Clearly if $u \neq x$ then $(\mathbf{Sc}(u), G(y)) \neq (\mathbf{Sc}(x), w)$, so that in this case $(\mathbf{Sc}(u), G(y)) \in \mathbf{W}$. On the other hand, if $u = x$ and $(\mathbf{Sc}(u), G(y)) = (\mathbf{Sc}(x), w)$, then $w = G(y)$, where $(x,y) \in \mathcal{F}$, contrary to the choice of $w$ henee $(\mathbf{Sc}(u), G(y)) \neq (\mathbf{Sc}(x), w)$, so again $(\mathbf{Sc}(u), G(y)) \in \mathbf{W}$. Thus whenever $(u,y) \in \mathbf{W}$, also $(\mathbf{Sc}(u), G(y)) \in \mathbf{W}$. Now that we have shown $\mathbf{W} \in \mathbf{M}$ we see by (2) that $\mathcal{F} \subseteq \mathbf{W}$ but this is false since $(\mathbf{Sc}(x), w) \in \mathcal{F}$ and $(\mathbf{Sc}(x), w) \notin \mathbf{W}$. Thus our hypothesis that (8) is incorrect has led to a contradiction, and now (8) is proved. Sinee (7) follows from (8), we have by *hyper infinite induction from (6) that* $A = \mathbb{N}^#$. Hence

(9) $\mathcal{F}$ is a function.

We have still to prove that $\mathcal{F}$ satisfies, condition (i); we must show that for each $x \in \mathbb{N}^#$ there is a $y$ with $(x,y) \in \mathcal{F}$. Since $\mathcal{F} \subseteq \mathbb{N}^# \times S$, it will then follow that $\mathbf{dom}(\mathcal{F}) = \mathbb{N}^#$ and $\mathbf{range}(\mathcal{F}) \subseteq S$. Let $B = \mathbf{dom}(\mathcal{F})$, that is, (10) $B = \{x|x \in \mathbb{N}^# \text{ and for some } y, (x,y) \in \mathcal{F}\}$.

We prove now by *hyper infinite induction* that $B = \mathbb{N}^#$. First, $1 \in B$, *sínce* $(1, c) \in \mathcal{F}$ by (3). Next, if $x \in B$, pick some $y$ with $(x,y) \in \mathcal{F}$; then by (4), $(\mathbf{Sc}(x), G(y)) \in \mathcal{F}$, and henee $\mathbf{Sc}(x) \in B$.

Thus concludes the first part of the proof, that there is at least one function $\mathcal{F}$ satisfying conditions (i)-(iii).

**Part 2.** We prove that there cannot be more than one such function.

Suppose that $\mathcal{F}_1$ and $\mathcal{F}_2$ both satisfy the conditions (i)-(iii) we wish to show $\mathcal{F}_1 = \mathcal{F}_2$, i.e., that for all $x \in \mathbb{N}^#$, $\mathcal{F}_1(x) = \mathcal{F}_2(x)$. Thus is proved by *hyper infinite induction* on $X$. By (ii), $\mathcal{F}_1(1) = c$ and $\mathcal{F}_2(1) = c$, so $\mathcal{F}_1(1) = \mathcal{F}_2(1)$. Suppose that $\mathcal{F}_1(x) = \mathcal{F}_2(x)$; then $\mathcal{F}_1(\mathbf{Sc}(x)) = G(\mathcal{F}_1(x))$ and $\mathcal{F}_2(\mathbf{Sc}(x)) = G(\mathcal{F}_2(x))$, so $\mathcal{F}_1(\mathbf{Sc}(x)) = \mathcal{F}_2(\mathbf{Sc}(x))$.

**Theorem 2.** Let $S$ be a set, $c \in S$ and $G: S \times \mathbb{N}^# \to S$ is a binary function with $\mathbf{dom}(G) = S \times \mathbb{N}^#$ and $\mathbf{range}(G) \subseteq S$.

Then there exists a function $\mathcal{F}: \mathbb{N}^# \to S$ such that:

(i) $\mathbf{dom}(\mathcal{F}) = \mathbb{N}^# \text{ and } \mathbf{range}(\mathcal{F}) \subseteq S$;

(ii) $\mathcal{F}(1) = c$;

(iii) for all $x \in \mathbb{N}^#$, $\mathcal{F}(\mathbf{Sc}(x)) = G(\mathcal{F}(x), x)$.

We omit the proof of the Theorem 3.4.2 since it can be given by simple modification of the proof to Theorem 3.4.1.

## §9. General associative and commutative laws.

**Definition C.1.** Suppose that $S$ is a set on which a binary operation $+$ is defined and under which $S$ is closed. Let $\{x_k\}_{k \in \mathbb{N}^#}$ be any hyper infinite sequence of terms of $S$. For every $n \in \mathbb{N}^#$ we denote by $\text{Ext}-\sum_{k=1}^{n} x_k$ the element of $S$ uniquely determined by the following conditions:

(i) $\text{Ext}-\sum_{k=1}^{1} x_k = x_1$; (ii) $\text{Ext}-\sum_{k=1}^{n+1} x_k = \text{Ext}-\sum_{k=1}^{n} x_k + x_{n+1}$ for all $n \in \mathbb{N}^#$.

**Remark 9.1.** This definition is justified on the following grounds. The sequence $\{x_k\}_{k \in \mathbb{N}^#}$ is a given external function $H$ with domain $\mathbb{N}^#$, $H(x_k) = x_k$ for every $k$. We seek a function $F$ with domain $\mathbb{N}^#$ whose value $F(n)$ is to be $\text{Ext}-\sum_{k=1}^{n} x_k$. Then the conditions
$$Ext-\prod_{k=1}^{n+m} x_k = \left(Ext-\prod_{k=1}^{n} x_k\right) \times Ext-\prod_{k=1}^{m} x_{n+k}. \quad (9.2)$$

**Proof.** We prove (3.5.1); the proof of (2) is completely similar. Let $n$ be fixed; we proceed by hyper infinite induction on $m$. For $m = 1$ from Eq.(3.8.1) we get

$$Ext-\sum_{k=1}^{n+1} x_k = \left(Ext-\sum_{k=1}^{n} x_k\right) + \left(Ext-\sum_{k=1}^{1} x_{n+k}\right). \quad (9.3)$$

By Definition 3.8.1(i) we obtain

$$Ext-\sum_{k=1}^{1} x_{n+k} = x_{n+1}. \quad (9.4)$$

Suppose Eq.(3.8.1) is true for $m \in \mathbb{N}^#$. We show that is true for $m+1$, i.e., that

$$Ext-\sum_{k=1}^{n+(m+1)} x_k = \left(Ext-\sum_{k=1}^{n} x_k\right) + \left(Ext-\sum_{k=1}^{m+1} x_{n+k}\right). \quad (9.4')$$

By associativity + on $\mathbb{N}^#$ we get

$$Ext-\sum_{k=1}^{n+(m+1)} x_k = Ext-\sum_{k=1}^{(n+m)+1} x_k. \quad (9.6)$$

From Eq.(3.8.6) by Definition 3.8.1(ii) we obtain

$$Ext-\sum_{k=1}^{(n+m)+1} x_k = Ext-\sum_{k=1}^{n+m} x_k + x_{(n+m)+1} = Ext-\sum_{k=1}^{n+m} x_k + x_{n+(m+1)}. \quad (9.7)$$

From Eq.(3.8.7) by induction hypothesis we obtain

$$Ext-\sum_{k=1}^{n+m} x_k + x_{n+(m+1)} = \left(Ext-\sum_{k=1}^{n} x_k + Ext-\sum_{k=n}^{m} x_k\right) + x_{n+(m+1)}. \quad (9.8)$$

From Eq.(3.8.8) by associativity + on $S$ we get

$$\left(Ext-\sum_{k=1}^{n} x_k + Ext-\sum_{k=n}^{m} x_k\right) + x_{n+(m+1)} = Ext-\sum_{k=1}^{n} x_k + \left(Ext-\sum_{k=n}^{m} x_k + x_{n+(m+1)}\right). \quad (9.9)$$

From Eq.(3.8.9) by Definition 3.8.1(ii) we obtain

$$Ext-\sum_{k=1}^{n} x_k + \left(Ext-\sum_{k=n}^{m} x_k + x_{n+(m+1)}\right) = Ext-\sum_{k=1}^{n} x_k + Ext-\sum_{k=n}^{m+1} x_k. \quad (9.10)$$

This equality completes the inductive step and hence the proof of the theorem.

**Definition 9.3.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^# \setminus \mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^#$.

Then $Ext-\sum_{k=m}^{n} x_k$ and $Ext-\prod_{k=m}^{n} x_k$ are defined for any $n, m \in \mathbb{N}^#$ by the recursions

(i) $Ext-\sum_{k=m}^{n} x_k = 0$ and $Ext-\prod_{k=m}^{n} x_k = 1$ if $n < m$;

(ii) $Ext-\sum_{k=m}^{n} x_k = \left(Ext-\sum_{k=m}^{n-1} x_k\right) + x_n$ and

(iii) $Ext-\prod_{k=m}^{n} x_k = x_n \times \left(Ext-\prod_{k=m}^{n-1} x_k\right)$ if $m < n$.

The condition (ii) of the above definition is justified by recursive definition, see Appendix B.

**Definition 9.4.** Let $\langle x_1, \dots, x_j, \dots \rangle, j \in \mathbb{N}$ be a countable sequence of elements of $\mathbb{R}_c^#$.
(i), (ii) above correspond to the following conditions on $F$ :

(i') $F(1) = H(1)$; (ii') $F(n+1) = F(n) + H(n+1)$, for all $n \in \mathbb{N}^#$.

Let (1) $c = H(1)$; (2) $G(n, z) = z + H(n + 1)$.

Thus the conditions (i') and (ii') are equivalent to

(i'') $F(1) = c$;

(ii'') $F(n+1) = G(n, F(n))$ for all $n \in \mathbb{N}^#$.

Given the function $H$, the element $c$ of $S$ and the function $G$ are well-defined by (1)-(2).

Then by Theorem B.1 we see that there is a unique function $F$ satisfying (1)-(2) with $\text{dom}(F) = \mathbb{N}^# \text{ and range}(F) \subseteq S$. Thus (i')-(ii') is just another form of recursive definition.

(Hence it should be expected that various properties of $\text{Ext}-\sum_{k=1}^{n} x_k$ will have to be verified by hyper infinite induction on $n \in \mathbb{N}^#$.)

**Definition 9.2.** Suppose that $S$ is a set on which a binary operation $\times$ is defined and under which $S$ is dosed. Let $\{x_k\}_{k \in \mathbb{N}^#}$ be an hyper infinite sequence of terms of $S$. For every $n \in \mathbb{N}^#$ we denote by $\text{Ext}-\prod_{k=1}^{n} x_k$ the element of $S$ uniquely determined by the following conditions:

(i) $\text{Ext}-\prod_{k=1}^{n} x_k = x_1$; (ii) $\text{Ext}-\prod_{k=1}^{n+1} x_k = (\text{Ext}-\prod_{k=1}^{n} x_k) \times x_{n+1}$ for all $n \in \mathbb{N}^#$.

**Theorem 1.** (1) Suppose that $S$ is a set closed under a binary operation $+$ and that $+$ is associative on $S$, i.e., for all $x, y, z \in S, x + (y + z) = (x + y) + z$. Let $\{x_k\}_{k \in \mathbb{N}^#}$ be any hyper infinite sequence of terms in $S$. Then for any $n, m \in \mathbb{N}^#$. we have

$$ \text{Ext}-\sum_{k=1}^{n+m} x_k = \left( \text{Ext}-\sum_{k=1}^{n} x_k \right) + \left( \text{Ext}-\sum_{k=1}^{m} x_{n+k} \right). \qquad (9.1) $$

(2) Suppose that $S$ is a set closed under a binary operation $\times$ and that $\times$ is associative on $S$, i.e., for all $x, y, z \in S, x \times (y \times z) = (x \times y) \times z$. Let $\{x_k\}_{k \in \mathbb{N}^#}$ be any hyper infinite sequence of terms in $S$. Then for any $n, m \in \mathbb{N}^#$. we have

$$ \text{Ext}-\prod_{k=1}^{n+m} x_k = \left( \text{Ext}-\prod_{k=1}^{n} x_k \right) \times \text{Ext}-\prod_{k=1}^{m} x_{n+k}. \qquad (9.2) $$

**Proof.** We prove (3.5.1); the proof of (2) is completely similar. Let $n$ be fixed; we proceed by hyper infinite induction on $m$. For $m = 1$ from Eq.(3.8.1) we get

$$ \text{Ext}-\sum_{k=1}^{n+1} x_k = \left( \text{Ext}-\sum_{k=1}^{n} x_k \right) + \left( \text{Ext}-\sum_{k=1}^{1} x_{n+k} \right). \qquad (9.3) $$

By Definition 3.8.1(i) we obtain

$$ \text{Ext}-\sum_{k=1}^{1} x_{n+k} = x_{n+1}. \qquad (9.4) $$

Suppose Eq.(3.8.1) is true for $m \in \mathbb{N}^#$. We show that is true for $m+1$, i.e., that

$$ \text{Ext}-\sum_{k=1}^{n+(m+1)} x_k = \left( \text{Ext}-\sum_{k=1}^{n} x_k \right) + \left( \text{Ext}-\sum_{k=1}^{m+1} x_{n+k} \right). \qquad (9.4') $$

By associativity + on $\mathbb{N}^# \text{ we get}$
Then $\omega$-sum $Ext-\sum_{j=m}^{\omega} x_k$ and $\omega$-product $Ext-\prod_{j=m}^{\omega} x_k$ are defined for any $m \in \mathbb{N}$ by

(iv) $Ext-\sum_{j=m}^{\omega} x_j \triangleq Ext-\sum_{j=m}^{n} y_j$, where $\langle y_1, \dots, y_j, \dots, y_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ is a hyperfinite sequence
such that $x_j = y_j$ for all $j \in \mathbb{N}$ and $y_j = 0$ for all $j \in \mathbb{N}^\#\mathbb{N}$;

(v) $Ext-\prod_{j=m}^{\omega} x_j \triangleq Ext-\prod_{j=m}^{\omega} y_j$, where $\langle y_1, \dots, y_j, \dots, y_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ is a hyperfinite sequence
such that $x_j = y_j$ for all $j \in \mathbb{N}$ and $y_j = 1$ for all $j \in \mathbb{N}^\#\mathbb{N}$.

**Theorem 9.2.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^\#$.
Then we have

$$ Ext-\sum_{k=m}^{n} x_k = Ext-\sum_{k=m}^{n-m+q} x_{k+m-q} \quad (9.11) $$

and

$$ z \times \left( Ext-\sum_{k=m}^{n} x_k \right) = Ext-\sum_{k=m}^{n} z \times x_k, \quad (9.12) $$

$z \in \mathbb{R}_c^\#$.

