is equal to $\Psi_n \times 0_{\mathbb{R}^n} = 0_{\mathbb{R}^n}$ for $n \le N$, and equals $\Psi_n \times \Phi_n = \Psi_n \times 1_{\mathbb{R}^n}/\Psi_n = 1_{\mathbb{R}^n}$ for $n > N$.

Well, then, if we look at the hyper infinite sequence

$$1_{\mathbb{R}^n} = (1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, \dots), \qquad (14.1.41)$$

we have $(1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, \dots) - (\Psi_n \times \Phi_n)$ is the hyper infinite sequence which is $1_{\mathbb{R}^n} - 0_{\mathbb{R}^n} = 1_{\mathbb{R}^n}$ for $n \le N$ and equals $1_{\mathbb{R}^n} - 1_{\mathbb{R}^n} = 0_{\mathbb{R}^n}$ for $n > N$. Since this hyper infinite sequence is eventually equal to $0_{\mathbb{R}^n}$, it #-converges to $0_{\mathbb{R}^n}$ as $n \to \infty^\#$, and so $[\{\Psi_n \times \Phi_n\}_{n=0}^\infty] = [(1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, 1_{\mathbb{R}^n}, \dots)] = 1_{\mathbb{R}^n} \in \tilde{\mathbb{R}}^n$ and similarly $[\{\Psi_n \times \Phi_n\}_{n=0}^\infty] = \tilde{I}_{\mathbb{R}^n} \in \tilde{\mathbb{R}}^n$. This shows that $t = [\{\Phi_n\}_{n=0}^\infty]$ is a multiplicative inverse (and similarly quasi inverse) to $s = [\{\Psi_n\}_{n=0}^\infty]$.

**Definition 14.2.41.** Let $s \in \tilde{\mathbb{R}}^n$. Say that $s$ is positive if $s \neq 0_{\tilde{\mathbb{R}}^n}$, and if $s = [\{\Psi_n\}_{n=0}^\infty]$ for some Cauchy hyper infinite sequence such that for some $N$, $\Psi_n > 0_{\tilde{\mathbb{R}}^n}$ for all $n > N$. Given two hyperreal numbers $s, t \in \tilde{\mathbb{R}}^n$, say that $s > t$ if $s - t$ is positive.

**Theorem 14.1.7.** Let $s, t \in \tilde{\mathbb{R}}^n$ be hyperreal numbers such that $s > t$, and let $r \in \tilde{\mathbb{R}}^n$. Then $s + r > t + r$.

**Proof.** Let $s = [\{\Psi_n\}_{n=0}^\infty]$, $t = [\{\Phi_n\}_{n=0}^\infty]$, and $r = [\{\Theta_n\}_{n=0}^\infty]$. Since $s > t$, i.e. $s - t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have $\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\tilde{\mathbb{R}}^n}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\tilde{\mathbb{R}}^n}$ as $n \to \infty$, by the assumption that $s - t > 0_{\tilde{\mathbb{R}}^n}$. Thus, by Definition 14.2.41, this means that:

$$s + r = [\{\Psi_n + \Theta_n\}_{n=0}^\infty] > [\{\Phi_n + \Theta_n\}_{n=0}^\infty] = t + r.$$

**Definition 14.1.42.** There is canonical imbeding

$$\tilde{\mathbb{R}}^n \hookrightarrow \tilde{\mathbb{R}}^n \qquad (14.1.42)$$

defined by

$$a \mapsto \tilde{a} \qquad (14.1.43)$$

where $\tilde{a}$ is hyper infinite sequence $\tilde{a} = (a, a, \dots) \in \tilde{\mathbb{R}}^n, a \in \mathbb{R}^n$.

**Notation 14.1.5.** $\hat{a} = (a, a, \dots) \in \tilde{\mathbb{R}}^n, a \in \mathbb{R}^n$.

**Definition 14.1.43.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\tilde{\mathbb{R}}^n, \{a_n\}_{n=0}^k \subset \tilde{\mathbb{R}}^n$. We define external hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^k} \subset \tilde{\mathbb{R}}^n$ by

$$\{A_n; k\}_{n=0}^{\infty} = \overbrace{\{a_n\}_{n=0}^k}^{k} = \qquad (14.1.44) \\ = (a_0, a_1, \dots, a_m, \dots, a_{k-1}, a_k, \hat{a}_k).$$

(ii) Let $\{a_n\}_{n=0}^\infty$ be countable sequence in $\tilde{\mathbb{R}}^n : \{a_n\}_{n=0}^\infty \subset \tilde{\mathbb{R}}^n$.
Hence, we have $|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}|$.
Now, from Theorem 14.1.9, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in \mathbb{N}^#$. Taking some number $R$ (for example $R = M+L$) which is bigger than both, we have

$$|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}| \le R(|\Psi_n - \Psi_{1,n}| + |\Phi_n - \Phi_{1,n}|).$$

Now, noting that both $\Psi_n - \Psi_{1,n}$ and $\Phi_n - \Phi_{1,n}$ #-tend to $0_{\mathbb{R}^#}$ and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2R$, we see that

$$\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to_{\#} 0_{\mathbb{R}^#}$$

**Theorem 14.2.11.** Given any hyperreal number $s \in \tilde{\mathbb{R}}^#$, $s \ne 0_{\tilde{\mathbb{R}}^#}$, there is a hyperreal number $t \in \tilde{\mathbb{R}}^#$ such that $s \times t = 1_{\tilde{\mathbb{R}}^#}$ or $s \times t = \tilde{1}_{\tilde{\mathbb{R}}^#}$.

**Proof.** First we must properly understand what the theorem says. The premise is that $s$ is nonzero, which means that $s$ is not in the equivalence class of

$$0_{\tilde{\mathbb{R}}^#} = (0_{\mathbb{R}^#}, 0_{\mathbb{R}^#}, 0_{\mathbb{R}^#}, 0_{\mathbb{R}^#}, \dots). \qquad (14.1.40)$$

In other words, $s = [\{\Psi_n\}_{n=0}^{\infty}]$ where $\Psi_n - 0_{\tilde{\mathbb{R}}^#}$ does not #-converge to $0_{\tilde{\mathbb{R}}^#}$.

From this, we are to deduce the existence of a hyperreal number $t = [\{\Phi_n\}_{n=0}^{\infty}]$ such that $s \times t = [\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}]$ is the same equivalence class as $1_{\tilde{\mathbb{R}}^#} = [(1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, \dots)]$ or as some $\tilde{1}_{\tilde{\mathbb{R}}^#}$. Doing so is actually an easy consequence of the fact that nonzero hyperreal numbers from $\tilde{\mathbb{R}}^#$ have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{\Psi_n\}_{n=0}^{\infty}$ does not #-tend to $0_{\tilde{\mathbb{R}}^#}$ as $n \to \infty$), there's no reason any number of the terms in $\{\Psi_n\}_{n=0}^{\infty}$ can't equal $0_{\tilde{\mathbb{R}}^#}$. However, it turns out that eventually, $\Psi_n \ne 0_{\tilde{\mathbb{R}}^#}$.

That is,

**Lemma 14.1.2.** If $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence which does not #-tends to $0_{\tilde{\mathbb{R}}^#}$, then there is an $N \in \mathbb{N}^#$ such that, for $n > N$, $\Psi_n \ne 0_{\tilde{\mathbb{R}}^#}$. We will now use it to complete the proof of Theorem 14.2.11.