**Proof.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^\#$.

Consider now any hyperfinite nonnegative integers

$n_1, n_2, \dots, n_i, \dots, n_t, n_i \in \mathbb{N}^\#\mathbb{N}, 1 \le i \le t,$

and set

$$ n = n_1 + n_2 + \dots + n_t. \quad (9.13) $$

Given $x_1, \dots, x_n$, we can group these as:

$$ x_1, \dots, x_{n_1}; \ x_{n_1+1}, \dots, x_{n_1+n_2}; \ x_{n_1+n_2+1}, \dots, x_{n_1+n_2+n_3}; \dots x_{n_1+n_2+\dots+n_{i+1}}, \dots, x_{n_1+n_2+\dots+n_{i+1}}; \dots \quad (9.14) $$

Here, if $n_i = 0$, the corresponding subsequence is regarded as being empty.

**Theorem 9.3.** Let $\langle x_1, \dots, x_k, \dots \rangle$ be an hyper infinite sequence of elements of $\mathbb{R}_c^\#$.

Let $\langle n_1, \dots, n_t \rangle$ be a sequence of nonnegative integers. For each $i = 1, \dots, t \in \mathbb{N}^\#$,

let $m_i = \sum_{j=1}^{i-1} n_j$ and let $n = m_t + n_t$. Then

$$ Ext-\sum_{k=1}^{n} x_k = \sum_{i=1}^{t} \left( Ext-\sum_{k=1}^{n_i} x_{m_i+k} \right) \quad (9.15) $$

and

$$ Ext-\prod_{k=1}^{n} x_k = \prod_{i=1}^{t} \left( Ext-\prod_{k=1}^{n_i} x_{m_i+k} \right). \quad (9.16) $$

**Proof.** By hyper infinite induction.

**Definition 9.5.** A function $F$ is said to be a permutation of a set $S$ if it is one-to-one and $\text{dom}(F) = \text{range}(F) = S$.

**Definition 9.6.** Let $[1, n]$ a set $\{k|k \in \mathbb{N}^\# \wedge (1 \le k \le n)\}$

**Theorem 9.4.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ be an hyperfinite external sequence of elements of $\mathbb{R}_c^\#$. Then for any $n \in \mathbb{N}^\#$ and any permutation $F$ of $[1, n]$ following holds

$$ Ext-\sum_{k=1}^{n} x_k = Ext-\sum_{k=1}^{n} x_{F(k)}. \quad (9.17) $$

The same holds if we replace $Ext-\sum$ by $Ext-\prod$.
$$ Ext- \sum_{k=1}^{n+(m+1)} x_k = Ext- \sum_{k=1}^{(n+m)+1} x_k. \qquad (9.6) $$

From Eq.(3.8.6) by Definition 3.8.1(ii) we obtain

$$ Ext- \sum_{k=1}^{(n+m)+1} x_k = Ext- \sum_{k=1}^{n+m} x_k + x_{(n+m)+1} = Ext- \sum_{k=1}^{n+m} x_k + x_{n+(m+1)}. \qquad (9.7) $$

From Eq.(3.8.7) by induction hypothesis we obtain

$$ Ext- \sum_{k=1}^{n+m} x_k + x_{n+(m+1)} = \left( Ext- \sum_{k=1}^{n} x_k + Ext- \sum_{k=n}^{m} x_k \right) + x_{n+(m+1)}. \qquad (9.8) $$

From Eq.(3.8.8) by associativity + on S we get

$$ \left( Ext- \sum_{k=1}^{n} x_k + Ext- \sum_{k=n}^{m} x_k \right) + x_{n+(m+1)} = Ext- \sum_{k=1}^{n} x_k + \left( Ext- \sum_{k=n}^{m} x_k + x_{n+(m+1)} \right). \qquad (9.9) $$

From Eq.(3.8.9) by Definition 3.8.1(ii) we obtain

$$ Ext- \sum_{k=1}^{n} x_k + \left( Ext- \sum_{k=n}^{m} x_k + x_{n+(m+1)} \right) = Ext- \sum_{k=1}^{n} x_k + Ext- \sum_{k=n}^{m+1} x_k. \qquad (9.10) $$

This equality completes the inductive step and hence the proof of the theorem.

**Definition 9.3.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^\#$.

Then $Ext- \sum_{k=m}^{n} x_k$ and $Ext- \prod_{k=m}^{n} x_k$ are defined for any $n, m \in \mathbb{N}^\#$ by the recursions

(i) $Ext- \sum_{k=m}^{n} x_k = 0$ and $Ext- \prod_{k=m}^{n} x_k = 1$ if $n < m$;

(ii) $Ext- \sum_{k=m}^{n} x_k = \left( Ext- \sum_{k=m}^{n-1} x_k \right) + x_n$ and

(iii) $Ext- \prod_{k=m}^{n} x_k = x_n \times \left( Ext- \prod_{k=m}^{n-1} x_k \right)$ if $m < n$.

The condition (ii) of the above definition is justified by recursive definition, see Appendix B.

**Definition 9.4.** Let $\langle x_1, \dots, x_j, \dots \rangle, j \in \mathbb{N}$ be a countable sequence of elements of $\mathbb{R}_c^\#$.

Then $\omega$-sum $Ext- \sum_{j=m}^{\omega} x_k$ and $\omega$-product $Ext- \prod_{j=m}^{\omega} x_k$ are defined for any $m \in \mathbb{N}$ by

(iv) $Ext- \sum_{j=m}^{\omega} x_j \triangleq Ext- \sum_{j=m}^{n} y_j$, where $\langle y_1, \dots, y_j, \dots, y_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ is a hyperfinite sequence

such that $x_j = y_j$ for all $j \in \mathbb{N}$ and $y_j = 0$ for all $j \in \mathbb{N}^\#\mathbb{N}$;

(v) $Ext- \prod_{j=m}^{\omega} x_j \triangleq Ext- \prod_{j=m}^{\omega} y_j$, where $\langle y_1, \dots, y_j, \dots, y_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ is a hyperfinite sequence

such that $x_j = y_j$ for all $j \in \mathbb{N}$ and $y_j = 1$ for all $j \in \mathbb{N}^\#\mathbb{N}$.

**Theorem 9.2.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\#\mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^\#$.

Then we have

$$ Ext- \sum_{k=m}^{n} x_k = Ext- \sum_{k=m}^{n-m+q} x_{k+m-q} \qquad (9.11) $$

and
**Proof.** The proof is by hyper infinite induction on $n \in \mathbb{N}^#$. For $n = 1$ it is trivial. Suppose that it is true for $n$. Let $\mathbf{G}$ be a permutation of $[1, n+1]$. Then $\mathbf{G}(m) = m+1$ for a unique $m$, such that $1 \le m \le n+1$. Then by Eq.(3.5.15)

$$ \mathrm{Ext}-\sum_{k=1}^{n+1} x_{\mathbf{G}(k)} = \mathrm{Ext}-\sum_{k=1}^{m-1} x_{\mathbf{G}(k)} + x_{n+1} + \mathrm{Ext}-\sum_{k=m+1}^{n+1} x_{\mathbf{G}(k)} \quad (9.18) $$

and by Eq.(3.8.18)

$$ \mathrm{Ext}-\sum_{k=1}^{m-1} x_{\mathbf{G}(k)} + x_{n+1} + \mathrm{Ext}-\sum_{k=m+1}^{n+1} x_{\mathbf{G}(k)} = \mathrm{Ext}-\sum_{k=1}^{m-1} x_{\mathbf{G}(k)} + \mathrm{Ext}-\sum_{k=m}^{n} x_{\mathbf{G}(k+1)} + x_{n+1}. \quad (9.19) $$

Thus by Eq.(3.8.11) we obtain

$$ \mathrm{Ext}-\sum_{k=1}^{n+1} x_{\mathbf{G}(k)} = \mathrm{Ext}-\sum_{k=1}^{m-1} x_{\mathbf{G}(k)} + \mathrm{Ext}-\sum_{k=m}^{n} x_{\mathbf{G}(k+1)} + x_{n+1}. \quad (9.20) $$

To reduce this to the inductive hypothesis, we wish to rewrite the external sum of the first

two terms as $\mathrm{Ext}-\sum_{k=1}^{n} x_{\mathbf{F}(k)}$ for suitable $\mathbf{F}$. Define $\mathbf{F}$ by

$$ \mathbf{F}(k) = \begin{cases} \mathbf{G}(k) & \text{if } 1 \le k < m \\ \mathbf{G}(k+1) & \text{if } m \le k \le n \end{cases} \quad (9.21) $$

Since all values of $\mathbf{G}(k)$ for $k \ne m$, we have for all $k \le n$

$$ 1 \le \mathbf{F}(k) \le n \quad (9.22) $$

Now we claim that

$$ \mathbf{F} \text{ is a permutation of } [1, n]. \quad (9.23) $$

By (3.8.21) and (3.8.22) we need only check that $\mathbf{F}$ is one-to one. Suppose that $\mathbf{F}(k_1) = \mathbf{F}(k_2)$.

If both $k_1, k_2$ are $< m$ or both are $\ge m$, it follows from (3.8.21) and the fact that $\mathbf{G}$ is a permutation that $k_1 = k_2$. If, say, $k_1 < m \le k_2$, we have $\mathbf{G}(k_1) = \mathbf{G}(k_2 + 1)$, hence $k_1 = k_2 + 1$, which contradicts our assumption. Thus neither this case nor, by symmetry, the case $k_2 < m \le k_1$ can occur. We have from (3.8.20) and (3.8.21) that

$$ \mathrm{Ext}-\sum_{k=1}^{m+1} x_{\mathbf{G}(k)} = \mathrm{Ext}-\sum_{k=1}^{m-1} x_{\mathbf{F}(k)} + \mathrm{Ext}-\sum_{k=m}^{n} x_{\mathbf{F}(k)} + x_{n+1} = \mathrm{Ext}-\sum_{k=1}^{n} x_{\mathbf{F}(k)} + x_{n+1} \quad (9.24) $$

by (3.8.23) and inductive hypothesis

$$ \mathrm{Ext}-\sum_{k=1}^{n} x_{\mathbf{F}(k)} + x_{n+1} = \mathrm{Ext}-\sum_{k=1}^{n} x_{k} + x_{n+1} = \mathrm{Ext}-\sum_{k=1}^{n+1} x_{k} \quad (9.25) $$

This equality completes the inductive step and hence the proof of the theorem.

# §10. Hyperrationalem Q#.

Now that we have the hypernatural numbers $\mathbb{N}^#$, defining hyperintegers and hyperrational numbers is well within reach [2].

**Definition 10.1.** Let $Z^{#'} = \mathbb{N}^# \times \mathbb{N}^#$. We can define an equivalence relation $\approx$ on $Z^{#'}$ by $(a, b) \approx (c, d)$ if and only if $a+d = b+c$. Then we denote the set of all hyperintegers by $\mathbb{Z}^# = Z^{#'} / \approx$ (The set of all equivalence classes of $Z^{#'}$ modulo $\approx$).
$$z \times \left( \mathrm{Ext} - \sum_{k=m}^{n} x_k \right) = \mathrm{Ext} - \sum_{k=m}^{n} z \times x_k, \qquad (9.12)$$

$z \in \mathbb{R}_c^#.$

**Proof.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\# \setminus \mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^#$.

Consider now any hyperfinite nonnegative integers

$n_1, n_2, \dots, n_i, \dots, n_t, n_i \in \mathbb{N}^\# \setminus \mathbb{N}, 1 \le i \le t,$

and set

$$n = n_1 + n_2 + \dots + n_t. \qquad (9.13)$$

Given $x_1, \dots, x_n$, we can group these as:

$$x_1, \dots, x_{n_1}; \quad x_{n_1+1}, \dots, x_{n_1+n_2}; \quad x_{n_1+n_2+1}, \dots, x_{n_1+n_2+n_3}; \dots, x_{n_1+n_2+\dots+n_{i+1}}, \dots, x_{n_1+n_2+\dots+n_{i+1}}; \dots \qquad (9.14)$$

Here, if $n_i = 0$, the corresponding subsequence is regarded as being empty.

**Theorem 9.3.** Let $\langle x_1, \dots, x_k, \dots \rangle$ be an hyper infinite sequence of elements of $\mathbb{R}_c^#$.

Let $\langle n_1, \dots, n_t \rangle$ be a sequence of nonnegative integers. For each $i = 1, \dots, t \in \mathbb{N}^#,$

let $m_i = \sum_{j=1}^{i-1} n_j$ and let $n = m_t + n_t$. Then

$$\mathrm{Ext} - \sum_{k=1}^{n} x_k = \sum_{i=1}^{t} \left( \mathrm{Ext} - \sum_{k=1}^{n_i} x_{m_i+k} \right) \qquad (9.15)$$

and

$$\mathrm{Ext} - \prod_{k=1}^{n} x_k = \prod_{i=1}^{t} \left( \mathrm{Ext} - \prod_{k=1}^{n_i} x_{m_i+k} \right). \qquad (9.16)$$

**Proof.** By hyper infinite induction.

**Definition 9.5.** A function $F$ is said to be a permutation of a set $S$ if it is one-to-one and $\mathrm{dom}(F) = \mathrm{range}(F) = S$.

**Definition 9.6.** Let $[1, n]$ a set $\{k | k \in \mathbb{N}^# \wedge (1 \le k \le n)\}$

**Theorem 9.4.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\# \setminus \mathbb{N}$ be an hyperfinite external sequence of elements of $\mathbb{R}_c^#$. Then for any $n \in \mathbb{N}^#$ and any permutation $F$ of $[1, n]$ following holds

$$\mathrm{Ext} - \sum_{k=1}^{n} x_k = \mathrm{Ext} - \sum_{k=1}^{n} x_{F(k)}. \qquad (9.17)$$

The same holds if we replace $\mathrm{Ext} - \sum$ by $\mathrm{Ext} - \prod$.

**Proof.** The proof is by hyper infinite induction on $n \in \mathbb{N}^#$. For $n = 1$ it is trivial.

Suppose that it is true for $n$. Let $G$ be a permutation of $[1, n+1]$. Then $G(m) = n+1$ for a unique $m$, such that $1 \le m \le n+1$. Then by Eq.(3.5.15)

$$\mathrm{Ext} - \sum_{k=1}^{n+1} x_{G(k)} = \mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + x_{n+1} + \mathrm{Ext} - \sum_{k=m+1}^{n+1} x_{G(k)} \qquad (9.18)$$

and by Eq.(3.8.18)

$$\mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + x_{n+1} + \mathrm{Ext} - \sum_{k=m+1}^{n+1} x_{G(k)} = \mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + \mathrm{Ext} - \sum_{k=m}^{n} x_{G(k+1)} + x_{n+1}. \qquad (9.19)$$

Thus by Eq.(3.8.11) we obtain

$$\mathrm{Ext} - \sum_{k=1}^{n+1} x_{G(k)} = \mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + \mathrm{Ext} - \sum_{k=m}^{n} x_{G(k+1)} + x_{n+1}. \qquad (9.20)$$

To reduce this to the inductive hypothesis, we wish to rewrite the external sum of the
**Definition 10.2.** Let $Q^{#} = \mathbb{Z}^{#} \times (\mathbb{Z}^{#} - \{0\}) = \{(a,b) \in \mathbb{Z}^{#} \times \mathbb{Z}^{#} | b \neq 0\}$. We can define an equivalence relation $\approx$ on $Q^{#}$ by $(a,b) \approx (c,d)$ if and only if $a \times d = b \times c$. Then we denote

the set of all hyperrational numbers by $\mathbb{Q}^# = Q'/\approx$ (The set of all equivalence classes of $Q'$ modulo $\approx$).

**Definition 10.3.** A linearly ordered set $(P, <)$ is called dense if for any $a, b \in P$ such that $a < b$, there exists $z \in P$ such that $a < z < b$.

**Lemma 10.1.** $(\mathbb{Q}^#<\emptyset)$ is dense.

**Proof.** Let $x = (a,b)$, $y = (c,d) \in \mathbb{Q}^#$ be such that $x < y$. Consider $z = (ad + bc, 2bd) \in \mathbb{Q}^#$. It is easily shown that $x < z < y$.

# 11. External Cauchy hyperreals $\mathbb{R}_c^#$ via Cauchy completion.

**Definition 11.1.** A hyper infinite sequence of hyperrational numbers (or for the sake of brevity simply hyperrational sequence) is a function from the hypernatural numbers $\mathbb{N}^#$ into the hyperrational numbers $\mathbb{Q}^#$. We usually denote such a function by $n \mapsto a_n$, or by $a : n \to a_n$, so the terms in the sequence are written $\{a_1, a_2, a_3, \dots, a_n\dots\}$. To refer to the whole hyper infinite sequence, we will write $\{a_n\}_{n=1}^{\infty}$, or $\{a_n\}_{n \in \mathbb{N}^#}$, or for the sake of brevity simply $\{a_n\}$.

**Definition 11.2.** Let $\{a_n\}$ be a hyperrational sequence. Say that $\{a_n\}$ #-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^#\mathbb{N}$, $N = N(\varepsilon)$ such that, after $N$ (i.e. for all $n > N$), $|a_n| \le \varepsilon$. We often denote this symbolically by $a_n \xrightarrow{\#} 0$. We can also, at this point, define what it means for a hyperrational sequence #-tends to any given number $q \in \mathbb{Q}^# : \{a_n\}$ #-tends to $q$ if the hyperrational sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \xrightarrow{\#} 0$.

**Definition 11.3.** Let $\{a_n\}$ be a hyper infinite hyperrational sequence. We call $\{a_n\}$ a Cauchy hyperrational sequence if the difference between its terms #-tends to 0. To be precise: given any hyperrational number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 11.1.** If $\{a_n\}$ is a #-convergent hyperrational sequence (that is, $a_n \xrightarrow{\#} q$ for some hyperrational number $q \in \mathbb{Q}^#$), then $\{a_n\}$ is a Cauchy hyperrational sequence.

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

**Theorem 11.2.** If $\{a_n\}$ is a Cauchy hyperrational sequence, then it is bounded or hyperbolic; that is, there is some $M \in \mathbb{Q}^#$ finite or hyperfinite such that $|a_n| \le M$ for all $n \in \mathbb{N}^#$.

**Proof.** Since $\{a_n\}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N \in \mathbb{N}^#\mathbb{N}$ such that $|a_m - a_n| < 1$ whenever $m, n > N$. Thus, $|a_{N+1} - a_n| < 1$ for $n > N$. We can rewrite this as $a_{N+1} - 1 < a_n < a_{N+1} + 1$. This means that $|a_n|$ is less than the maximum of $|a_{N+1} - 1|$
first

two terms as $Ext-\sum_{k=1}^{n}x_{F(k)}$ for suitable $F$. Define $F$ by

$$ F(k) = \begin{cases} G(k) & \text{if } 1 \le k < m \\ G(k+1) & \text{if } m \le k \le n \end{cases} \qquad (9.21) $$

Since all values of $G(k)$ for $k \ne m$, we have for all $k \le n$

$$ 1 \le F(k) \le n \qquad (9.22) $$

Now we claim that

$$ F \text{ is a permutation of } [1,n]. \qquad (9.23) $$

By (3.8.21) and (3.8.22) we need only check that $F$ is one-to one. Suppose that $F(k_1) = F(k_2)$.

If both $k_1, k_2$ are $< m$ or both are $\ge m$, it follows from (3.8.21) and the fact that $G$ is a permutation that $k_1 = k_2$. If, say, $k_1 < m \le k_2$, we have $G(k_1) = G(k_2 + 1)$, hence $k_1 = k_2 + 1$, which contradicts our assumption. Thus neither this case nor, by symmetry, the case $k_2 < m \le k_1$ can occur. We have from (3.8.20) and (3.8.21) that

$$ Ext-\sum_{k=1}^{m+1} x_{G(k)} = Ext-\sum_{k=1}^{m-1} x_{F(k)} + Ext-\sum_{k=m}^{n} x_{F(k)} + x_{n+1} = Ext-\sum_{k=1}^{n} x_{F(k)} + x_{n+1} \qquad (9.24) $$

by (3.8.23) and inductive hypothesis

$$ Ext-\sum_{k=1}^{n} x_{F(k)} + x_{n+1} = Ext-\sum_{k=1}^{n} x_{k} + x_{n+1} = Ext-\sum_{k=1}^{n+1} x_{k} \qquad (9.25) $$

This equality completes the inductive step and hence the proof of the theorem.

# §10. Hyperrationalem Q#.

Now that we have the hypernatural numbers $\mathbb{N}^#$, defining hyperintegers and hyperrational numbers is well within reach [2].

**Definition 10.1.** Let $Z^# = \mathbb{N}^# \times \mathbb{N}^#$. We can define an equivalence relation $\approx$ on $Z^#$ by $(a,b) \approx (c,d)$ if and only if $a+d = b+c$. Then we denote the set of all hyperintegers by $\mathbb{Z}^# = Z^#/ \approx$ (The set of all equivalence classes of $Z^#$ modulo $\approx$).

**Definition 10.2.** Let $Q^# = \mathbb{Z}^# \times (\mathbb{Z}^# - \{0\}) = \{(a,b) \in \mathbb{Z}^# \times \mathbb{Z}^# | b \ne 0\}$. We can define an equivalence relation $\approx$ on $Q^#$ by $(a,b) \approx (c,d)$ if and only if $a \times d = b \times c$. Then we denote

the set of all hyperrational numbers by $Q^# = Q'/ \approx$ (The set of all equivalence classes of $Q^#$ modulo $\approx$).

**Definition 10.3.** A linearly ordered set $(P,<)$ is called dense if for any $a,b \in P$ such that

$a < b$, there exists $z \in P$ such that $a < z < b$.

**Lemma 10.1.** $(Q^#, <)$ is dense.

**Proof.** Let $x = (a,b), y = (c,d) \in Q^#$ be such that $x < y$. Consider $z = (ad+bc, 2bd) \in Q^#$.

It is easily shown that $x < z < y$.
and $|a_{N+1} + 1|$. So, set $M$ equal to the maximum number in the following list:
$\{|a_0|, |a_1|, \dots, |a_N|, |a_{N+1} - 1|, |a_{N+1} + 1|\}$. Then for any term $a_n$, if $n \le N$, then $|a_n|$ appears in the list and so $|a_n| \le M$; if $n > N$, then (as shown above) $|a_n|$ is less than at least one of the last two entries in the list, and so $|a_n| \le M$. Hence, $M$ is a bound for the sequence.

**Definition 11.4.** Let $S$ be a set. A relation $x \sim y$ among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

Reflexivity: for any $s \in S, s \sim s$.

Symmetry: for any $s, t \in S$, if $s \sim t$ then $t \sim s$.

Transitivity: for any $s, t, r \in S$, if $s \sim t$ and $t \sim r$, then $s \sim r$.

**Theorem 11.3.** Let $S$ be a set, with an equivalence relation $(\cdot \sim \cdot)$ on pairs of elements.
For $s \in S$, denote by $cl[s]$ the set of all elements in $S$ that are related to $s$. Then for
any $s, t \in S$, either $cl[s] = cl[t]$ or $cl[s]$ and $cl[t]$ are disjoint.

The hyperreal numbers $\mathbb{R}_c^\#$ will be constructed as equivalence classes of Cauchy hyperrational sequences. Let $\mathcal{F}_{\mathbb{Q}^\#}$ denote the set of all Cauchy hyperrational sequences of hyperrational numbers. We define the equivalence relation on $\mathcal{F}_{\mathbb{Q}^\#}$.

**Definition 11.5.** Let $\{a_n\}$ and $\{b_n\}$ be in $\mathcal{F}_{\mathbb{Q}^\#}$. Say they are \#-equivalent if $a_n - b_n \xrightarrow{\#} 0$ i.e., if and only if the hyperrational sequence $\{a_n - b_n\}$ tends to 0.

**Theorem 11.4.** Definition 11.4 yields an equivalence relation on $\mathcal{F}_{\mathbb{Q}^\#}$.

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

**Reflexive:** $a_n - a_n = 0$, and the sequence all of whose terms are 0 clearly \#-converges to 0. So $\{a_n\}$ is related to $\{a_n\}$.

**Symmetric:** Suppose $\{a_n\}$ is related to $\{b_n\}$, so $a_n - b_n \xrightarrow{\#} 0$.

But $b_n - a_n = -(a_n - b_n)$, and since only the absolute value $|a_n - b_n| = |b_n - a_n|$ comes into play in Definition 11.2, it follows that $b_n - a_n \xrightarrow{\#} 0$ as well. Hence, $\{b_n\}$ is related to $\{a_n\}$.

**Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 11.1. Suppose $\{a_n\}$ is related to $\{b_n\}$, and $\{b_n\}$ is related to $\{c_n\}$. This means that $a_n - b_n \xrightarrow{\#} 0$ and $b_n - c_n \xrightarrow{\#} 0$. To be fully precise, let us fix $\varepsilon > 0$, $\varepsilon \approx 0$; then there exists an $N \in \mathbb{N}^\#$ such that for all $n > N$, $|a_n - b_n| < \varepsilon/2$; also, there exists an $M \in \mathbb{N}^\#$ such that for all $n > M$, $|b_n - c_n| < \varepsilon/2$. Well, then, as long as $n$ is bigger than both $N$ and $M$, we have that $|a_n - c_n| = |(a_n - b_n) + (b_n - c_n)| \le |a_n - b_n| + |b_n - c_n| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|a_n - c_n| < \varepsilon$. This means that $a_n - c_n \xrightarrow{\#} 0$ i.e. $\{a_n\}$ is related to $\{c_n\}$.

**Definition 11.6.** The hyperreal numbers $\mathbb{R}_c^\#$ are the equivalence classes $cl[\{a_n\}]$ of Cauchy sequences of hyperrational numbers, as per Definition 11.5. That is, each such equivalence class is a hyperreal number.

**Definition 11.7.** Given any hyperrational number $q \in \mathbb{Q}^\#$, define a hyperreal number $q^\#$ to be the equivalence class of the sequence $q^\# = (q, q, q, q, \dots)$ consisting entirely of $q$. So we view $\mathbb{Q}^\#$ as being inside $\mathbb{R}_c^\#$ by thinking of each hyperrational number $q \in \mathbb{Q}^\#$ as its associated equivalence class $q^\#$. It is standard to abuse this notation, and simply refer to the equivalence class as $q$ as well.

**Definition 11.8.** Let $s, t \in \mathbb{R}_c^\#$, so there are Cauchy sequences $\{a_n\}, \{b_n\}$ of hyperrational numbers with $s = cl[\{a_n\}]$ and $t = cl[\{b_n\}]$.

(a) Define $s + t$ to be the equivalence class of the sequence $\{a_n + b_n\}$.

(b) Define $s \times t$ to be the equivalence class of the sequence $\{a_n \times b_n\}$.
# 11. External Cauchy hyperreals $\mathbb{R}_c^\#$ via Cauchy completion.

**Definition 11.1.** A hyper infinite sequence of hyperrational numbers (or for the sake of brevity simply hyperrational sequence) is a function from the hypernatural numbers $\mathbb{N}^\#$ into the hyperrational numbers $\mathbb{Q}^\#$. We usually denote such a function by $n \mapsto a_n$, or by $a : n \to a_n$, so the terms in the sequence are written $\{a_1, a_2, a_3, \dots, a_n\}$. To refer to the whole hyper infinite sequence, we will write $\{a_n\}_{n=1}^\infty$, or $\{a_n\}_{n \in \mathbb{N}^\#}$, or for the sake of brevity simply $\{a_n\}$.

**Definition 11.2.** Let $\{a_n\}$ be a hyperrational sequence. Say that $\{a_n\}$ #-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^\#\mathbb{N}$, $N = N(\varepsilon)$ such that, after $N$ (i.e.for all $n > N$), $|a_n| \le \varepsilon$. We often denote this symbolically by $a_n \xrightarrow{\#} 0$. We can also, at this point, define what it means for a hyperrational sequence #-tends to any given number $q \in \mathbb{Q}^\# : \{a_n\}$ #-tends to $q$ if the hyperrational sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \xrightarrow{\#} 0$.

**Definition 11.3.** Let $\{a_n\}$ be a hyper infinite hyperrational sequence. We call $\{a_n\}$ a Cauchy hyperrational sequence if the difference between its terms #-tends to 0. To be precise: given any hyperrational number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 11.1.** If $\{a_n\}$ is a #-convergent hyperrational sequence (that is, $a_n \xrightarrow{\#} q$ for some hyperrational number $q \in \mathbb{Q}^\#$), then $\{a_n\}$ is a Cauchy hyperrational sequence.

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

**Theorem 11.2.** If $\{a_n\}$ is a Cauchy hyperrational sequence, then it is bounded or hyper bounded; that is, there is some $M \in \mathbb{Q}^\#$ finite or hyperfinite such that $|a_n| \le M$ for all $n \in \mathbb{N}^\#$.

**Proof.** Since $\{a_n\}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N \in \mathbb{N}^\#/\mathbb{N}$ such that $|a_m - a_n| < 1$ whenever $m, n > N$. Thus, $|a_{N+1} - a_n| < 1$ for $n > N$. We can rewrite this as $a_{N+1} - 1 < a_n < a_{N+1} + 1$. This means that $|a_n|$ is less than the maximum of $|a_{N+1} - 1|$ and $|a_{N+1} + 1|$. So, set $M$ equal to the maximum number in the following list: $\{|a_0|, |a_1|, \dots, |a_N|, |a_{N+1} - 1|, |a_{N+1} + 1|\}$. Then for any term $a_n$, if $n \le N$, then $|a_n|$ appears in the list and so $|a_n| \le M$; if $n > N$, then (as shown above) $|a_n|$ is less than at least one of the last two entries in the list, and so $|a_n| \le M$. Hence, $M$ is a bound for the sequence.