Let $N \in \mathbb{N}^#$ be such that $\Psi_n \ne 0_{\tilde{\mathbb{R}}^#}$ for $n > N$. Define hyper infinite sequence $\Phi_n$ of hyperreal numbers from $\tilde{\mathbb{R}}^#$ as follows:

for $n \le N$, $\Phi_n = 0_{\tilde{\mathbb{R}}^#}$, and for $n > N$, $\Phi_n = 1_{\tilde{\mathbb{R}}^#}/\Psi_n$ :

$$\{\Phi_n\}_{n=0}^{\infty} = (0_{\tilde{\mathbb{R}}^#}, 0_{\tilde{\mathbb{R}}^#}, \dots, 0_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}/\Psi_{N+1}, 1_{\tilde{\mathbb{R}}^#}/\Psi_{N+2}, \dots).$$

This makes sense since, for $n > N$, an is a nonzero hyperreal number, so $1_{\tilde{\mathbb{R}}^#}/\Psi_n$ exists.

Then $\Psi_n \times \Phi_n$ is equal to $\Psi_n \times 0_{\tilde{\mathbb{R}}^#} = 0_{\tilde{\mathbb{R}}^#}$ for $n \le N$, and equals

$$\Psi_n \times \Phi_n = \Psi_n \times 1_{\tilde{\mathbb{R}}^#}/\Psi_n = 1_{\tilde{\mathbb{R}}^#} \text{ for } n > N$$

Well, then, if we look at the hyper infinite sequence

$$1_{\tilde{\mathbb{R}}^#} = (1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, \dots), \qquad (14.1.41)$$

we have $(1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, 1_{\tilde{\mathbb{R}}^#}, \dots) - (\Psi_n \times \Phi_n)$ is the hyper infinite sequence which is $1_{\tilde{\mathbb{R}}^#} - 0_{\tilde{\mathbb{R}}^#} = 1_{\tilde{\mathbb{R}}^#}$ for $n \le N$ and equals $1_{\tilde{\mathbb{R}}^#} - 1_{\tilde{\mathbb{R}}^#} = 0_{\tilde{\mathbb{R}}^#}$ for $n > N$. Since this hyper infinite sequence is eventually equal to $0_{\tilde{\mathbb{R}}^#}$, it #-converges to $0_{\tilde{\mathbb{R}}^#}$ as $n \to \infty$,
We define hyper infinite sequence ${A_n}_{n=0}^{\infty} = \overbrace{\{a_n\}}^{\sim} \subset \tilde{\mathbb{R}}_c^\#$ by

$$ \begin{aligned} \{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}}^{\sim} = \\ &= (\hat{a}_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}}^{\sim}). \end{aligned} \quad (14.1.45) $$

(iii) Let ${a_n}_{n=0}^N, N \in \mathbb{N}^\# \mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$. We define hyper infinite sequence $\overbrace{\{a_n\}}^{\sim}$ by

$$ \begin{aligned} \{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}}^{\sim} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{N-1}a_N, \hat{a}_N). \end{aligned} \quad (14.1.46) $$

**Definition 14.1.44.** (i) Let ${a_n}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\tilde{\mathbb{R}}_c^\#$, ${a_n}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$. We define external finite sum $\text{Ext}-\sum_{n=0}^{n=k} a_n$ by

$$ \text{Ext}-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}}^{\sim} = (c_0, c_1, \dots, c_k, \hat{c}_k) \quad (14.1.47) $$

where $c_0 = a_0, c_k = \text{Ext}-\sum_{n=0}^{n=j} a_n, 0 \le j \le k$.

(ii) Let ${a_n}_{n=0}^\infty$ be countable sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^\infty \subset \tilde{\mathbb{R}}_c^\#$. We define external countable sum $\text{Ext}-\sum_{n=0}^{n=\infty} a_n$ by

$$ \begin{aligned} \text{Ext}-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}}^{\sim} = \\ &= (c_0, c_1, \dots, c_k, \dots, \{c_n\}_{n=0}^{\infty}, \overbrace{\{c_n\}}^{\sim}) \in [\overbrace{\{c_n\}}^{\sim}] \end{aligned} \quad (14.1.48) $$

where $c_0 = a_0, c_k = \text{Ext}-\sum_{n=0}^{n=k} a_n, k \in \mathbb{N}$.

(iii) Let ${a_n}_{n=0}^{n=N}, N \in \mathbb{N}^\# \mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$. We define external hyperfinite sum $\text{Ext}-\sum_{n=0}^{n=N} a_n$ by

$$ \text{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}}^{\sim} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \quad (14.1.49) $$

where $c_0 = a_0, c_k = \text{Ext}-\sum_{n=0}^{n=k} a_n, 0 \le k \le N, c_N = \text{Ext}-\sum_{n=0}^{n=N} a_n$.

(iv) Let ${a_n}_{n=0}^{n=N}, N \in \mathbb{N}^\#$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$ such that $a_n = 0$ for all $n \in \mathbb{N}^\# \mathbb{N}$. We assume that

$$ \text{Ext}-\sum_{n=0}^{n=N} a_n = \text{Ext}-\sum_{n=0}^{n=\infty} a_n. \quad (14.1.50) $$

**Example 14.1.3.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^\#$, $\alpha \in \tilde{\mathbb{R}}_c^\#$, $r \in \tilde{\mathbb{R}}_c^\#$ be the first term and the ratio of the G.P respectively. Then for any $N \in \mathbb{N}^\#$ by Proposition 14.1.6 and Definition 14.1.44 one obtains that

$$ \text{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{\overbrace{1-r^N}^{\tilde{\mathbb{R}}_c^\#}}{\overbrace{1-r}^{\tilde{\mathbb{R}}_c^\#}} = \alpha \frac{\overbrace{1-r}^{\tilde{\mathbb{R}}_c^\#}}{\overbrace{1-r}^{\tilde{\mathbb{R}}_c^\#}} - \alpha \frac{\overbrace{r^N}^{\tilde{\mathbb{R}}_c^\#}}{\overbrace{1-r}^{\tilde{\mathbb{R}}_c^\#}}. \quad (14.1.51) $$

and
and so $[\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}] = [(1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#},...)] = 1_{\tilde{\mathbb{R}}_c^\#} \in \tilde{\mathbb{R}}_c^\#$ and similarly

$[\{\Psi_n \times \Phi_n\}_{n=0}^{\infty}] = \tilde{1}_{\tilde{\mathbb{R}}_c^\#} \in \tilde{\mathbb{R}}_c^\#$. This shows that $t = [\{\Phi_n\}_{n=0}^{\infty}]$ is a multiplicative inverse (and similarly quasi inverse) to $s = [\{\Psi_n\}_{n=0}^{\infty}]$.

**Definition 14.2.41.** Let $s \in \tilde{\mathbb{R}}_c^\#$. Say that $s$ is positive if $s \neq 0_{\tilde{\mathbb{R}}_c^\#}$, and if $s = [\{\Psi_n\}_{n=0}^{\infty}]$ for some Cauchy hyper infinite sequence such that for some $N$, $\Psi_n > 0_{\tilde{\mathbb{R}}_c^\#}$ for all $n > N$. Given two hyperreal numbers $s, t \in \tilde{\mathbb{R}}_c^\#$, say that $s > t$ if $s - t$ is positive.

**Theorem 14.1.7.** Let $s, t \in \tilde{\mathbb{R}}_c^\#$ be hyperreal numbers such that $s > t$, and let $r \in \tilde{\mathbb{R}}_c^\#$. Then $s + r > t + r$.

**Proof.** Let $s = [\{\Psi_n\}_{n=0}^{\infty}]$, $t = [\{\Phi_n\}_{n=0}^{\infty}]$, and $r = [\{\Theta_n\}_{n=0}^{\infty}]$. Since $s > t$, i.e. $s - t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have $\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\tilde{\mathbb{R}}_c^\#}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\tilde{\mathbb{R}}_c^\#}$ as $n \to \infty$, by the assumption that $s - t > 0_{\tilde{\mathbb{R}}_c^\#}$. Thus, by Definition 14.2.41, this means that:

$$s + r = [\{\Psi_n + \Theta_n\}_{n=0}^{\infty}] > [\{\Phi_n + \Theta_n\}_{n=0}^{\infty}] = t + r.$$

**Definition 14.1.42.** There is canonical imbeding

$$\tilde{\mathbb{R}}_c^\# \hookrightarrow \tilde{\mathbb{R}}_c^\# \qquad (14.1.42)$$

defined by

$$a \mapsto \tilde{a} \qquad (14.1.43)$$

where $\tilde{a}$ is hyper infinite sequence $\tilde{a} = (a, a, \dots) \in \tilde{\mathbb{R}}_c^\#$, $a \in \tilde{\mathbb{R}}_c^\#$.

**Notation 14.1.5.** $\hat{a} = (a, a, \dots) \in \tilde{\mathbb{R}}_c^\#$, $a \in \tilde{\mathbb{R}}_c^\#$.

**Definition 14.1.43.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\tilde{\mathbb{R}}_c^\#$, $\{a_n\}_{n=0}^k \subset \tilde{\mathbb{R}}_c^\#$.

We define external hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^k} \subset \tilde{\mathbb{R}}_c^\#$ by

$$\{A_n; k\}_{n=0}^{\infty} = \overbrace{\{a_n\}_{n=0}^{k}}^{\infty} = \qquad (14.1.44) \\ = (a_0, a_1, \dots, a_m, \dots, a_{k-1}, a_k, \hat{a}_k).$$

(ii) Let $\{a_n\}_{n=0}^\infty$ be countable sequence in $\tilde{\mathbb{R}}_c^\#$: $\{a_n\}_{n=0}^\infty \subset \tilde{\mathbb{R}}_c^\#$.

We define hyper infinite sequence $\{A_n\}_{n=0}^{\infty} = \overbrace{\{a_n\}_{n=0}^{\infty}}^{\infty} \subset \tilde{\mathbb{R}}_c^\#$ by

$$\{A'_n; \infty\}_{n=0}^{\infty} = \overbrace{\{a_n\}_{n=0}^{\infty}}^{\infty} = \qquad (14.1.45) \\ = (\hat{a}_0, a_1, \dots, a_k, \dots \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}_{n=0}^{\infty}}^{\infty}).$$

(iii) Let $\{a_n\}_{n=0}^N, N \in \mathbb{N}^\# \mathbb{N}$ be external hyperfinite sequence in $\tilde{\mathbb{R}}_c^\#$: $\{a_n\}_{n=0}^N \subset \tilde{\mathbb{R}}_c^\#$.

We define hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^N} \subset \tilde{\mathbb{R}}_c^\#$ by
$$ \operatorname{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{\widehat{1_{\mathbb{R}_c^{\#}}}}{1_{\mathbb{R}_c^{\#}} - r} - \alpha \left\{ \frac{\widehat{r^n}}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty}. \quad (14.1.52) $$

**Example 14.1.4.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in \mathbb{N}^{\#}, \alpha \in \tilde{\mathbb{R}}_{c}^{\#}, r \in \tilde{\mathbb{R}}_{c}^{\#}, r < 0_{\tilde{\mathbb{R}}_{c}^{\#}}, |r| < 1$. Note that

$$ \begin{aligned} \alpha \frac{\widehat{1_{\mathbb{R}_c^{\#}}}}{1_{\mathbb{R}_c^{\#}} - r} &= \operatorname{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \\ &= \operatorname{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} + \operatorname{Ext}-\sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \\ &= \alpha \frac{\widehat{1_{\mathbb{R}_c^{\#}}}}{1_{\mathbb{R}_c^{\#}} - r} - \alpha \left\{ \frac{\widehat{r^n}}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty} + \operatorname{Ext}-\sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \end{aligned} \quad (14.1.53) $$

From (14.1.53) we obtain

$$ \begin{aligned} \operatorname{Ext}-\sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{\widehat{1_{\mathbb{R}_c^{\#}}}}{1_{\mathbb{R}_c^{\#}} - r} - \alpha \frac{\widehat{1_{\mathbb{R}_c^{\#}}}}{1_{\mathbb{R}_c^{\#}} - r} + \alpha \left\{ \frac{\widehat{r^n}}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty} = \\ & \qquad \alpha \left\{ \frac{\widehat{(-1_{\mathbb{R}_c^{\#}})}^n |r|^n}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty} - \alpha \frac{\widehat{r^N}}{1_{\mathbb{R}_c^{\#}} - r}. \end{aligned} \quad (14.1.54) $$

Assume that: (i) $r < 0_{\tilde{\mathbb{R}}_{c}^{\#}}, |r| < 1$ then from (14.1.54) we obtain

$$ \operatorname{Ext}-\sum_{(n \in \mathbb{N}^{\#}\mathbb{N}) \wedge (n \le N-1)} \alpha (-1_{\tilde{\mathbb{R}}_{c}^{\#}})^{n-1} |r|^{n-1} \neq 0_{\tilde{\mathbb{R}}_{c}^{\#}}. \quad (14.1.55) $$

## §14.2. External non-Archimedean field $\tilde{\mathbb{R}}_{c}^{\#}$ via special extension of non-Archimedean field $\mathbb{R}_{c}^{\#}$

**Notation 14.2.3.** Let $\Delta \subset \mathbb{R}_{c}^{\#}$ and $\Delta \neq \{0\}$. Then we write $\Delta > 0$ iff $a \in \Delta \Rightarrow a > 0$.

**Definition 14.2.13.** Let $\Delta \subset \mathbb{R}_{c}^{\#}$ and $\Delta > 0$. Assume that: $a, b \in \Delta \Rightarrow a + b \in \Delta$. Then we say that $\Delta$ is a positive idempotent in $\mathbb{R}_{c}^{\#}$.

**Notation 14.2.4.** We will denote by $\mathbb{R}_{c+,\text{fin}}^{\#}$ a set of the all positive finite number in $\mathbb{R}_{c}^{\#}$ except infinitesimals in $\mathbb{R}_{c}^{\#}$.

**Remark 14.2.6.** Note that a set $\mathbb{R}_{c+,\text{fin}}^{\#}\setminus\{0\} \subset \mathbb{R}_{c}^{\#}$ is a positive idempotent in $\mathbb{R}_{c}^{\#}$.

**Proposition 14.2.1.** Let $\Delta \subset \mathbb{R}_{c}^{\#}$ is a positive idempotent in $\mathbb{R}_{c}^{\#}$. Then the following are equivalent. [In what follows assume $a, b > 0$].

(i) $a \in \Delta \Rightarrow 2a \in \Delta$,

(ii) $a \in \Delta \Rightarrow na \in \Delta$ for all standard integers $n \in \mathbb{N}$,

(iii) $a \in \Delta \Rightarrow ra \in \Delta$ for all finite $r \in \mathbb{R}_{c+}^{\#}$.