**Definition 11.4.** Let $S$ be a set. A relation $x \sim y$ among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:
* Reflexivity: for any $s \in S$, $s \sim s$.
* Symmetry: for any $s, t \in S$, if $s \sim t$ then $t \sim s$.
* Transitivity: for any $s, t, r \in S$, if $s \sim t$ and $t \sim r$, then $s \sim r$.

**Theorem 11.3.** Let $S$ be a set, with an equivalence relation $(\cdot \sim \cdot)$ on pairs of elements. For $s \in S$, denote by $\text{cl}[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $\text{cl}[s] = \text{cl}[t]$ or $\text{cl}[s]$ and $\text{cl}[t]$ are disjoint.

The hyperreal numbers $\mathbb{R}_c^\#$ will be constructed as equivalence classes of Cauchy
**Theorem 11.5.** The operations $+, \times$ in Definition 8.8 (a),(b) are well-defined.

**Proof.** Suppose that $\mathrm{cl}[\{a_n\}] = \mathrm{cl}[\{c_n\}]$ and $\mathrm{cl}[\{b_n\}] = \mathrm{cl}[\{d_n\}]$. Thus means that $a_n - c_n \to_\# 0$ and $b_n - d_n \to_\# 0$. Then $(a_n + b_n) - (c_n + d_n) = (a_n - c_n) + (b_n - d_n)$.

Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to 0, and so $\mathrm{cl}[\{a_n + b_n\}] = \mathrm{cl}[\{c_n + d_n\}]$.

Multiplication is a little trickier; this is where we will use Theorem 11.2. We will also use another ubiquitous technique: adding 0 in the form of $s - s$. Again, suppose that $\mathrm{cl}[(a_n) = \mathrm{cl}[(c_n)]$ and $\mathrm{cl}[\{b_n\}] = \mathrm{cl}[\{d_n\}]$; we wish to show that $\mathrm{cl}[\{a_n \times b_n\}] = \mathrm{cl}[\{c_n \times d_n\}]$, or, in other words, that $a_n \times b_n - c_n \cdot d_n \to_\# 0$. Well, we add and subtract one of the other cross terms, say

$$b_n \times c_n : a_n \times b_n - c_n \times d_n = a_n \times b_n + (b_n \times c_n - b_n \times c_n) - c_n \times d_n = \\ = (a_n \times b_n - b_n \times c_n) + (b_n \times c_n - c_n \times d_n) = b_n \times (a_n - c_n) + c_n \times (b_n - d_n).$$

Hence, we have $|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n|$. Now, from Theorem 11.2, there are numbers $M$ and $L$ such that $|b_n| \le M$ and $|c_n| \le L$ for all $n \in \mathbb{N}^\#$. Taking some number $K$ which is bigger than both, we have

$$|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n| \le K(|a_n - c_n| + |b_n - d_n|).$$

Now, noting that both $a_n - c_n$ and $b_n - d_n$ tend to 0 and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2K$), we see that $a_n \times b_n - c_n \times d_n \to_\# 0$.

**Theorem 11.6.** Given any hyperreal number $s \neq 0$, there is a hyperreal number $t$ such that $s \times t = 1$.

**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 $\{0, 0, 0, 0, ...\}$. In other words, $s = \mathrm{cl}[\{a_n\}]$ where $a_n \to 0$ does not $\#-\text{converge}$ to 0. From this, we are to deduce the existence of a hyperreal number $t = \mathrm{cl}[\{b_n\}]$ such that $s \times t = \mathrm{cl}[\{a_n \times b_n\}]$ is the same equivalence class as $\mathrm{cl}[\{1, 1, 1, 1, ...\}]$. Doing so is actually an easy consequence of the fact that nonzero rational numbers have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{a_n\}$ does not tend to 0), there's no reason any number of the terms in $\{a_n\}$ can't equal 0. However, it turns out that eventually, $a_n \neq 0$.

That is,

**Lemma 11.1.** If $\{a_n\}$ is a Cauchy hyper infinite sequence which does not $\#$-tend to 0, then there is an $N \in \mathbb{N}^\#/\mathbb{N}$ such that, for $n > N, a_n \neq 0$.

We will now use Lemma 11.1 to complete the proof of Theorem 11.7.

Let $N$ be such that $a_n \neq 0$ for $n > N$. Define a hyper infinite sequence $b_n$ of hyperrational numbers as follows:

for $n \le N, b_n = 0$, and for $n > N, b_n = 1/a_n$; $\{b_n\} = (0, 0, ..., 0, 1/a_{N+1}, 1/a_{N+2}, ...)$.

This makes sense since, for $n > N$, an is a nonzero hyperrational number, so $1/a_n$ exists. Then $a_n \cdot b_n$ is equal to $a_n \cdot 0 = 0$ for $n \le N$, and equals $a_n \cdot b_n = a_n \cdot 1/a_n = 1$ for $n > N$. Well, then, if we

look at the hyper infinite sequence $(1, 1, 1, 1, ...)$, we have $(1, 1, 1, 1, ...)$ – $(a_n \cdot b_n)$ is the

hyper infinite sequence which is $1 - 0 = 1$ for $n \le N$ and equals $1 - 1 = 0$ for $n > N$.

Since this sequence is eventually equal to 0, it $\#$-converges to 0, and so

$$\mathrm{cl}[\{a_n \cdot b_n\}] = \mathrm{cl}[(1, 1, 1, 1, ...)] = 1 \in \mathbb{R}_c^\#.$$ This shows that $t = \mathrm{cl}[\{b_n\}]$ is a
hyperrational sequences. Let $\mathcal{F}_{\mathbb{Q}^#}$ denote the set of all Cauchy hyperrational sequences of hyperrational numbers. We define the equivalence relation on $\mathcal{F}_{\mathbb{Q}^#}$.

**Definition 11.5.** Let $\{a_n\}$ and $\{b_n\}$ be in $\mathcal{F}_{\mathbb{Q}^#}$. Say they are #-equivalent if $a_n - b_n \to_{\#} 0$ i.e., if and only if the hyperrational sequence $\{a_n - b_n\}$ tends to 0.

**Theorem 11.4.** Definition 11.4 yields an equivalence relation on $\mathcal{F}_{\mathbb{Q}^#}$.

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

**Reflexive:** $a_n - a_n = 0$, and the sequence all of whose terms are 0 clearly #-converges to 0. So $\{a_n\}$ is related to $\{a_n\}$.

**Symmetric:** Suppose $\{a_n\}$ is related to $\{b_n\}$, so $a_n - b_n \to_{\#} 0$.

But $b_n - a_n = -(a_n - b_n)$, and since only the absolute value $|a_n - b_n| = |b_n - a_n|$ comes into play in Definition 11.2, it follows that $b_n - a_n \to_{\#} 0$ as well. Hence, $\{b_n\}$ is related to $\{a_n\}$.

**Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 11.1. Suppose $\{a_n\}$ is related to $\{b_n\}$, and $\{b_n\}$ is related to $\{c_n\}$. This means that $a_n - b_n \to_{\#} 0$ and $b_n - c_n \to_{\#} 0$. To be fully precise, let us fix $\varepsilon > 0$, $\varepsilon \approx 0$; then there exists an $N \in \mathbb{N}^#$ such that for all $n > N$, $|a_n - b_n| < \varepsilon/2$; also, there exists an $M \in \mathbb{N}^#$ such that for all $n > M$, $|b_n - c_n| < \varepsilon/2$. Well, then, as long as $n$ is bigger than both $N$ and $M$, we have that $|a_n - c_n| = |(a_n - b_n) + (b_n - c_n)| \le |a_n - b_n| + |b_n - c_n| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|a_n - c_n| < \varepsilon$. This means that $a_n - c_n \to_{\#} 0$ i.e. $\{a_n\}$ is related to $\{c_n\}$.

**Definition 11.6.** The hyperreal numbers $\mathbb{R}_c^#$ are the equivalence classes $\text{cl}[\{a_n\}]$ of Cauchy sequences of hyperrational numbers, as per Definition 11.5. That is, each such equivalence class is a hyperreal number.

**Definition 11.7.** Given any hyperrational number $q \in \mathbb{Q}^#$, define a hyperreal number $q^#$ to be the equivalence class of the sequence $q^# = (q, q, q, q, \dots)$ consisting entirely of $q$. So we view $\mathbb{Q}^#$ as being inside $\mathbb{R}_c^#$ by thinking of each hyperrational number $q \in \mathbb{Q}^#$ as its associated equivalence class $q^#$. It is standard to abuse this notation, and simply refer to the equivalence class as $q$ as well.

**Definition 11.8.** Let $s, t \in \mathbb{R}_c^#$, so there are Cauchy sequences $\{a_n\}, \{b_n\}$ of hyperrational numbers with $s = \text{cl}[\{a_n\}]$ and $t = \text{cl}[\{b_n\}]$.

(a) Define $s + t$ to be the equivalence class of the sequence $\{a_n + b_n\}$.

(b) Define $s \times t$ to be the equivalence class of the sequence $\{a_n \times b_n\}$.

**Theorem 11.5.** The operations $+, \times$ in Definition 8.8 (a),(b) are well-defined.

**Proof.** Suppose that $\text{cl}[\{a_n\}] = \text{cl}[\{c_n\}]$ and $\text{cl}[\{b_n\}] = \text{cl}[\{d_n\}]$. Thus means that $a_n - c_n \to_{\#} 0$ and $b_n - d_n \to_{\#} 0$. Then $(a_n + b_n) - (c_n + d_n) = (a_n - c_n) + (b_n - d_n)$.

Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to 0, and so $\text{cl}[\{a_n + b_n\}] = \text{cl}[\{c_n + d_n\}]$.

Multiplication is a little trickier; this is where we will use Theorem 11.2. We will also use another ubiquitous technique: adding 0 in the form of $s - s$. Again, suppose that $\text{cl}[(a_n) = \text{cl}[(c_n)]$ and $\text{cl}[(b_n) = \text{cl}[(d_n)]$; we wish to show that $\text{cl}[(a_n \times b_n) = \text{cl}[(c_n \times d_n)]$, or, in other words, that $a_n \times b_n - c_n \cdot d_n \to_{\#} 0$. Well, we add and subtract one of the other cross terms, say

$$b_n \times c_n : a_n \times b_n - c_n \times d_n = a_n \times b_n + (b_n \times c_n - b_n \times c_n) - c_n \times d_n = \\ = (a_n \times b_n - b_n \times c_n) + (b_n \times c_n - c_n \times d_n) = b_n \times (a_n - c_n) + c_n \times (b_n - d_n).$$
multiplicative inverse to $s = \text{cl}[\{a_n\}]$.

**Definition 11.9.** Let $s \in \mathbb{R}_c^#$. Say that $s$ is positive if $s \neq 0$, and if $s = \text{cl}[\{a_n\}]$ for some Cauchy sequence of hyperrational numbers such that for some $N \in \mathbb{N}^#$, $a_n > 0$ for all $n > N$. Given two hyperreal numbers $s, t$, say that $s > t$ if $s - t$ is positive.

**Theorem 11.7.** Let $s, t$ be hyperreal numbers such that $s > t$, and let $r \in \mathbb{R}_c^#$. Then $s + r > t + r$.

**Proof.** Let $s = \text{cl}[\{a_n\}]$, $t = \text{cl}[\{b_n\}]$, and $r = \text{cl}[\{c_n\}]$. Since $s > t$ i.e., $s - t > 0$, we know that there is an $N \in \mathbb{N}^#$ such that, for $n > N$, $a_n - b_n > 0$. So $a_n > b_n$ for $n > N$. Now, adding $c_n$ to both sides of this inequality (as we know we can do for hyperrational numbers), we have $a_n + c_n > b_n + c_n$ for $n > N$, or $(a_n + c_n) - (b_n + c_n) > 0$ for $n > N$. Note also that $(a_n + c_n) - (b_n + c_n) = a_n - b_n$ does not #-converge to 0, by the assumption that $s - t > 0$. Thus, by Definition 11.8, this means that $s + r = \text{cl}[\{a_n + c_n\}] > \text{cl}[\{b_n + c_n\}] = t + r$.

**Theorem 11.8.** (Generalized Archimedean property) Let $s, t > 0$ be hyperreal numbers. Then there is $m \in \mathbb{N}^#$ such that $m \times s > t$.

**Proof.** Let $s, t > 0$ be hyperreal numbers. We need to find a hypernatural number $m$ so that $m \times s > t$. First, recall that, by $m$ in this context, we mean $\text{cl}[\{m, m, m, m, \dots\}]$. So, letting $s = \text{cl}[\{a_n\}]$ and $t = \text{cl}[\{b_n\}]$, what we need to show is that there exists $m \in \mathbb{N}^#$ with $\text{cl}[\{m, m, m, m, \dots\}] \times \text{cl}[\{a_1, a_2, a_3, a_4, \dots\}] = \text{cl}[\{m \times a_1, m \times a_2, m \times a_3, m \times a_4, \dots\}] > \text{cl}[\{b_1, b_2, b_3, b_4, \dots\}]$.

Now, to say that $\text{cl}[\{m \times a_n\}] > \text{cl}[\{b_n\}]$, or $\text{cl}[\{m \times a_n - b_n\}]$ is positive, is, by Definition 11.9, just to say that there is $N \in \mathbb{N}^#$ such that $m \times a_n - b_n > 0$ for all $n > N$, while $m \times a_n - b_n \not\to_# 0$. To be precise, the first statement is: There exist $m, N \in \mathbb{N}^#$ so that $m \times a_n > b_n$ for all $n > N$.

To produce a contradiction, we assume this is not the case; assume that (#) for every $m$ and $N$, there exists an $n > N$ so that $m \times a_n \leq b_n$.

Now, since $\{b_n\}$ is a Cauchy sequence, by Theorem 11.2 it is hyperbounded – there is a hyperrational number $M \in \mathbb{Q}^#$ such that $b_n \leq M$ for all $n$. Now, by the properties for the hyperrational numbers $\mathbb{Q}^#$, given any hyperrational number $\varepsilon > 0$, $\varepsilon \approx 0$, there is an $m \in \mathbb{N}^#$ such that $M/m < \varepsilon/2$. Fix such an $m$. Then if $m \times a_n \leq b_n$, we have $a_n \leq b_n/m \leq M/m < \varepsilon/2$.

Now, $\{a_n\}$ is a Cauchy sequence, and so there exists $N$ so that for $n, k > N$, $|a_n - a_k| < \varepsilon/2$.

By Asumption (#), we also have an $n > N$ such that $m \times a_n \leq b_n$, which means that $a_n < \varepsilon/2$. But then for every $k > N$, we have that $a_k - a_n < \varepsilon/2$, so $a_k < a_n + \varepsilon/2 < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Hence, $a_k < \varepsilon$ for all $k > N$. This proves that $a_k \to_# 0$, which by Definition 11.9 contradicts the fact that $\text{cl}[\{a_n\}] = s > 0$.

Thus, there is indeed some $m \in N$ so that $m \times a_n - b_n > 0$ for all sufficiently infinite large $n \in \mathbb{N}^# \setminus \mathbb{N}$. To conclude the proof, we must also show that $m \times a_n - b_n \not\to 0$.

Actually, it is possible that $m \times a_n - b_n \to 0$ (for example if $\{a_n\} = \{1, 1, 1, \dots\}$ and $\{b_n\} = \{m, m, m, \dots\}$). But that's okay: then we can simply choose a larger $m$. That is: let $m$ be a hypernatural number constructed as above, so that $m \times a_n - b_n > 0$ for all sufficiently large $n \in \mathbb{N}^# \setminus \mathbb{N}$. If it happens to be true that $m \times a_n - b_n \not\to 0$, then the proof is complete.
Hence, we have $|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n|$. Now, from Theorem 11.2, there are numbers $M$ and $L$ such that $|b_n| \le M$ and $|c_n| \le L$ for all $n \in \mathbb{N}^\#$. Taking some number $K$ which is bigger than both, we have

$$|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n| \le K(|a_n - c_n| + |b_n - d_n|).$$

Now, noting that both $a_n - c_n$ and $b_n - d_n$ tend to 0 and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2K$), we see that $a_n \times b_n - c_n \times d_n \to_\# 0$.

**Theorem 11.6.** Given any hyperreal number $s \ne 0$, there is a hyperreal number $t$ such that $s \times t = 1$.

**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 $\{0,0,0,0,\dots\}$. In other words, $s = \text{cl}[\{a_n\}]$ where $a_n = 0$ does not #-converge to 0. From this, we are to deduce the existence of a hyperreal number $t = \text{cl}[\{b_n\}]$ such that $s \times t = \text{cl}[\{a_n \times b_n\}]$ is the same equivalence class as $\text{cl}[\{1,1,1,1,\dots\}]$. Doing so is actually an easy consequence of the fact that nonzero rational numbers have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{a_n\}$ does not tend to 0), there's no reason any number of the terms in $\{a_n\}$ can't equal 0. However, it turns out that eventually, $a_n \ne 0$.

That is,

**Lemma 11.1.** If $\{a_n\}$ is a Cauchy hyper infinite sequence which does not #-tend to 0, then there is an $N \in \mathbb{N}^\#/\mathbb{N}$ such that, for $n > N, a_n \ne 0$.

We will now use Lemma 11.1 to complete the proof of Theorem 11.7.

Let $N$ be such that $a_n \ne 0$ for $n > N$. Define a hyper infinite sequence $b_n$ of hyperrational numbers as follows:

for $n \le N, b_n = 0$, and for $n > N, b_n = 1/a_n; \{b_n\} = (0,0,\dots,0, 1/a_{N+1}, 1/a_{N+2},\dots)$.

This makes sense since, for $n > N$, an is a nonzero hyperrational number, so $1/a_n$ exists. Then $a_n \cdot b_n$ is equal to $a_n \cdot 0 = 0$ for $n \le N$, and equals $a_n \cdot b_n = a_n \cdot 1/a_n = 1$ for $n > N$. Well, then, if we

look at the hyper infinite sequence $(1,1,1,1,\dots)$, we have $(1,1,1,1,\dots) - (a_n \cdot b_n)$ is the

hyper infinite sequence which is $1 - 0 = 1$ for $n \le N$ and equals $1 - 1 = 0$ for $n > N$.

Since this sequence is eventually equal to 0, it #-converges to 0, and so

$$\text{cl}[\{a_n \cdot b_n\}] = \text{cl}[(1,1,1,1,\dots)] = 1 \in \mathbb{R}^\#_c. \text{ This shows that } t = \text{cl}[\{b_n\}] \text{ is a multiplicative inverse to } s = \text{cl}[\{a_n\}].$$

**Definition 11.9.** Let $s \in \mathbb{R}^\#_c$. Say that $s$ is positive if $s \ne 0$, and if $s = \text{cl}[\{a_n\}]$ for some Cauchy sequence of hyperrational numbers such that for some $N \in \mathbb{N}^\#$, $a_n > 0$ for all $n > N$. Given two hyperreal numbers $s, t$, say that $s > t$ if $s - t$ is positive.

**Theorem 11.7.** Let $s, t$ be hyperreal numbers such that $s > t$, and let $r \in \mathbb{R}^\#_c$. Then $s + r > t + r$.

**Proof.** Let $s = \text{cl}[\{a_n\}], t = \text{cl}[\{b_n\}],$ and $r = \text{cl}[\{c_n\}]$. Since $s > t$ i.e., $s - t > 0$, we know that there is an $N \in \mathbb{N}^\#$ such that, for $n > N, a_n - b_n > 0$. So $a_n > b_n$ for $n > N$. Now, adding $c_n$ to both sides of this inequality (as we know we can do for hyperrational numbers), we have $a_n + c_n > b_n + c_n$ for

$n > N$, or $(a_n + c_n) - (b_n + c_n) > 0$ for $n > N$. Note also that $(a_n + c_n) - (b_n + c_n) = a_n - b_n$ does not #-converge to 0, by the assumption that $s - t > 0$. Thus, by Definition 11.8, this means that

$$s + r = \text{cl}[\{a_n + c_n\}] > \text{cl}[\{b_n + c_n\}] = t + r.$$
If, on the other hand, it turned out that $m \times a_n - b_n \to 0$, then take instead the integer $m+1$. Since $s = \text{cl}[\{a_n\}] > 0$, we have a $n > 0$ for all infinite large $n$, so $(m+1) \times a_n - b_n = m \times a_n - b_n + a_n > a_n > 0$ for all infinite large $n$, so $m+1$ works just as

well as $m$ did in this regard; and since $m \times a_n - b_n \to 0$, we have

$$(m+1) \times a_n - b_n = (m \times a_n - b_n) + a_n \to 0 \text{ since } s = \text{cl}[\{a_n\}] > 0 \text{ (so } a_n \to 0).$$

It will be handy to have one more Theorem about how the hyperrationals $\mathbb{Q}^\#$ and hyperreals $\mathbb{R}_c^\#$ compare before we proceed. This theorem is known as the density of $\mathbb{Q}^\#$ in $\mathbb{R}_c^\#$, and it follows almost immediately from the construction of the $\mathbb{R}_c^\#$ from $\mathbb{Q}^\#$.

**Theorem 11.9.** Given any hyperreal number $r \in \mathbb{R}_c^\#$, and any hyperrational number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hyperrational number $q \in \mathbb{Q}^\#$ such that $|r-q| < \varepsilon$.

**Proof.** The hyperreal number $r$ is represented by a Cauchy hyperrational sequence $\{a_n\}$.

Since this sequence is Cauchy, given $\varepsilon > 0$, $\varepsilon \approx 0$, there is $N \in \mathbb{N}^\#$ so that for all $m, n > N$,

$|a_n - a_m| < \varepsilon$. Picking some fixed $l > N$, we can take the hyperrational number $q$ given by $q = \text{cl}[\{a_l, a_{l+1}, a_{l+2}, \dots\}]$. Then we have $r-q = \text{cl}[\{a_n - a_l\}_{n \in \mathbb{N}^\#}]$, and $q-r = \text{cl}[\{a_l - a_n\}_{n \in \mathbb{N}^\#}]$.

Now, since $l > N$, we see that for $n > N$, $a_n - a_l < \varepsilon$ and $a_l - a_n < \varepsilon$, which means by Definition 11.9 that $r-q < \varepsilon$ and $q-r < \varepsilon$; hence, $|r-q| < \varepsilon$.

**Definition 11.10.** 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 11.1.** The order $\le$ given by Definition 11.9 obviously is $\le$-incomplete.

**Definition 11.11.** 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 exists $\alpha \in S$ 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 exists $\alpha \in S$ and $\beta \in L(S)$ such that $\alpha - \beta \le \varepsilon \approx 0$.

**Theorem 11.10.** (i) Any $\le$-admissible above subset $S \subset \mathbb{R}_c^\#$ has the least upper bound property. (ii) Any $\le$-admissible below subset $S \subset \mathbb{R}_c^\#$ has the greatest lower bound property.

**Proof.** Let $S \subset \mathbb{R}_c^\#$ be a nonempty subset, and let $M$ be an upper bound for $S$. We are going to construct two sequences of hyperreal numbers, $\{u_n\}$ and $\{l_n\}$. First, since $S$ is nonempty, there is some element $s_0 \in S$. Now, we go through the following hyperinductive procedure to produce numbers $u_0, u_1, u_2, \dots, u_n, \dots$ and $l_1, l_2, l_3, \dots, l_n, \dots$

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

(ii) Suppose that we have already defined $u_n$ and $l_n$. Consider the number $m_n = (u_n + l_n)/2$, 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 $l_{n+1} = l_n$.
**Theorem 11.8.** (Generalized Archimedean property) Let $s, t > 0$ be hyperreal numbers.
Then there is $m \in \mathbb{N}^\#$ such that $m \times s > t$.

**Proof.** Let $s, t > 0$ be hyperreal numbers. We need to find a hypernatural number $m$ so that $m \times s > t$. First, recall that, by $m$ in this context, we mean $\text{cl}[\{m, m, m, m, \dots\}]$. So, letting $s = \text{cl}[\{a_n\}]$ and $t = \text{cl}[\{b_n\}]$, what we need to show is that there exists $m \in \mathbb{N}^\#$ with $\text{cl}[\{m, m, m, m, \dots\}] \times \text{cl}[\{a_1, a_2, a_3, a_4, \dots\}] = \text{cl}[\{m \times a_1, m \times a_2, m \times a_3, m \times a_4, \dots\}] > \text{cl}[\{b_1, b_2, b_3, b_4, \dots\}]$.

Now, to say that $\text{cl}[\{m \times a_n\}] > \text{cl}[\{b_n\}]$, or $\text{cl}[\{m \times a_n - b_n\}]$ is positive, is, by Definition 11.9, just to say that there is $N \in \mathbb{N}^\#$ such that $m \times a_n - b_n > 0$ for all $n > N$, while $m \times a_n - b_n \not\to_\# 0$. To be precise, the first statement is:
There exist $m, N \in \mathbb{N}^\#$ so that $m \times a_n > b_n$ for all $n > N$.

To produce a contradiction, we assume this is not the case; assume that
(#) for every $m$ and $N$, there exists an $n > N$ so that $m \times a_n \le b_n$.

Now, since $\{b_n\}$ is a Cauchy sequence, by Theorem 11.2 it is hyperbounded – there is a hyperrational number $M \in \mathbb{Q}^\#$ such that $b_n \le M$ for all $n$. Now, by the properties for the hyperrational numbers $\mathbb{Q}^\#$, given any hyperrational number $\varepsilon > 0$, $\varepsilon \approx 0$, there is an $m \in \mathbb{N}^\#$ such that $M/m < \varepsilon/2$. Fix such an $m$. Then if $m \times a_n \le b_n$, we have $a_n \le b_n/m \le M/m < \varepsilon/2$.

Now, $\{a_n\}$ is a Cauchy sequence, and so there exists $N$ so that for $n, k > N$, $|a_n - a_k| < \varepsilon/2$.

By Asumption (#), we also have an $n > N$ such that $m \times a_n \le b_n$, which means that $a_n < \varepsilon/2$. But then for every $k > N$, we have that $a_k - a_n < \varepsilon/2$, so $a_k < a_n + \varepsilon/2 < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Hence, $a_k < \varepsilon$ for all $k > N$. This proves that $a_k \to_\# 0$, which by Definition 11.9 contradicts the fact that $\text{cl}[\{a_n\}] = s > 0$.

Thus, there is indeed some $m \in N$ so that $m \times a_n - b_n > 0$ for all sufficiently infinite large $n \in \mathbb{N}^\#\mathbb{N}$. To conclude the proof, we must also show that $m \times a_n - b_n \not\to 0$. Actually, it is possible that $m \times a_n - b_n \to 0$ (for example if $\{a_n\} = \{1, 1, 1, \dots\}$ and $\{b_n\} = \{m, m, m, \dots\}$). But that's okay: then we can simply choose a larger $m$. That is: let $m$ be a hypernatural number constructed as above, so that $m \times a_n - b_n > 0$ for all sufficiently large $n \in \mathbb{N}^\#\mathbb{N}$. If it happens to be true that $m \times a_n - b_n \not\to 0$, then the proof is complete.

If, on the other hand, it turned out that $m \times a_n - b_n \to 0$, then take instead the integer $m+1$. Since $s = \text{cl}[\{a_n\}] > 0$, we have a $n > 0$ for all infinite large $n$, so $(m+1) \times a_n - b_n = m \times a_n - b_n + a_n > a_n > 0$ for all infinite large $n$, so $m+1$ works just as

well as $m$ did in this regard; and since $m \times a_n - b_n \to 0$, we have

$$ (m+1) \times a_n - b_n = (m \times a_n - b_n) + a_n \not\to 0 \text{ since } s = \text{cl}[\{a_n\}] > 0 \text{ (so } a_n \not\to 0). $$

It will be handy to have one more Theorem about how the hyperrationals $\mathbb{Q}^\#$ and hyperreals $\mathbb{R}_c^\#$ compare before we proceed. This theorem is known as the density of $\mathbb{Q}^\#$ in $\mathbb{R}_c^\#$, and it follows almost immediately from the construction of the $\mathbb{R}_c^\#$ from $\mathbb{Q}^\#$.

**Theorem 11.9.** Given any hyperreal number $r \in \mathbb{R}_c^\#$, and any hyperrational number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hyperrational number $q \in \mathbb{Q}^\#$ such that $|r - q| < \varepsilon$.

**Proof.** The hyperreal number $r$ is represented by a Cauchy hyperrational sequence $\{a_n\}$.
(2) If $m_n$ is not an upper bound for $S$, define $u_{n+1} = u_n$ and $l_{n+1} = m_n$.

Since $s < M$, it is easy to prove by hyper infinite induction that (i) $\{u_n\}$ is a non-increasing sequence: $u_{n+1} \le u_n, n \in \mathbb{N}^#(ii) \{l_n\}$ is a non-decreasing sequence $l_{n+1} \ge l_n, n \in \mathbb{N}^# $ and (iii) $u_n - l_n = 2^{-n}(M-s)$.

This gives us the following lemma.

**Lemma 11.2.** $\{u_n\}$ and $\{l_n\}$ are Cauchy sequences of hyperreal numbers.

**Proof.** Note that each $l_n \le M$ for all $n \in \mathbb{N}^#$. Since $\{l_n\}$ is non-decreasing and $u_n - l_n = 2^{-n}(M-s)$, it follows directly that $\{l_n\}$ is Cauchy.

For $\{u_n\}$, we have $u_n \ge s_0$ for all $n \in \mathbb{N}^# $, and so $-u_n \le -s_0$.

Since $\{u_n\}$ is non-increasing, $\{-u_n\}$ is non-decreasing, and so as above, $\{-u_n\}$ is Cauchy. It is easy to verify that, therefore, $\{u_n\}$ is Cauchy.

The following Lemma shows that $\{u_n\}$ does tend to a hyperreal number.