**Proof.** All parts are immediate from the Definition 14.2.13.

**Notation 14.2.4.** $\Delta_{\approx}^{\#} \triangleq \{\delta \in \mathbb{R}_{c+}^{\#} | \delta > 0, \delta \approx 0\}$, i.e. $\Delta_{\approx}^{\#}$ is a set of the all positive infinitesimals in $\mathbb{R}_{c+}^{\#}$; $\Delta_{\approx}^{-} \triangleq \{\delta \in \mathbb{R}_{c+}^{\#} | \delta < 0, \delta \approx 0_{\tilde{\mathbb{R}}_{c}^{\#}}\}$, i.e. $\Delta_{\approx}^{-} = \Delta_{\approx}^{\#}$ is a set of the all negative infinitesimals in $\mathbb{R}_{c}^{\#}$. Note that $\Delta_{\approx}^{-} = -\Delta_{\approx}^{\#}$.
$$
\begin{align}
\{A_n; N\}_{n=0}^{*\infty} &= \overbrace{\{a_n\}_{n=0}^N}^{\overbrace{a_n}^{\overline{a_n}}} = \tag{14.1.46} \\
&= (a_0, a_1, \dots, a_m, \dots, a_{N-1}a_N, \widehat{a_N}). \nonumber
\end{align}
$$

**Definition 14.1.44.** (i) Let {$a_n$}$_{n=0}^k$, $k \in \mathbb{N}$ be finite sequence in $\widetilde{\mathbb{R}}_c^\#$, {$a_n$}$_{n=0}^N \subset \widetilde{\mathbb{R}}_c^\#$.
We define external finite sum $\mathrm{Ext}-\sum_{n=0}^{n-k} a_n$ by

$$
\mathrm{Ext}-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k}^{\overline{c_n}} = (c_0, c_1, \dots, c_k, \hat{c}_k) \quad (14.1.47)
$$

where $c_0 = a_0, c_j = \mathrm{Ext}-\sum_{n=0}^{n=j} a_n, 0 \le j \le k$.

(ii) Let {$a_n$}$_{n=0}^\infty$ be countable sequence in $\widetilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^\infty \subset \widetilde{\mathbb{R}}_c^\#$. We define external countable sum $\mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n$ by

$$
\begin{align}
\mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}_{n=0}^{\infty}}^{\overline{c_n}} = \nonumber \\
&= \left( c_0, c_1, \dots, c_k, \dots \{c_n\}_{n=0}^{\infty}, \overbrace{\{c_n\}_{n=0}^{\infty}}^{\overline{c_n}} \right) \in \left[ \overbrace{\{c_n\}_{n=0}^{\infty}}^{\overline{c_n}} \right] \tag{14.1.48}
\end{align}
$$

where $c_0 = a_0, c_k = \mathrm{Ext}-\sum_{n=0}^{n=k} a_n, k \in \mathbb{N}$.

(iii) Let {$a_n$}$_{n=0}^{n=N}$, $N \in *\mathbb{N}\mathbb{N}$ be external hyperfinite sequence in $^{*\widetilde{\mathbb{R}}_c^\#} : \{a_n\}_{n=0}^N \subset ^{*\widetilde{\mathbb{R}}_c^\#}$.
We define external hyperfinite sum $\mathrm{Ext}-\sum_{n=0}^{n=N} a_n$ by

$$
\mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{N}}^{\overline{c_n}} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \quad (14.1.49)
$$

where $c_0 = a_0, c_k = \mathrm{Ext}-\sum_{n=0}^{n=k} a_n, 0 \le k \le N, c_N = \mathrm{Ext}-\sum_{n=0}^{n=N} a_n$.

(iv) Let {$a_n$}$_{n=0}^{n=N}$, $N \in \mathbb{N}^\#$ be external hyperfinite sequence in $\widetilde{\mathbb{R}}_c^\#$ : {$a_n$}$_{n=0}^N \subset \widetilde{\mathbb{R}}_c^\#$ such that $a_n \equiv 0$ for all $n \in \mathbb{N}^\# \setminus \mathbb{N}$. We assume that

$$
\mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n. \qquad (14.1.50)
$$

**Example 14.1.3.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in \mathbb{N}^{\#}, \alpha \in \widetilde{\mathbb{R}}_{c}^{\#}$,
$r \in \widetilde{\mathbb{R}}_{c}^{\#}$ be the first term and the ratio of the G.P respectively. Then for any
$N \in \mathbb{N}^{\#}$ by Proposition 14.1.6 and Definition 14.1.44 one obtains that

$$
\mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{\overline{1_{\mathbb{R}_c^\#} - r^N}}{\overline{1_{\mathbb{R}_c^\#} - r}} = \alpha \frac{\overline{1_{\mathbb{R}_c^\#}}}{\overline{1_{\mathbb{R}_c^\#} - r}} - \alpha \frac{\overline{r^N}}{\overline{1_{\mathbb{R}_c^\#} - r}}. \quad (14.1.51)
$$

and

$$
\mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{\overline{1_{\mathbb{R}_c^\#}}}{\overline{1_{\mathbb{R}_c^\#} - r}} - \alpha \left\{ \frac{\overline{r^n}}{\overline{1_{\mathbb{R}_c^\#} - r}} \right\}_{n=1}^{\infty}. \quad (14.1.52)
$$

**Example 14.1.4.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in \mathbb{N}^{\#}, \alpha \in \widetilde{\mathbb{R}}_{c}^{\#}, r \in \widetilde{\mathbb{R}}_{c}^{\#}, r < 0_{\widetilde{\mathbb{R}}_{c}^{\#}}, |r| < 1$. Note that
**Remark 14.2.7.** Note that a set $\Delta_{\approx}^{\#+} \subset *\mathbb{R}_c^{\#}$ is a positive idempotent in $*\mathbb{R}_c^{\#}$ and $\Delta_{\approx}^{-\#}$ is a negative idempotent in $*\mathbb{R}_c^{\#}$.

**Definition 14.2.14.** Let $\{a_n\}_{n=0}^{\infty}$ be $*\mathbb{R}_{c+,fin}^{\#}$-valued countable sequence $a : \mathbb{N} \to *\mathbb{R}_{c+,fin}^{\#}$ such that:

(i) there is $M \in \mathbb{N}$ such that $\{a_n\}_{n=M}^{\infty}$ is monotonically decreasing $*\mathbb{R}_{c+,fin}^{\#}$-valued countable sequence $a : \mathbb{N} \to *\mathbb{R}_{c+,fin}^{\#} \setminus \{0_{\mathbb{R}_c^{\#}}\}$

(ii) there is $M \in \mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{*\mathbb{R}_c^{\#}}$ [it follows from (ii)]

(iii) for all $n \in \mathbb{N}$, $a_n \not\approx 0_{*\mathbb{R}_c^{\#}}$ and for any $\epsilon > 0$, $\epsilon \not\approx 0_{*\mathbb{R}_c^{\#}}$, $\epsilon \in *\mathbb{R}_{c+,fin}^{\#}$ there is $N \in \mathbb{N}$ such that for all $n > N$ : $a_n < \epsilon$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\epsilon}$.

We define a set $\Delta_{\omega}^{-10}$ by $c_n \in \Delta_{\omega}^{-10} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$. Note that $\Delta_{\omega}^{-10} = -\Delta_{\omega}^{+10}$.

**Remark 14.2.8.** Note that a set $\Delta_{\omega}^{+10}$ is a positive idempotent in $*\mathbb{R}_c^{\#}$ and a set $\Delta_{\omega}^{-10}$ is a negative idempotent in $*\mathbb{R}_c^{\#}$.

**Proposition 14.2.2.** (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then:

$$ (i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} $$

$$ (ii) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \cup \Delta_{\omega}^{-10} \cup \Delta_{\approx}^{+1} \cup \Delta_{\approx}^{-1} \cup \{0_{*\mathbb{R}_c^{\#}}\}_{n=0}^{\infty} $$ 
where $\{0_{*\mathbb{R}_c^{\#}}\}_{n=0}^{\infty}$ is a countable $0_{*\mathbb{R}_c^{\#}}$-valued sequence.

$$ (iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} $$

(2) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then we define

$$ (i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10} $$

$$ (ii) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \cup \Delta_{\omega}^{-10} $$

$$ (iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} $$

(3) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \cup \Delta_{\omega}^{-10}$ and $x, y \in \mathbb{R}_c^{\#}$ then we define

$$ (iv) x + y \{a_n\}_{n=0}^{\infty} \triangleq \{x + ya_n\}_{n=0}^{\infty} $$

**Proof.** Immediately by definitions and by Definition 14.2.14.

**Definition 14.2.15.** We define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{+10}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N$ : $a_n \le b_n$ and similarly we define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{-10}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N$ : $a_n \le b_n$

**Definition 14.2.16.** (1) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega}^{+1} \times *\mathbb{R}_{c+,fin}^{\#}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $x \in *\mathbb{R}_{c+,fin}^{\#}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N$ : $a_n < x$.

(2) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega}^{+1} \times \Delta_{\omega}^{+1}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $x \in \Delta_{\omega}^{+1}$ then $x < \{a_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N$ : $x < a_n$.

(3) Let $\{a_n\}_{n=0}^{\infty}$ be $\Delta_{\approx}^{+1}$-valued countable sequence $a : \mathbb{N} \to \Delta_{\approx}^{+1}$, and we denote a set of the all these sequences by $\Delta_{\approx,\omega}^{+1}$.

We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\approx,\omega}^{+1} \times \Delta_{\omega}^{+1}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\approx,\omega}^{+1}$ and $x \in \Delta_{\approx}^{+1}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N$ : $a_n < x$.

**Proposition 14.2.2.** Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \{a_n\}_{n=0}^{\infty} \neq 0_{*\mathbb{R}_c^{\#}}$ then there is $N \in \mathbb{N}$ such that $0_{*\mathbb{R}_c^{\#}} < \Delta_{\approx}^{+1} < \{a_n\}_{n=0}^{\infty} < *\mathbb{R}_{c+,fin}^{\#} \setminus \{0_{*\mathbb{R}_c^{\#}}\}$.

**Proof.** Immediately by definitions and by Definition 14.2.16.

**Remark 14.2.9.** Note that it follows from Proposition 14.2.2 that
$$
\begin{align}
\alpha \frac{\widehat{1_{\mathbb{R}_c^#} - r^N}}{1_{\mathbb{R}_c^#} - r} &= \mathrm{Ext} - \sum_{n=1}^{n=N-1} \alpha r^{n-1} = \nonumber \\
&= \mathrm{Ext} - \sum_{n=1}^{\infty} \alpha r^{n-1} + \mathrm{Ext} - \sum_{(n \in \mathbb{N}^# \mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \tag{14.1.53} \\
&= \alpha \frac{\widehat{1_{\mathbb{R}_c^#}}}{1_{\mathbb{R}_c^#} - r} - \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext} - \sum_{(n \in \mathbb{N}^# \mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \nonumber
\end{align}
$$

From (14.1.53) we obtain

$$
\begin{equation} \tag{14.1.54}
\begin{split}
\mathrm{Ext} - \sum_{(n \in \mathbb{N}^# \mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{1 - \widehat{\mathbb{R}_c^#} r^N}{1_{\mathbb{R}_c^#} - r} - \alpha \frac{1}{\widehat{\mathbb{R}_c^#} - r} + \alpha \left\{ \frac{r^n}{1_{\widehat{\mathbb{R}_c^#}} - r} \right\}_{n=1}^{\infty} \\
&\quad - \alpha \frac{\widehat{r^N}}{1_{\widehat{\mathbb{R}_c^#}} - r}.
\end{split}
\end{equation}
$$

Assume that: (i) $r < 0_{\mathbb{R}_c^#}$, $|r| < 1$ then from (14.1.54) we obtain

$$
\mathrm{Ext} - \sum_{(n \in \mathbb{N}^# \mathbb{N}) \wedge (n \le N-1)} \alpha (-1_{\mathbb{R}_c^#})^{n-1} |r|^{n-1} \ne 0_{\mathbb{R}_c^#} \quad (14.1.55)
$$

§14.2. External non-Archimedean field *R\#c via special extension of non-Archimedean field *R\#c

**Notation 14.2.3.** Let $\Delta \subset *\mathbb{R}_c^\#$ and $\Delta \ne \{0\}$. Then we write $\Delta > 0$ iff $a \in \Delta \Rightarrow a > 0$.

**Definition 14.2.13.** Let $\Delta \subset *\mathbb{R}_c^\#$ and $\Delta > 0$. Assume that: $a, b \in \Delta \Rightarrow a + b \in \Delta$. Then we say that $\Delta$ is a positive idempotent in $*\mathbb{R}_c^\#$.

**Notation 14.2.4.** We will denote by $*\mathbb{R}_{c+,\text{fin}}^\#$ a set of the all positive finite number in $\mathbb{R}_c^\#$ except infinitesimals in $*\mathbb{R}_c^\#$.

**Remark 14.2.6.** Note that a set $*\mathbb{R}_{c+,\text{fin}}^\# \setminus \{0\} \subset *\mathbb{R}_c^\#$ is a positive idempotent in $*\mathbb{R}_c^\#$.

**Proposition 14.2.1.** Let $\Delta \subset *\mathbb{R}_c^\#$ is a positive idempotent in $*\mathbb{R}_c^\#$. Then the following are equivalent. [In what follows assume $a, b > 0$].

(i) $a \in \Delta \Rightarrow 2a \in \Delta$,

(ii) $a \in \Delta \Rightarrow na \in \Delta$ for all standard integers $n \in \mathbb{N}$,

(iii) $a \in \Delta \Rightarrow ra \in \Delta$ for all finite $r \in *\mathbb{R}_{c+}^\#$.

**Proof.** All parts are immediate from the Definition 14.2.13.

**Notation 14.2.4.** $\Delta_{\approx}^{\#+} \triangleq \{\delta \in *\mathbb{R}_{c+}^{\#} | \delta > 0, \delta \approx 0\}$, i.e. $\Delta_{\approx}^{\#+}$ is a set of the all positive infinitesimals in $*\mathbb{R}_{c+}^{\#}$; $\Delta_{\approx}^{\#-} \triangleq \{\delta \in *\mathbb{R}_{c+}^{\#} | \delta < 0, \delta \approx 0_{\mathbb{R}_c^{\#}}\}$, i.e. $\Delta_{\approx}^{\#-} = \Delta_{\approx}^{\#+}$ is a set of the all negative infinitesimals in $*\mathbb{R}_c^{\#}$. Note that $\Delta_{\approx}^{\#-} = -\Delta_{\approx}^{\#+}$.

**Remark 14.2.7.** Note that a set $\Delta_{\approx}^{\#+} \subset *\mathbb{R}_c^\#$ is a positive idempotent in $*\mathbb{R}_c^\#$ and $\Delta_{\approx}^{\#-}$ is a negative idempotent in $*\mathbb{R}_c^\#$.

**Definition 14.2.14.** Let $\{a_n\}_{n=0}^{\infty}$ be $*\mathbb{R}_{c+,\text{fin}}^\#$-valued countable sequence
$a : \mathbb{N} \to *\mathbb{R}_{c+,\text{fin}}^\#$ such that:

(i) there is $M \in \mathbb{N}$ such that $\{a_n\}_{n=M}^{\infty}$ is monotonically decreasing $*\mathbb{R}_{c+,\text{fin}}^\#$$-\text{valued}
$$0_{*\mathbb{R}_c^#} < \Delta_{\approx}^{\#+} < \Delta_{\omega}^{+\downarrow 0} < *\mathbb{R}_{c+,\text{fin}}^# \setminus \{0_{*\mathbb{R}_c^#}\}. \quad (14.2.9)$$

**Definition 14.2.17.** Let {$a_n$}$_{n=0}^\infty$ be monotonically increasing *$\mathbb{R}_{c+,\text{fin}}^#$* - valued countable sequence $a: \mathbb{N} \to *\mathbb{R}_{c+,\text{fin}}^# \Delta_{\approx}^\#$ such that:

(i) there is $M \in \mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{*\mathbb{R}_c^#}$

(ii) there is $N \in \mathbb{N}$ such that for all $n > N$ and for any $\xi > 0_{*\mathbb{R}_c^#}$, $\xi \in *\mathbb{R}_{c+,\text{fin}}^#$, $a_n > \xi$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\downarrow\infty}$. We define a set $\Delta_{\omega}^{-\downarrow\infty}$ by $c_n \in \Delta_{\omega}^{-\downarrow\infty} \Leftrightarrow \{-c_n\}_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty}$. Note that $\Delta_{\omega}^{-\downarrow\infty} = -\Delta_{\omega}^{+\downarrow\infty}$.