**Lemma 11.3.** There is a hyperreal number $u$ such that $u_n \to_{\#} u$.

**Proof.** Fix a term $u_n$ in the sequence $\{u_n\}$. By Theorem 11.9, there is a hyperrational number $q_n$ such that $|u_n - q_n| < 1/n$. Consider the sequence $\{q_1, q_2, q_3, \dots, q_n, \dots\}$ of hyperrational numbers. We will show this hypersequence is Cauchy. Fix $\varepsilon > 0, \varepsilon \approx 0$. By the Theorem 11.8, we can choose $N \in \mathbb{N}^# $ so that $1/N < \varepsilon/3$. We know, since $\{u_n\}$ is Cauchy, that there is an $M \in \mathbb{N}^# $ such that for $n, m > M, |u_n - u_m| < \varepsilon/3$. Then, so long as $n, m > \max\{N, M\}$, we have

$$|q_n - q_m| = |(q_n - u_n) + (u_n - u_m) + (u_m - q_m)| \leq \\ \leq |q_n - u_n| + |u_n - u_m| + |u_m - q_m| < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon.$$

Thus, $\{q_n\}$ is a Cauchy sequence of hyperrational numbers, and so it represents a hyperreal number $u = \text{cl}[\{q_n\}]$. We must show that $u_n - u \to_{\#} 0$, but this is practically built into the definition of $u$. To be precise, letting $q_n^*$ be the hyperreal number $\text{cl}[\{q_n, q_n, q_n, \dots\}]$, we see immediately that $q_n^* - u \to_{\#} 0$ (this is precisely equivalent to the statement that $\{q_n\}$ is Cauchy). But $u_n - q_n^* < 1/n$ by construction; it is easily verify that the assertion that if a sequence $q_n^* \to_{\#} u$ and $u_n - q_n^* \to_{\#} 0$, then $u_n \to_{\#} u$. So $\{u_n\}$, a non-increasing sequence of upper bounds for $S$, tends to a hyperreal number $u$.

As you've guessed, $u$ is the least upper bound of our set $S$. To prove this, we need one more lemma.

**Lemma 11.4.** $l_n \to_{\#} u$.

**Proof.** First, note in the first case above, we have that

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

In the second case, we also have

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

Now, this means that $u_1 - l_1 = \frac{1}{2}(M-s)$, and so $u_2 - l_2 = \frac{1}{2}(u_1 - l_1) = \frac{1}{2^2}(M-s)$, and in general by hyper infinite induction, $u_n - l_n = 2^{-n}(M-s)$. Since $M > s$ so $L - s > 0$, and since $2^{-n} < 1/n$, by the Theorem 11.8, we have for any $\varepsilon > 0$ that $2^{-n}(L-s) < \varepsilon$ for all sufficiently large $n \in \mathbb{N}^#/\mathbb{N}$. Thus, $u_n - l_n = 2^{-n}(L-s) < \varepsilon$ as well, and so $u_n - l_n \to_{\#} 0$. Again, it is easily verify that, since $u_n \to_{\#} u$, we have $l_n \to_{\#} u$ as well.

**Proof of Theorem 11.10.** First, we show that $u$ is an upper bound. Well, suppose
Since this sequence is Cauchy, given $\varepsilon > 0, \varepsilon \approx 0$, there is $N \in \mathbb{N}^\#$ so that for all $m, n > N$,

$|a_n - a_m| < \varepsilon$. Picking some fixed $l > N$, we can take the hyperrational number $q$ given by
$$q = \text{cl}[\{a_l, a_{l+1}, a_{l+2}, \dots\}].$$
Then we have $r - q = \text{cl}[\{a_n - a_l\}_{n \in \mathbb{N}^\#}]$, and
$$q - r = \text{cl}[\{a_l - a_n\}_{n \in \mathbb{N}^\#}].$$

Now, since $l > N$, we see that for $n > N, a_n - a_l < \varepsilon$ and $a_l - a_n < \varepsilon$, which means by Definition 11.9 that $r - q < \varepsilon$ and $q - r < \varepsilon$; hence, $|r - q| < \varepsilon$.

**Definition 11.10.** 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 11.1.** The order $\le$ given by Definition 11.9 obviously is $\le$-incomplete.

**Definition 11.11.** 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 exst $\alpha \in S$ and $\beta \in A(S)$ such that $\beta - \alpha \le \varepsilon \approx 0$.

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

(i) $S$ bounded belov;

(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 exst $\alpha \in S$ and $\beta \in L(S)$ such that $\alpha - \beta \le \varepsilon \approx 0$.

**Theorem 11.10.** (i) Any $\le$-admissible above subset $S \subset \mathbb{R}_c^\#$ has the least upper bound property.(ii) Any $\le$-admissible below subset $S \subset \mathbb{R}_c^\#$ has the greatest lower bound property.

**Proof.** Let $S \subset \mathbb{R}_c^\#$ be a nonempty subset, and let $M$ be an upper bound for $S$. We are going to construct two sequences of hyperreal numbers, $\{u_n\}$ and $\{l_n\}$. First, since $S$ is nonempty, there is some element $s_0 \in S$. Now, we go through the following hyperinductive procedure to produce numbers $u_0, u_1, u_2, \dots, u_n, \dots$ and $l_1, l_2, l_3, \dots, l_n, \dots$

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

(ii) Suppose that we have already defined $u_n$ and $l_n$. Consider the number $m_n = (u_n + l_n)/2$, 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 $s < M$, it is easy to prove by hyper infinite induction that (i) $\{u_n\}$ is a non-increasing sequence: $u_{n+1} \le u_n, n \in \mathbb{N}^\#$(ii) $\{l_n\}$ is a non-decreasing sequence $l_{n+1} \ge l_n, n \in \mathbb{N}^\#$ and (iii) $u_n - l_n = 2^{-n}(M-s)$.

This gives us the following lemma.

**Lemma 11.2.** $\{u_n\}$ and $\{l_n\}$ are Cauchy sequences of hyperreal numbers.

**Proof.** Note that each $l_n \le M$ for all $n \in \mathbb{N}^\#$. Since $\{l_n\}$ is non-decreasing and $u_n - l_n = 2^{-n}(M-s)$, it follows directly that $\{l_n\}$ is Cauchy.

For $\{u_n\}$, we have $u_n \ge s_0$ for all $n \in \mathbb{N}^\#$, and so $-u_n \le -s_0$. Since $\{u_n\}$ is non-increasing, $\{-u_n\}$ is non-decreasing, and so as above, $\{-u_n\}$ is Cauchy. It is easy to verify that, therefore, $\{u_n\}$ is Cauchy.

The following Lemma shows that $\{u_n\}$ does tend to a hyperreal number.

**Lemma 11.3.** There is a hyperreal number $u$ such that $u_n \to \# u$.

**Proof.** Fix a term $u_n$ in the sequence $\{u_n\}$. By Theorem 11.9, there is a hyperrational
it is not, so that $u < s$ for some $s \in S$. Then $\varepsilon = s - u$ is $> 0$, and since $u_n \to u$ and is non-increasing, there must be an $n$ so that $u_n - u < \varepsilon$, meaning that $u_n < u + \varepsilon = u + (s - u) = s$. Since $u_n$ is an upper bound for $S$, however, this is a contradiction. Hence, $u$ is an upper bound for $S$.

Now, we also know that, for each $n, l_n$ is not an upper bound, meaning that for each $n$, there is an $s_n \in S$ so that $l_n \le s_n$. Lemma 11.4 tells us that $l_n \to_{\#} u$, and since the sequence $\{l_n\}$ is non-decreasing, this means that for each $\varepsilon > 0$, there is an $N \in \mathbb{N}^\#/\mathbb{N}$ so that for $n > N, l_n > u - \varepsilon$. Hence, for $n > N, s_n \ge l_n > u - \varepsilon$ as well. In particular, for each $\varepsilon > 0$, there is an element $s \in S$ such that $s > u - \varepsilon$. This means that no number smaller than $u$ can be an upper bound for $S$. Hence, $u$ is the least upper bound for $S$.

**Remark 11.2.** Note that assumption in Theorem 11.10 that $S$ is $\le$-admissible above subset of $\mathbb{R}_c^\#$ is necessarily, othervice Theorem 11.10 is not holds. For example let $\Delta = \{\varepsilon|\varepsilon \ge 0 \land \varepsilon \approx 0\}$. Obviously a set $\Delta$ is not $\le$-admissible above subset of $\mathbb{R}_c^\#$. It is clear that Theorem 11.10 is not holds for a set $\Delta$.

**Theorem 11.11. (Generalized Nested Intervals Theorem)**

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

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

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

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

**Proof.** Note that: (a) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is bounded or hyperbouded above by $b_1$ and (b) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is $\le$-admissible above subset of $\mathbb{R}_c^\#$.

By Theorem 11.10 there exists $\sup A$. Let $\xi = \sup A$.

Since $I_n$ are nested, for any positive hyperintegers $m$ and $n$ we have $a_m \le a_{m+n} \le b_{m+n} \le b_n$, so that $\xi \le b_n$ for each $n \in \mathbb{N}^\#$. Since we obviously have $a_n \le \xi$ for each $n \in \mathbb{N}^\#$, we have $a_n \le \xi \le b_n$ for all $n \in \mathbb{N}^\#$, which implies $\xi \in \cap_{n=1}^\infty I_n$. Finally, if $\xi, \eta \in \cap_{n=1}^\infty I_n$, with $\xi \le \eta$, then we get $0 \le \eta - \xi \le b_n - a_n$, for all $n \in \mathbb{N}^\#$, so that $0 \le \eta - \xi \le \inf_{n \in \mathbb{N}^\#} |b_n - a_n| = 0$.

**Theorem 11.12. (Generalized Squeeze Theorem)**

Let $\{a_n\}$, $\{c_n\}$ be two hyper infinite sequences $\#$-converging to $L$, and $\{b_n\}$ a hyper infinite sequence. If $\forall n \ge K, K \in \mathbb{N}^\#$ we have $a_n \le b_n \le c_n$, then $\{b_n\}$ also $\#$-converges to $L$.

**Proof.** Choose an $\varepsilon > 0, \varepsilon \approx 0$. By definition of the $\#$-limit, there is an $N_1 \in \mathbb{N}^\#$ such that for all $n > N_1$ we have $|a_n - L| < \varepsilon$, in other words $L - \varepsilon < a_n < L + \varepsilon$. Similarly, there is an $N_2 \in \mathbb{N}^\#$ such that for all $n > N_2$ we have $L - \varepsilon < c_n < L + \varepsilon$. Denote $N = \max(N_1, N_2, K)$. Then for $n > N, L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon$, in other words $|b_n - L| < \varepsilon$. Since $\varepsilon > 0, \varepsilon \approx 0$ was arbitrary, by definition of the $\#$-limit this says that $\#\text{-lim}_{n \to \infty^\#} b_n = L$.

**Theorem 11.13. (Corollary of the Generalized Squeeze Theorem).**

If $\#\text{-lim}_{n \to \infty^\#} |a_n| = 0$ then $\#\text{-lim}_{n \to \infty^\#} a_n = 0$.

**Proof.** We know that $-|a_n| \le a_n \le |a_n|$. We want to apply the Generalized Squeeze Theorem. We are given that $\#\text{-lim}_{n \to \infty^\#} |a_n| = 0$. This also implies that $\#\text{-lim}_{n \to \infty^\#} (-|a_n|) = 0$. So by the Generalized Squeeze Theorem, $\#\text{-lim}_{n \to \infty^\#} a_n = 0$.

**Theorem 11.14. (Generalized Bolzano-Weierstrass Theorem)**

Every hyperbounded hyperinfinite sequence has a $\#$-convergent hyper infinite
number $q_n$ such that $|u_n - q_n| < 1/n$. Consider the sequence $\{q_1, q_2, q_3, \dots, q_n, \dots\}$ of hyperrational numbers. We will show this hypersequence is Cauchy. Fix $\varepsilon > 0$, $\varepsilon \approx 0$. By the Theorem 11.8, we can choose $N \in \mathbb{N}^\#$ so that $1/N < \varepsilon/3$. We know, since $\{u_n\}$ is Cauchy, that there is an $M \in \mathbb{N}^\#$ such that for $n, m > M$, $|u_n - u_m| < \varepsilon/3$. Then, so long as $n, m > \max\{N, M\}$, we have

$$ \begin{aligned} |q_n - q_m| &= |(q_n - u_n) + (u_n - u_m) + (u_m - q_m)| \le \\ &\le |q_n - u_n| + |u_n - u_m| + |u_m - q_m| < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon. \end{aligned} $$

Thus, $\{q_n\}$ is a Cauchy sequence of hyperrational numbers, and so it represents a hyperreal number $u = \text{cl}[\{q_n\}]$. We must show that $u_n - u \to_\# 0$, but this is practically built into the definition of $u$. To be precise, letting $q_n^*$ be the hyperreal number $\text{cl}[\{q_n, q_n, q_n, \dots\}]$, we see immediately that $q_n^* - u \to_\# 0$ (this is precisely equivalent to the statement that $\{q_n\}$ is Cauchy). But $u_n - q_n^* < 1/n$ by construction; it is easily verify that the assertion that if a sequence $q_n^* \to_\# u$ and $u_n - q_n^* \to_\# 0$, then $u_n \to_\# u$. So $\{u_n\}$, a non-increasing sequence of upper bounds for $S$, tends to a hyperreal

number $u$. As you've guessed, $u$ is the least upper bound of our set $S$. To prove this, we

need one more lemma.

**Lemma 11.4.** $l_n \to_\# u$.

**Proof.** First, note in the first case above, we have that

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

In the second case, we also have

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

Now, this means that $u_1 - l_1 = \frac{1}{2}(M - s)$, and so $u_2 - l_2 = \frac{1}{2}(u_1 - l_1) = \frac{1}{2^2}(M - s)$, and in general by hyperinfinite induction, $u_n - l_n = 2^{-n}(M - s)$. Since $M > s$ so $L - s > 0$, and since $2^{-n} < 1/n$, by the Theorem 11.8, we have for any $\varepsilon > 0$ that $2^{-n}(L - s) < \varepsilon$ for all sufficiently large $n \in \mathbb{N}^\#/\mathbb{N}$. Thus, $u_n - l_n = 2^{-n}(L - s) < \varepsilon$ as well, and so $u_n - l_n \to_\# 0$. Again, it is easily verify that, since $u_n \to_\# u$, we have $l_n \to_\# u$ as well.

**Proof of Theorem 11.10.** First, we show that $u$ is an upper bound. Well, suppose it is not, so that $u < s$ for some $s \in S$. Then $\varepsilon = s - u$ is $> 0$, and since $u_n \to u$ and is non-increasing, there must be an $n$ so that $u_n - u < \varepsilon$, meaning that $u_n < u + \varepsilon = u + (s - u) = s$. Since $u_n$ is an upper bound for $S$, however, this is a contradiction. Hence, $u$ is an upper bound for $S$.