**Proposition 14.2.3. (1)** Let {$a_n$}$_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow\infty}$ and {$b_n$}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty}$ then we define

(i) {$a_n$}$_{n=0}^\infty + \{b_n\}$_{n=0}^\infty \triangleq \{a_n + b_n\}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty}$

(ii) {$a_n$}$_{n=0}^\infty - \{b_n\}$_{n=0}^\infty \triangleq \{a_n - b_n\}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty} \cup \Delta_{\omega}^{-\downarrow\infty} \cup \Delta_{\approx}^{\#+} \cup \Delta_{\approx}^{\#-} \cup \Delta_{\omega}^{+\downarrow 0} \cup \cup \Delta_{\omega}^{-\downarrow 0} \cup \{0_{*\mathbb{R}_c^#}\}$_{n=0}^\infty$ where {$0_{*\mathbb{R}_c^#}$}$_{n=0}^\infty$ is a countable $0_{*\mathbb{R}_c^#}$- valued sequence.

(iii) {$a_n$}$_{n=0}^\infty \times \{b_n\}$_{n=0}^\infty \triangleq \{a_n \times b_n\}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty}$.

(2) Let {$a_n$}$_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow\infty}$ and {$b_n$}$_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow\infty}$ then we define

(i) {$a_n$}$_{n=0}^\infty + \{b_n\}$_{n=0}^\infty \triangleq \{a_n + b_n\}$_{n=0}^\infty \in \Delta_{\omega}^{-\downarrow\infty}$

(ii) {$a_n$}$_{n=0}^\infty - \{b_n\}$_{n=0}^\infty \triangleq \{a_n - b_n\}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty} \cup \Delta_{\omega}^{-\downarrow\infty}$

(iii) {$a_n$}$_{n=0}^\infty \times \{b_n\}$_{n=0}^\infty \triangleq \{a_n \times b_n\}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty}$.

(3) Let {$a_n$}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty}$ and $x, y \in \mathbb{R}_c^#$ then we define

(iv) $x_n + y_n \{a_n\}_{n=0}^\infty \triangleq \{x_n + y_n a_n\}_{n=0}^\infty$ and we denote a set of the all these sequences by $\{\Delta_{\omega}^{+\downarrow\infty}, \{x_n\}_{n=0}^\infty, \{y_n\}_{n=0}^\infty\}$.

**Proof.** Immediately by definitions and by Definition 14.2.16.

**Remark 14.2.10.** Note that {$a_n$}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow\infty} \Leftrightarrow \{a_n^{-1}\}$_{n=N}^\infty \in \Delta_{\omega}^{+\downarrow 0}$.

**Definition 14.2.18. (1)** Let {$a_n$}$_{n=0}^\infty \in \Delta_{\omega}^{+\downarrow 0}$ and let {$A_n$}$_{n=0}^{*\infty} = \overline{\{a_n\}$_{n=0}^\infty}$ be a hyper infinite sequence

$$\{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots) \quad (14.2.10)$$

i.e. for any infinite $m \in *\mathbb{N}\mathbb{N}$, $A_m \equiv \{a_n\}_{n=0}^\infty$. We will denote a set of the all these hyper infinite sequences by $\overline{\Delta_{\omega}^{+\downarrow 0}}$ and a set of the all hyper infinite sequences

$\overline{\{-a_n\}$_{n=0}^\infty}$ by $\overline{\Delta_{\omega}^{-\downarrow 0}}$. (2) Let {$x_n + y_n a_n$}$_{n=0}^\infty \in \{\Delta_{\omega}^{+\downarrow 0}, \{x_n\}_{n=0}^\infty, \{y_n\}_{n=0}^\infty\}$ and let

$$\{x_n + y_n A_n\}_{n=0}^{*\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} \qquad (14.2.11) \\ (x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, x_k + y_k a_k, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots),$$

i.e. for any infinite $m \in *\mathbb{N}\mathbb{N}$, $A_m \equiv \{x_n + y_n a_n\}_{n=0}^\infty$. We will denote a set of the all these hyper infinite sequences by $\overline{\{\Delta_{\omega}^{+\downarrow 0}, \{x_n\}_{n=0}^\infty, \{y_n\}_{n=0}^\infty\}}$.

**Definition 14.2.19.** Let {$A_n$}$_{n=0}^{*\infty} = \overline{\{a_n\}$_{n=0}^\infty}$ and {$B_n$}$_{n=0}^{*\infty} = \overline{\{b_n\}$_{n=0}^\infty}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$.

Then we define:

(i) {$A_n$}$_{n=0}^* + \{B_n\}^* = \overline{\{a_n\}^*_{n=0}} + \overline{\{b_n\}^*_{n=0}} \triangleq \overline{\{a_n + b_n\}^*_{n=0}} = \{A_n + B_n\}^*_{n=0} \in \overline{\Delta_{\omega}^{+\downarrow 0}}

(ii) {$A_n$}$_{n=0}^* - \{B_n\}^* = \overline{\{a_n\}^*_{n=0}} - \overline{\{b_n\}^*_{n=0}} \triangleq \overline{\{a_n - b_n\}^*_{n=0}} = \{A_n - B_n\}^*_{n=0} \in \overline{\Delta_{\omega}^{+\downarrow 0}} \cup \overline{\Delta_{\omega}^{-\downarrow 0}} \cup \{0_{*\mathbb{R}_c^#}\}^*_{n=0}

(iii) {$A_n$}$_{n=0}^* \times \{B_n\}^* = \overline{\{a_n\}^*_{n=0}} \times \overline{\{b_n\}^*_{n=0}} \triangleq \overline{\{a_n \times b_n\}^*_{n=0}} = \{A_n \times B_n\}^*_{n=0} \in \overline{\Delta_{\omega}^{+\downarrow 0}}
countable sequence $a : \mathbb{N} \to ^{*}\mathbb{R}_{c+,fin}^{\#} \setminus \{0_{\mathbb{R}^{\#}}\}$

(ii) there is $M \in \mathbb{N}$ such that for all $n > M, a_n \neq 0_{*\mathbb{R}^\#_c}$

(iii) for all $n \in \mathbb{N}$, $a_n \not\approx 0_{*\mathbb{R}^\#_c}$ and for any $\epsilon$ such that $\epsilon > 0$, $\epsilon \not\approx 0_{\mathbb{R}^\#_c}$, $\epsilon \in ^*\mathbb{R}^\#_{c+,fin}$ there is $N \in \mathbb{N}$ such that for all $n > N : a_n < \epsilon$ and we denote a set of the all these countable sequences by $\Delta_{\omega}^{+10}$.

We define a set $\Delta_{\omega}^{-10}$ by $c_n \in \Delta_{\omega}^{-10} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$. Note that $\Delta_{\omega}^{-10} = -\Delta_{\omega}^{+10}$.

**Remark 14.2.8.** Note that a set $\Delta_{\omega}^{+10}$ is a positive idempotent in $^*\mathbb{R}^\#_c$ and a set $\Delta_{\omega}^{-10}$ is a negative idempotent in $^*\mathbb{R}^\#_c$.