Now, we also know that, for each $n, l_n$ is not an upper bound, meaning that for each $n$, there is an $s_n \in S$ so that $l_n \le s_n$. Lemma 11.4 tells us that $l_n \to_\# u$, and since the sequence $\{l_n\}$ is non-decreasing, this means that for each $\varepsilon > 0$, there is an $N \in \mathbb{N}^\#/\mathbb{N}$ so that for $n > N, l_n > u - \varepsilon$. Hence, for $n > N, s_n \ge l_n > u - \varepsilon$ as well. In particular, for each $\varepsilon > 0$, there is an element $s \in S$ such that $s > u - \varepsilon$. This means that no number smaller than $u$ can be an upper bound for $S$. Hence, $u$ is the least upper bound for $S$.

**Remark 11.2.** Note that assumption in Theorem 11.10 that $S$ is $\le$-admissible above subset of $\mathbb{R}_c^\#$ is necessarily, othervice Theorem 11.10 is not holds. For example let $\Delta = \{\varepsilon | \varepsilon \ge 0 \wedge \varepsilon \approx 0\}$. Obviously a set $\Delta$ is not $\le$-admissible above subset of $\mathbb{R}_c^\#$. It is clear that Theorem 11.10 is not holds for a set $\Delta$.
subsequence.

**Proof.** Let {$w_n$}$_{n \in \mathbb{N}^\#}$ be a hyperbounded hyper infinite sequence. Then, there exists an interval $[a_1, b_1]$ such that $a_1 \le w_n \le b_1$ for all $n \in \mathbb{N}^\#$.

Either $[a_1, \frac{a_1+b_1}{2}]$ or $[\frac{a_1+b_1}{2}, b_1]$ contains hyper infinitely many terms of $\{w_n\}$. That is, there exists hyperinfinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[a_1, \frac{a_1+b_1}{2}]$ or there exists hyper infinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[\frac{a_1+b_1}{2}, b_1]$. If $[a_1, \frac{a_1+b_1}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_2, b_2] = [a_1, \frac{a_1+b_1}{2}]$. Otherwise, let $[a_2, b_2] = [\frac{a_1+b_1}{2}, b_1]$. Either $[a_2, \frac{a_2+b_2}{2}]$ or $[\frac{a_2+b_2}{2}, b_2]$ contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^\#}$. If $[a_2, \frac{a_2+b_2}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_3, b_3] = [a_2, \frac{a_2+b_2}{2}]$. Otherwise, let $[a_3, b_3] = [\frac{a_2+b_2}{2}, b_2]$. By hyper infinite induction, we can continue this construction and obtain hyper infinite sequence of intervals $\{[a_n, b_n]\}_{n \in \mathbb{N}^\#}$ such that:

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

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

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

Then generalized nested intervals theorem implies that the intersection of all of the intervals $[a_n, b_n]$ is a single point $w$. We will now construct a hyper infinite subsequence of $\{w_n\}_{n \in \mathbb{N}^\#}$ which will #-converge to $w$.

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

**Definition 11.12.** Let {$a_n$} be a $\mathbb{R}^\#$-valued hyper infinite sequence i.e., $a_n \in \mathbb{R}^\#$, $n \in \mathbb{N}^\#$. Say that {$a_n$} #-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that, for all $n > N$, $|a_n| \le \varepsilon$. We often denote this symbolically by $a_n \to_\# 0$.

We can also, at this point, define what it means for a hyperreal sequence #-tends to a given number $q \in \mathbb{R}^\#$ : {$a_n$} #-tends to $q$ if the hyperreal sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \to_\# 0$.

**Definition 11.13.** Let {$a_n$}, $n \in \mathbb{N}^\#$ be a hyperreal sequence. We call {$a_n$} a Cauchy hyperreal sequence if the difference between its terms #-tends to 0. To be precise: given any hyperreal number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 11.15.** If {$a_n$} is a #-convergent hyperreal sequence (that is, $a_n \to_\# b$ for some hyperreal number $b \in \mathbb{R}^\#$), then {$a_n$} is a Cauchy hyperreal sequence.

**Theorem 11.16.** If {$a_n$} is a Cauchy hyperreal sequence, then it is bounded or hyper bounded; that is, there is some $M \in \mathbb{R}^\#$ such that $|a_n| \le M$ for all $n \in \mathbb{N}^\#$.

**Theorem 11.17.** Any Cauchy hyperreal sequence {$a_n$} has a #-limit in $\mathbb{R}^\#$ i.e., there exists $b \in \mathbb{R}^\#$ such that $a_n \to_\# b$.
**Theorem 11.11. (Generalized Nested Intervals Theorem)**

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

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

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

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

**Proof.** Note that: (a) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is bounded or hyperbounded above by $b_1$ and (b) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is $\le$-admissible above subset of $\mathbb{R}_c^\#$.

By Theorem 11.10 there exists $\sup A$. Let $\xi = \sup A$.

Since $I_n$ are nested, for any positive hyperintegers $m$ and $n$ we have
$a_m \le a_{m+n} \le b_{m+n} \le b_n$, so that $\xi \le b_n$ for each $n \in \mathbb{N}^\#$. Since we obviously have $a_n \le \xi$
for each $n \in \mathbb{N}^\#$, we have $a_n \le \xi \le b_n$ for all $n \in \mathbb{N}^\#$, which implies $\xi \in \cap_{n=1}^\infty I_n$. Finally, if
$\xi, \eta \in \cap_{n=1}^\infty I_n$, with $\xi \le \eta$, then we get $0 \le \eta - \xi \le b_n - a_n$, for all $n \in \mathbb{N}^\#$, so that
$0 \le \eta - \xi \le \inf_{n \in \mathbb{N}^\#} |b_n - a_n| = 0$.

**Theorem 11.12. (Generalized Squeeze Theorem)**

Let {$a_n$}, {$c_n$} be two hyper infinite sequences #-converging to $L$, and {$b_n$} a hyper
infinite sequence. If $\forall n \ge K, K \in \mathbb{N}^\#$ we have $a_n \le b_n \le c_n$, then {$b_n$} also
#-converges to $L$.

**Proof.** Choose an $\varepsilon > 0, \varepsilon \approx 0$. By definition of the #-limit, there is an $N_1 \in \mathbb{N}^\#$ such that for all $n > N_1$ we have $|a_n - L| < \varepsilon$, in other words $L - \varepsilon < a_n < L + \varepsilon$. Similarly, there is an $N_2 \in \mathbb{N}^\#$ such that for all $n > N_2$ we have $L - \varepsilon < c_n < L + \varepsilon$. Denote $N = \max(N_1, N_2, K)$. Then for $n > N, L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon$, in other words $|b_n - L| < \varepsilon$. Since $\varepsilon > 0, \varepsilon \approx 0$ was arbitrary, by definition of the #-limit this says that $\#\text{-lim}_{n \to \infty}^# b_n = L$.

**Theorem 11.13. (Corollary of the Generalized Squeeze Theorem).**

If $\#\text{-lim}_{n \to \infty}^# |a_n| = 0$ then $\#\text{-lim}_{n \to \infty}^# a_n = 0$.

**Proof.** We know that $-|a_n| \le a_n \le |a_n|$. We want to apply the Generalized Squeeze Theorem. We are given that $\#\text{-lim}_{n \to \infty}^# |a_n| = 0$. This also implies that $\#\text{-lim}_{n \to \infty}^# (-|a_n|) = 0$. So by the Generalized Squeeze Theorem, $\#\text{-lim}_{n \to \infty}^# a_n = 0$.

**Theorem 11.14. (Generalized Bolzano-Weierstrass Theorem)**

Every hyperbounded hyperinfinite sequence has a #-convergent hyper infinite
subsequence.

**Proof.** Let {$w_n$}$_{n \in \mathbb{N}^\#}$ be a hyperbounded hyper infinite sequence. Then, there exists an interval $[a_1, b_1]$ such that $a_1 \le w_n \le b_1$ for all $n \in \mathbb{N}^\#$.

Either $[a_1, \frac{a_1+b_1}{2}]$ or $[\frac{a_1+b_1}{2}, b_1]$ contains hyper infinitely many terms of $\{w_n\}$. That is, there exists hyperinfinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[a_1, \frac{a_1+b_1}{2}]$ or there exists hyper infinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[\frac{a_1+b_1}{2}, b_1]$. If $[a_1, \frac{a_1+b_1}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_2, b_2] = [a_1, \frac{a_1+b_1}{2}]$. Otherwise, let $[a_2, b_2] = [\frac{a_1+b_1}{2}, b_1]$. Either $[a_2, \frac{a_2+b_2}{2}]$ or $[\frac{a_2+b_2}{2}, b_2]$ contains hyper infinitely many terms of $\{w_n\}$$_{n \in \mathbb{N}^\#}$. If $[a_2, \frac{a_2+b_2}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_3, b_3] = [a_2, \frac{a_2+b_2}{2}]$. Otherwise, let $[a_3, b_3] = [\frac{a_2+b_2}{2}, b_2]$. By hyper infinite induction, we can continue this construction and obtain hyper infinite sequence of intervals $\{[a_n, b_n]\}$$_{n \in \mathbb{N}^\#}$ such that:
**Proof.** By Definition 11.13 given $\varepsilon > 0, \varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $n, n' > N$,

$$|a_n - a_{n'}| < \varepsilon. \qquad (11.1)$$

From (11.1) for any $n, n' > N$ we get

$$a_{n'} - \varepsilon < a_n < a_n + \varepsilon. \qquad (11.2)$$

The generalized Bolzano-Weierstrass theorem implies there is a #-convergent hyper infinite subsequence $\{a_{n_k}\} \subset \{a_n\}$ such that $a_{n_k} \to_{\#} b$ for some hyperreal number $b \in \mathbb{R}_c^\#$. Let us show that the sequence $\{a_n\}$ also #-convergent to this $b \in \mathbb{R}_c^\#$. We can choose $k \in \mathbb{N}^\#$ so large that $n_k > N$ and

$$|a_{n_k} - b| < \varepsilon. \qquad (11.3)$$

We choose now in (11.1) $n' = n_k$ and therefore

$$|a_n - a_{n_k}| < \varepsilon. \qquad (11.4)$$

From (11.3) and (11.4) for any $n > N$ we get

$$|(a_{n_k} - b) + (a_n - a_{n_k})| = |a_n - b| < 2\varepsilon. \qquad (11.5)$$

Thus $a_n \to_{\#} b$ as well.

## 12. The Extended Hyperreal Number System $\mathbb{R}_c^\#$

**Definition 12.1.** (a) A set $S \subset \mathbb{N}^\#$ is hyperfinite if $card(S) = card(\{x|0 \le x \le n\})$, where $n \in \mathbb{N}^\# \mathbb{N}$. (b) A set $S \subseteq \mathbb{N}^\#$ is hyper infinite if $card(S) = card(\mathbb{N}^\#)$

**Notation 12.1.** If $F$ is an arbitrary collection of sets, then $\cup\{S|S \in F\}$ is the set of all elements that are members of at least one of the sets in $F$, and $\cap\{S|S \in F\}$ is the set of all elements that are members of every set in $F$. The union and intersection of finitely or hyper finitely many sets $S_k, 0 \le k \le n \in \mathbb{N}^\#$ are also written as $\cup_{k=0}^n S_k$ and $\cap_{k=0}^n S_k$. The union and intersection of an hyper infinite sequence $S_k, k \in \mathbb{N}^\#$ of sets are written as $\cup_{k=0}^\infty S$ or $\cup_{n \in \mathbb{N}^\#} S$ and $\cap_{k=0}^\infty S$ or $\cap_{n \in \mathbb{N}^\#} S$ correspondingly.

A nonempty set $S$ of hyperreal numbers $\mathbb{R}_c^\#$ is unbounded above if it has no hyperfinite upper bound, or unbounded below if it has no hyperfinite lower bound. It is convenient to adjoin to the hyperreal number system two points, $+\infty^\#$ (which we also write more simply as $\infty^\#$) and $-\infty^\#$, and to define the order relationships between them and any hyperreal number $x \in \mathbb{R}_c^\#$ by $-\infty^\# < x < \infty^\#$.

We call $-\infty^\#$ and $\infty^\#$ points at hyperinfinity. If $S$ is a nonempty set of hyperreals, we write $sup_S = \infty^\#$ to indicate that $S$ is hyper unbounded above, and $inf_S = -\infty^\#$ to indicate that $S$ is hyper unbounded below.

### 12.1. #-Open and #-Closed Sets on $\hat{\mathbb{R}}_c^\#$.

**Definition 12.2.** If $a$ and $b$ are in the extended hyperreals and $a < b$, then the #-open interval $(a, b)$ is defined by $(a, b) \triangleq \{x|a < x < b\}.$ :

The #-open intervals $(a, \infty^\#)$ and $(-\infty^\#, b)$ are semi-hyper infinite if $a$ and $b$ are finite or hyperfinite, and $(-\infty^\#, \infty^\#)$ is the entire hyperreal line.

If $-\infty^\# < a < b < \infty^\#$, the set $[a, b] \triangleq \{x|a \le x \le b\}$ is #-closed, since its complement is the union of the #-open sets $(-\infty^\#, a)$ and $(b, \infty^\#)$. We say that $[a, b]$ is a #-closed interval. Semi-hyper infinite #-closed intervals are sets of the form $[a, \infty) = \{x|a \le x\}$ and $(-\infty^\#, a] = \{x|x \le a\}$, where $a$ is finite or hyperfinite. They are #-closed sets,
(i) for each $n \in \mathbb{N}^{\#}, [a_n, b_n]$ contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^{\#}}$,

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

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

Then generalized nested intervals theorem implies that the intersection of all of the intervals $[a_n, b_n]$ is a single point $w$. We will now construct a hyper infinite subsequence of $\{w_n\}_{n \in \mathbb{N}^{\#}}$ which will #-converge to $w$.

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

contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^{\#}}$, there exists $k_3 \in \mathbb{N}^{\#}, k_3 > k_2$, such that $w_{k_3}$ is in $[a_3, b_3]$. Continuing this process by hyper infinite induction, we obtain hyper infinite sequence $\{w_{k_n}\}_{n \in \mathbb{N}^{\#}}$ such that $w_{k_n} \in [a_n, b_n]$ for each $n \in \mathbb{N}^{\#}$. The sequence $\{w_{k_n}\}_{n \in \mathbb{N}^{\#}}$ is a subsequence of $\{w_n\}_{n \in \mathbb{N}^{\#}}$ since $k_{n+1} > k_n$ for each $n \in \mathbb{N}^{\#}$.

Since $a_n \to_{\#} w$, and $a_n \le w_n \le b_n$ for each $n \in \mathbb{N}^{\#}$, the squeeze theorem implies that $w_{k_n} \to_{\#} w$.

**Definition 11.12.** Let $\{a_n\}$ be a $\mathbb{R}_c^{\#}$-valued hyper infinite sequence i.e., $a_n \in \mathbb{R}_c^{\#}, n \in \mathbb{N}^{\#}$. Say that $\{a_n\}_{n \in \mathbb{N}^{\#}}$ #-tends to 0 if, given any $\varepsilon > 0, \varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^{\#}\mathbb{N}, N = N(\varepsilon)$ such that, for all $n > N, |a_n| \le \varepsilon$. We often denote this symbolically by $a_n \to_{\#} 0$.