**Proposition 14.2.2.** (1) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then:

(i) $\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$

(ii) $\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \cup \Delta_{\omega}^{-10} \cup \Delta_{\approx}^{+1} \cup \Delta_{\approx}^{-1} \cup \{0_{*\mathbb{R}^\#_c}\}_{n=0}^{\infty}$
where $\{0_{*\mathbb{R}^\#_c}\}_{n=0}^{\infty}$ is a countable $0_{*\mathbb{R}^\#_c}$-valued sequence.

(iii) $\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$.

(2) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then we define

(i) $\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$

(ii) $\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \cup \Delta_{\omega}^{-10}$

(iii) $\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$

(3) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10} \cup \Delta_{\omega}^{+10}$ and $x, y \in \mathbb{R}^\#_c$ then we define

(iv) $x + y\{a_n\}_{n=0}^{\infty} \triangleq \{x + ya_n\}_{n=0}^{\infty}$

**Proof.** Immediately by definitions and by Definition 14.2.14.

**Definition 14.2.15.** We define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{+10}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N : a_n \le b_n$ and similarly we define the relation $(\cdot \le \cdot)$ on a set $\Delta_{\omega}^{-10}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-10}$ then $\{a_n\}_{n=0}^{\infty} \le \{b_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N : a_n \le b_n$

**Definition 14.2.16.** (1) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\omega}^{+1} \times ^*\mathbb{R}^\#_{c+,fin}$ by:

let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $x \in ^*\mathbb{R}^\#_{c+,fin}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N : a_n < x$.

(2) We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\approx}^{+1} \times \Delta_{\omega}^{+1}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10}$ and $x \in \Delta_{\approx}^{+1}$ then $x < \{a_n\}_{n=0}^{\infty}$ iff there is $N \in \mathbb{N}$ such that for all $n > N : x < a_n$.

(3) Let $\{a_n\}_{n=0}^{\infty}$ be $\Delta_{\approx}^{+1}$-valued countable sequence $a : \mathbb{N} \to \Delta_{\approx}^{+1}$, and we denote a set of the all these sequences by $\Delta_{\approx,\omega}^{+1}$.

We define the relation $(\cdot < \cdot)$ on a set $\Delta_{\approx,\omega}^{+1} \times \Delta_{\omega}^{+1}$ by: let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\approx,\omega}^{+1}$ and $x \in \Delta_{\approx}^{+1}$ then $\{a_n\}_{n=0}^{\infty} < x$ iff there is $N \in \mathbb{N}$ such that for all $n > N : a_n < x$.

**Proposition 14.2.2.** Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+10} \{a_n\}_{n=0}^{\infty} \neq 0_{*\mathbb{R}^\#_c}$ then there is $N \in \mathbb{N}$ such that $0_{*\mathbb{R}^\#_c} < \Delta_{\approx}^{+1} < \{a_n\}_{n=0}^{\infty} < ^*\mathbb{R}^\#_{c+,fin} \setminus \{0_{*\mathbb{R}^\#_c}\}$.

**Proof.** Immediately by definitions and by Definition 14.2.16.

**Remark 14.2.9.** Note that it follows from Proposition 14.2.2 that

$$0_{*\mathbb{R}^\#_c} < \Delta_{\approx}^{+1} < \Delta_{\omega}^{+10} < ^*\mathbb{R}^\#_{c+,fin} \setminus \{0_{*\mathbb{R}^\#_c}\}. \quad (14.2.9)$$

**Definition 14.2.17.** Let $\{a_n\}_{n=0}^{\infty}$ be monotonically increasing $^*\mathbb{R}^\#_{c+,fin}$-valued countable sequence $a : \mathbb{N} \to ^*\mathbb{R}^\#_{c+,fin} \Delta_{\approx}^{+1}$ such that:

(i) there is $M \in \mathbb{N}$ such that for all $n > M, a_n \neq 0_{*\mathbb{R}^\#_c}$
Let {$A_n$}$_{n=0}^{\infty}$ = {$\overline{a_n}$}$_{n=0}^{\infty}$ and {$B_n$}$_{n=0}^{\infty}$ = {$\overline{b_n}$}$_{n=0}^{\infty}$ be in {$\overline{\Delta_{\omega}^{+\downarrow 0}}$, {$x_{1,n}$}$_{n=0}^{\infty}$, {$y_{1n}$}$_{n=0}^{\infty}$, } and
{$B_n$}$_{n=0}^{\infty}$ = {$\overline{b_n}$}$_{n=0}^{\infty}$ be in {$\overline{\Delta_{\omega}^{+\downarrow 0}}$, {$x_{2,n}$}$_{n=0}^{\infty}$, {$y_{2,n}$}$_{n=0}^{\infty}$, } . Then we define:

$$
\begin{align*}
\text{(iv)} \quad \{A_n\}_{n=0}^{\infty} + \{B_n\}_{n=0}^{\infty} &= \overline{\{x_{1,n} + y_{1,n}a_n\}_{n=0}^{\infty} + \{x_{2,n} + y_{2,n}b_n\}_{n=0}^{\infty}} \\
&\triangleq \overline{\{x_{1,n} + x_{2,n} + y_{1,n}a_n + y_{2,n}b_n\}_{n=0}^{\infty}} = \{x_{1,n} + \overline{x_{2,n} + y_{1,n}A_n + y_{2,n}B_n}\}_{n=0}^{\infty}
\end{align*}
$$

**Definition 14.2.20.** Let {$\Psi_n$}$_{n=0}^{\infty}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$, i.e. for all $n \in *\mathbb{N}$, $\Psi_n \in \overline{\Delta_{\omega}^{+\downarrow 0}}$. Say {$\Psi_n$}$_{n=0}^{\infty}$ #-tends to $0_{*\mathbb{R}_c^#}$ as $n \to *\infty$ iff for any given $\varepsilon > 0_{*\mathbb{R}_c^#}$, $\varepsilon \approx 0_{*\mathbb{R}_c^#}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.2.21.** Let {$\Psi_n$}$_{n=0}^{\infty}$ be a hyper infinite sequence such that for all $n \in *\mathbb{N}$, $\Psi_n \in \overline{\Delta_{\omega}^{+\downarrow 0}}$. We call {$\Psi_n$}$_{n=0}^{\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to 0$_{*\mathbb{R}_c^#}$. To be precise: given any $\varepsilon > 0_{*\mathbb{R}_c^#}$, $\varepsilon \approx 0_{*\mathbb{R}_c^#}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.2.3.** Let {$\Psi_n$}$_{n=0}^{\infty}$ be in $\overline{\Delta_{\omega}^{+\downarrow 0}}$. If {$\Psi_n$}$_{n=0}^{\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_* \Phi$ as $n \to *\infty$ for some $\Phi \in \overline{\Delta_{\omega}^{+\downarrow 0}}$), then {$\Psi_n$}$_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Proof.**cWe know that $\Psi_n \to_* \Phi$. Here is a ubiquitous trick: instead of using $\varepsilon$ in the definition, start with an arbitrary infinitesimal $\varepsilon > 0_{*\mathbb{R}_c^#}$, $\varepsilon \approx 0_{*\mathbb{R}_c^#}$ and then choose $N$ so that $|\Psi_n - \Phi| < \varepsilon/2$ when $n > N$. Then if $m, n > N$, we have
$|\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.
This shows that {$\Psi_n$}$_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Theorem 14.2.4.** If {$\Psi_n$}$_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence, then it is bounded in $*\mathbb{R}_c^#$; that is, there is some number $M \in *\mathbb{R}_c^#$ such that $|\{\Psi_n\}_{n=0}^{\infty}| \le M$ for all $n \in *\mathbb{N}$.

**Proof.** Since {$\Psi_n$}$_{n=0}^{\infty}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N$ such that $|\Psi_m - \Psi_n| < 1$ whenever $m, n > N$. Thus, $|\Psi_{N+1} - \Psi_n| < 1$ for $n > N$. We can rewrite this as $\Psi_{N+1} - 1 < \Psi_n < \Psi_{N+1} + 1$. This means that $|\Psi_n|$ is less than the maximum of $|\Psi_{N+1} - 1|$ and $|\Psi_{N+1} + 1|$. So, set $M \in *\mathbb{R}_c^#$ larger than any number in the following list: $\{|\Psi_0|, |\Psi_1|, \dots, |\Psi_N|, |\Psi_{N+1} - 1|, |\Psi_{N+1} + 1|\}$.
Then for any term $\Psi_n$, if $n \le N$, then $|\Psi_n|$ appears in the list and so $|\Psi_n| \le M$; if $n > N$, then (as shown above) $|\Psi_n|$ is less than at least one of the last two entries in the list, and so $|\Psi_n| \le M$. Hence, $M$ is a bound for the sequence.

Let $\Xi$ denote the set of all Cauchy hyper infinite sequences We must define an
equivalence relation on $\Xi$.

**Definition 14.2.22.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

**Reflexivity:** for any $s \in S$, $s$ is related to $s$.

**Symmetry:** for any $s, t \in S$, if $s$ is related to $t$ then $t$ is related to $s$.

**Transitivity:** for any $s, t, r \in S$, if $s$ is related to $t$ and $t$ is related to $r$, then $s$ is related to $r$.

The following well known proposition goes most of the way to showing that an
equivalence relation divides a set into bins.
(ii) there is $N \in \mathbb{N}$ such that for all $n > N$ and for any $\xi > 0_{*\mathbb{R}^c^#}, \xi \in *\mathbb{R}^c^#_{c+,-\text{fin}}$, $a_n > \xi$ and we denote a set of the all these sequences by $\Delta_{\omega}^{+\downarrow\infty}$. We define a set $\Delta_{\omega}^{-\downarrow\infty}$ by $c_n \in \Delta_{\omega}^{-\downarrow\infty} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}$. Note that $\Delta_{\omega}^{-\downarrow\infty} = -\Delta_{\omega}^{+\downarrow\infty}$.

**Proposition 14.2.3. (1)** Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty}$ then we define

$$ (i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} $$

$$ (ii) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \cup \Delta_{\omega}^{-\downarrow\infty} \cup \Delta_{\omega}^{\#\downarrow} \cup \Delta_{\omega}^{\#\downarrow-} \cup \Delta_{\omega}^{+\downarrow0} \cup \Delta_{\omega}^{-\downarrow0} \cup \{0_{*\mathbb{R}^c^#}\}_{n=0}^{\infty} \text{ where } \{0_{\mathbb{R}^c^#}\}_{n=0}^{\infty} \text{ is a countable } 0_{\mathbb{R}^c^#} \text{-valued sequence.} $$

$$ (iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}. $$

(2) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty}$ and $\{b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty}$ then we define

$$ (i) \{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty} \triangleq \{a_n + b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{-\downarrow\infty} $$

$$ (ii) \{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty} \triangleq \{a_n - b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \cup \Delta_{\omega}^{-\downarrow\infty} $$

$$ (iii) \{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty} \triangleq \{a_n \times b_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} $$

(3) Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty}$ and $x, y \in \mathbb{R}^c^#$, then we define

$$ (iv) x + y\{a_n\}_{n=0}^{\infty} \triangleq \{x + ya_n\}_{n=0}^{\infty} \text{ and we denote a set of the all these sequences by } \{\Delta_{\omega}^{+\downarrow\infty}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}. $$

**Proof.** Immediately by definitions and by Definition 14.2.16.

**Remark 14.2.10.** Note that $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow\infty} \Leftrightarrow \{a_n^{-1}\}_{n=N}^{\infty} \in \Delta_{\omega}^{+\downarrow0}$.

**Definition 14.2.18. (1)** Let $\{a_n\}_{n=0}^{\infty} \in \Delta_{\omega}^{+\downarrow0}$ and let $\{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ be a hyper infinite sequence

$$ \{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots) \quad (14.2.10) $$

i.e. for any infinite $m \in *\mathbb{N}\mathbb{N}$, $A_m \equiv \{a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\overline{\Delta_{\omega}^{+\downarrow0}}$ and a set of the all hyper infinite sequences

$\overline{\{-a_n\}_{n=0}^{\infty}}$ by $\overline{\Delta_{\omega}^{-\downarrow0}}$. (2) Let $\{x_n + y_n a_n\}_{n=0}^{\infty} \in \{\Delta_{\omega}^{+\downarrow0}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}$ and let

$$ \{x_n + y_n A_n\}_{n=0}^{*\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} \quad (14.2.11) \\ (x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, x_k + y_k a_k, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots), $$

i.e. for any infinite $m \in *\mathbb{N}\mathbb{N}$, $A_m \equiv \{x_n + y_n a_n\}_{n=0}^{\infty}$. We will denote a set of the all these hyper infinite sequences by $\overline{\{\Delta_{\omega}^{+\downarrow0}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}}$.

**Definition 14.2.19.** Let $\{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ and $\{B_n\}_{n=0}^{*\infty} = \overline{\{b_n\}_{n=0}^{\infty}}$ be in $\overline{\Delta_{\omega}^{+\downarrow0}}$.

Then we define:

$$ (i) \{A_n\}_{n=0}^{*\infty} + \{B_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n + b_n\}_{n=0}^{\infty}} = \{A_n + B_n\}_{n=0}^{*\infty} \in \overline{\Delta_{\omega}^{+\downarrow0}} $$

$$ (ii) \{A_n\}_{n=0}^{*\infty} - \{B_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty}} \triangleq \{a_n - b_n\}_{n=0}^{\infty} = \\ = \{A_n - B_n\}_{n=0}^{*\infty} \in \overline{\Delta_{\omega}^{+\downarrow0} \cup \Delta_{\omega}^{-\downarrow0} \cup \{0_{*\mathbb{R}^c^#}\}_{n=0}^{*\infty}} $$

$$ (iii) \{A_n\}_{n=0}^{*\infty} \times \{B_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty} \times \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n \times b_n\}_{n=0}^{\infty}} = \{A_n \times B_n\}_{n=0}^{*\infty} \in \overline{\Delta_{\omega}^{+\downarrow0}} \\ \text{Let } \{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}} \text{ and } \{B_n\}_{n=0}^{*\infty} = \overline{\{b_n\}_{n=0}^{\infty}} \text{ be in } \{\Delta_{\omega}^{+\downarrow0}, \{x_{1,n}\}_{n=0}^{\infty}, \{y_{1,n}\}_{n=0}^{\infty}\} \text{ and } \{B_n\}_{n=0}^{*\infty} = \overline{\{b_n\}_{n=0}^{\infty}} \text{ be in } \{\Delta_{\omega}^{+\downarrow0}, \{x_{2,n}\}_{n=0}^{\infty}, \{y_{2,n}\}_{n=0}^{\infty}\}. \text{ Then we define:} $$

$$ (iv) \{A_n\}_{n=0}^{*\infty} + \{B_n\}_{n=0}^{*\infty} = \overline{\{x_{1,n} + y_{1,n}a_n\}_{n=0}^{\infty} + \{x_{2,n} + y_{2,n}b_n\}_{n=0}^{\infty}} \triangleq \\ \triangleq \overline{\{x_{1,n} + x_{2,n} + y_{1,n}a_n + y_{2,n}b_n\}_{n=0}^{\infty}} = \{x_{1,n} + x_{2,n} + y_{1,n}A_n + y_{2,n}B_n\}_{n=0}^{*\infty} $$