We can also, at this point, define what it means for a hyperreal sequence #-tends to a given number $q \in \mathbb{R}_c^{\#}: \{a_n\} \text{-tends to } q$ if the hyperreal sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \to_{\#} 0$.

**Definition 11.13.** Let $\{a_n\}, n \in \mathbb{N}^{\#}$ be a hyperreal sequence. We call $\{a_n\}$ a Cauchy hyperreal sequence if the difference between its terms #-tends to 0. To be precise: given any hyperreal number $\varepsilon > 0, \varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N, |a_n - a_m| < \varepsilon$.

**Theorem 11.15.** If $\{a_n\}$ is a #-convergent hyperreal sequence (that is, $a_n \to_{\#} b$ for some hyperreal number $b \in \mathbb{R}_c^{\#}$), then $\{a_n\}$ is a Cauchy hyperreal sequence.

**Theorem 11.16.** If $\{a_n\}$ is a Cauchy hyperreal sequence, then it is bounded or hyper bounded; that is, there is some $M \in \mathbb{R}_c^{\#}$ such that $|a_n| \le M$ for all $n \in \mathbb{N}^{\#}$.

**Theorem 11.17.** Any Cauchy hyperreal sequence $\{a_n\}$ has a #-limit in $\mathbb{R}_c^{\#}$ i.e., there exists $b \in \mathbb{R}_c^{\#}$ such that $a_n \to_{\#} b$.

**Proof.** By Definition 11.13 given $\varepsilon > 0, \varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $n, n' > N$,

$$|a_n - a_{n'}| < \varepsilon. \qquad (11.1)$$

From (11.1) for any $n, n' > N$ we get

$$a_{n'} - \varepsilon < a_n < a_n + \varepsilon. \qquad (11.2)$$

The generalized Bolzano-Weierstrass theorem implies there is a #-convergent hyper infinite subsequence $\{a_{n_k}\} \subset \{a_n\}$ such that $a_{n_k} \to_{\#} b$ for some hyperreal number $b \in \mathbb{R}_c^{\#}$. Let us show that the sequence $\{a_n\}$ also #-convergent to this $b \in \mathbb{R}_c^{\#}$. We can choose $k \in \mathbb{N}^{\#}$ so large that $n_k > N$ and

$$|a_{n_k} - b| < \varepsilon. \qquad (11.3)$$

We choose now in (11.1) $n' = n_k$ and therefore

$$|a_n - a_{n_k}| < \varepsilon. \qquad (11.4)$$
since their complements are the #-open intervals $(-\infty^{#}, a)$ and $(a, \infty^{#})$, respectively.

**Definition 12.3.** If $x_0 \in \mathbb{R}_c^#$ is a hyperreal number and $\varepsilon > 0, \varepsilon \approx 0$ then the #-open interval $(x_0 - \varepsilon, x_0 + \varepsilon)$ is an #-neighborhood of $x_0$. If a set $S \subset \mathbb{R}_c^#$ contains an #-neighborhood of $x_0$, then $S$ is a #-neighborhood of $x_0$, and $x_0$ is an #-interior point of $S$. The set of #-interior points of $S$ is the #-interior of $S$, denoted by $\text{\#-Int}(S)$.

(i) If every point of $S$ is an #-interior point (that is, $S = \text{\#-Int}(S)$), then $S$ is #-open.

(ii) A set $S$ is #-closed if $S^c = \mathbb{R}_c^# \setminus S$ is #-open.

**Example 12.1.** An open interval $(a, b)$ is an #-open set, because if $x_0 \in (a, b)$ and $\varepsilon \le \min \{x_0 - a; b - x_0\}$, then $(x_0 - \varepsilon, x_0 + \varepsilon) \subset (a, b)$.

**Remark 12.1.** The entire hyperline $\mathbb{R}_c^# = (-\infty^{#}, \infty^{#})$ is #-open, and therefore $\emptyset$ is #-closed. However, $\emptyset$ is also #-open, for to deny this is to say that $\emptyset$ contains a point that is not an #-interior point, which is absurd because $\emptyset$ contains no points. Since $\emptyset$ is #-open, $\hat{\mathbb{R}}_c^#$ is #-closed. Thus, $\hat{\mathbb{R}}_c^#$ and $\emptyset$ are both #-open and #-closed.

**Remark 12.2.** They are not the only subsets of $\hat{\mathbb{R}}_c^#$ with this property mentioned above.

**Definition 12.4.** A deleted #-neighborhood of a point $x_0$ is a set that contains every point of some #-neighborhood of $x_0$ except for $x_0$ itself. For example, $S = \{x|0 < |x - x_0| < \varepsilon\}$, where $\varepsilon \approx 0$, is a deleted #-neighborhood of $x_0$. We also say that it is a deleted $\varepsilon$-#-neighborhood of $x_0$.

**Theorem 12.1.** (a) The union of #-open sets is #-open:

(b) The #-intersection of #-closed sets is #-closed:

These statements apply to arbitrary collections, hyperfinite or hyperinfinite, of #-open and #-closed sets.

**Proof** (a) Let $L$ be a collection of #-open sets and $S = \bigcup \{G | G \in L\}$.

If $x_0 \in S$, then $x_0 \in G_0$ for some $G_0$ in $L$, and since $G_0$ is #-open, it contains some $\varepsilon$-#-neighborhood of $x_0$. Since $G_0 \subset S$, this $\varepsilon$-#-neighborhood is in $S$, which is consequently a #-neighborhood of $x_0$. Thus, $S$ is a #-neighborhood of each of its points, and therefore #-open, by definition.

(b) Let $F$ be a collection of #-closed sets and $T = \bigcap \{H | H \in F\}$. Then $T^c = \bigcup \{H^c | H \in F\}$ and, since each $H^c$ is #-open, $T^c$ is #-open, from (a). Therefore, $T$ is #-closed, by definition.

**Example 12.2.** If $-\infty^{#} < a < b < \infty^{#}$, the set $[a, b] = \{x|a \le x \le b\}$ is #-closed, since its complement is the union of the #-open sets $(-\infty^{#}a)$ and $(b, \infty^{#})$. We say that $[a, b]$ is a #-closed interval. The set $[a, b) = \{x|a \le x < b\}$ is a half-#-closed or half-#-open interval if $-\infty^{#} < a < b < \infty^{#}$, as is $(a, b] = \{x|a < x \le b\}$ however, neither of these sets is #-open or #-closed. Semi-infinite #-closed intervals are sets of the form $[a, \infty^{#}) = \{x|a \le x\}$ and $(-\infty^{#}, a] = \{x|x \le a\}$, where $a$ is hyperfinite. They are #-closed sets, since their complements are the #-open intervals $(-\infty^{#}, a)$ and

$(a, \infty^{#})$, respectively.

**Definition 12.5.** Let $S$ be a subset of $\hat{\mathbb{R}}_c^# = (-\infty^{#}, \infty^{#})$. Then

(a) $x_0$ is a #-limit point of $S$ if every deleted #-neighborhood of $x_0$ contains a point of $S$.

(b) $x_0$ is a boundary point of $S$ if every #-neighborhood of $x_0$ contains at least one point in $S$ and one not in $S$. The set of #-boundary points of $S$ is the #-boundary of $S$, denoted

by $\text{#-}\partial S$. The #-closure of $S$, denoted by $\text{#-}\bar{S}$, is $S \cup \text{\#-}\partial S$.

(c) $x_0$ is an #-isolated point of $S$ if $x_0 \in S$ and there is a #-neighborhood of $x_0$ that
From (11.3) and (11.4) for any $n > N$ we get

$$|(a_{n_k} - b) + (a_n - a_{n_k})| = |a_n - b| < 2\epsilon. \quad (11.5)$$

Thus $a_n \to_{\#} b$ as well.

## 12. The Extended Hyperreal Number System $\hat{\mathbb{R}}_c^\#$

**Definition 12.1.** (a) A set $S \subset \mathbb{N}^\#$ is hyperfinite if $card(S) = card(\{x|0 \le x \le n\})$, where $n \in \mathbb{N}^\# \setminus \mathbb{N}$. (b) A set $S \subset \mathbb{N}^\#$ is hyper infinite if $card(S) = card(\mathbb{N}^\#)$

**Notation 12.1.** If $F$ is an arbitrary collection of sets, then $\cup\{S|S \in F\}$ is the set of all elements that are members of at least one of the sets in $F$, and $\cap\{S|S \in F\}$ is the set of all elements that are members of every set in $F$. The union and intersection of finitely or hyper finitely many sets $S_k, 0 \le k \le n \in \mathbb{N}^\#$ are also written as $\cup_{k=0}^n S_k$ and $\cap_{k=0}^n S_k$. The union and intersection of an hyperinfinite sequence $S_k, k \in \mathbb{N}^\#$ of sets are written as $\cup_{k=0}^\infty S_k$ or $\cup_{n \in \mathbb{N}^\#} S$ and $\cap_{k=0}^\infty S_k$ or $\cap_{n \in \mathbb{N}^\#} S$ correspondingly.

A nonempty set $S$ of hyperreal numbers $\mathbb{R}_c^\#$ is unbounded above if it has no hyperfinite upper bound, or unbounded below if it has no hyperfinite lower bound. It is convenient to adjoin to the hyperreal number system two points, $+\infty^\#$ (which we also write more simply as $\infty^\#$) and $-\infty^\#$, and to define the order relationships between them and any hyperreal number $x \in \mathbb{R}_c^\#$ by $-\infty^\# < x < \infty^\#$.

We call $-\infty^\#$ and $\infty^\#$ points at hyperinfinity. If $S$ is a nonempty set of hyperreals, we write $sup_S = \infty^\#$ to indicate that $S$ is hyper unbounded above, and $inf_S = -\infty^\#$ to indicate that $S$ is hyper unbounded below.

### 12.1.\#-Open and \#-Closed Sets on $\hat{\mathbb{R}}_c^\#$.

**Definition 12.2.** If $a$ and $b$ are in the extended hyperreals and $a < b$, then the \#-open interval $(a, b)$ is defined by $(a, b) \triangleq \{x|a < x < b\}$. :

The \#-open intervals $(a, \infty^\#$) and $(-\infty^\#, b)$ are semi-hyper infinite if $a$ and $b$ are finite or hyperfinite, and $(-\infty^\#, \infty^\#)$ is the entire hyperreal line.

If $-\infty^\# < a < b < \infty^\#$, the set $[a, b] \triangleq \{x|a \le x \le b\}$ is \#-closed, since its complement is the union of the \#-open sets $(-\infty^\#, a)$ and $(b, \infty^\#)$. We say that $[a, b]$ is a \#-closed interval. Semi-hyper infinite \#-closed intervals are sets of the form $[a, \infty) = \{x|a \le x\}$ and $(-\infty^\#, a] = \{x|x \le a\}$, where $a$ is finite or hyperfinite. They are \#-closed sets, since their complements are the \#-open intervals $(-\infty^\#, a)$ and $(a, \infty^\#)$, respectively.

**Definition 12.3.** If $x_0 \in \mathbb{R}_c^\#$ is a hyperreal number and $\epsilon > 0, \epsilon \approx 0$ then the \#-open interval

$(x_0 - \epsilon, x_0 + \epsilon)$ is an \#-neighborhood of $x_0$. If a set $S \subset \mathbb{R}_c^\#$ contains an \#-neighborhood of $x_0$, then $S$ is a \#-neighborhood of $x_0$, and $x_0$ is an \#-interior point of $S$. The set of \#-interior points of $S$ is the \#-interior of $S$, denoted by $\text{\#-Int}(S)$.

(i) If every point of $S$ is an \#-interior point (that is, $S = \text{\#-Int}(S)$), then $S$ is \#-open.

(ii) A set $S$ is \#-closed if $S^c = \mathbb{R}_c^\# \setminus S$ is \#-open.

**Example 12.1.** An open interval $(a, b)$ is an \#-open set, because if $x_0 \in (a, b)$ and $\epsilon \le \min\{x_0 - a; b - x_0\}$, then $(x_0 - \epsilon, x_0 + \epsilon) \subset (a, b)$.

**Remark 12.1.** The entire hyperline $\hat{\mathbb{R}}_c^\# = (-\infty^\#, \infty^\#)$ is \#-open, and therefore $\emptyset$ is \#-closed. However, $\emptyset$ is also \#-open, for to deny this is to say that $\emptyset$ contains a point that is not an \#-interior point, which is absurd because $\emptyset$ contains no points. Since $\emptyset$ is \#-open, $\hat{\mathbb{R}}_c^\#$ is \#-closed. Thus, $\hat{\mathbb{R}}_c^\#$ and $\emptyset$ are both \#-open and \#-closed.
contains no other point of S.

(d) $x_0$ is #-exterior to S if $x_0$ is in the #-interior of $S^c$. The collection of such points is the #-exterior of S.

**Theorem 12.2.** A set $S$ is #-closed if and only if no point of $S^c$ is a #-limit point of $S$.

**Proof.** Suppose that $S$ is #-closed and $x_0 \in S^c$. Since $S^c$ is #-open, there is a #-neighborhood of $x_0$ that is contained in $S^c$ and therefore contains no points of $S$. Hence, $x_0$ cannot be a #-limit point of $S$. For the converse, if no point of $S^c$ is a #-limit point of $S$ then every point in $S^c$ must have a #-neighborhood contained in $S^c$. Therefore, $S^c$ is #-open and $S$ is #-closed.

**Corollary 12.1.** A set $S$ is #-closed if and only if it contains all its #-limit points. If $S$ is #-closed and hyper bounded, then $\inf(S)$ and $\sup(S)$ are both in $S$.

**Proposition 12.1.** If $S$ is #-closed and hyper bounded, then $\inf(S)$ and $\sup(S)$ are both in $S$.

## 12.2. #-Open Coverings

**Definition 12.6.** A collection $H$ of #-open sets of $\mathbb{R}_c^\#$ is an #-open covering of a set $S$ if every point in $S$ is contained in a set $H$ belonging to $H$; that is, if $S \subset \cup\{F|F \in H\}$.

**Definition 12.7.** A set $S \subset \mathbb{R}_c^\#$ is called #-compact (or hyper compact) if each of its #-open covers has a finite or hyperfinite subcover.

**Theorem 12.3. (Generalized Heine–Borel Theorem)** If $H$ is an #-open covering of a #-closed and hyper bounded subset $S$ of the hyperreal line $\mathbb{R}_c^\#$ (or of the $\mathbb{R}_c^{\#n}, n \in \mathbb{N}^\#$) then $S$ has an #-open covering $\tilde{H}$ consisting of hyper finite many #-open sets belonging to $H$.

**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 an 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 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
**Remark 12.2.** They are not the only subsets of $\mathbb{R}_c^\#$ with this property mentioned above.

**Definition 12.4.** A deleted #-neighborhood of a point $x_0$ is a set that contains every point of some #-neighborhood of $x_0$ except for $x_0$ itself. For example, $S = \{x|0 < |x-x_0| < \varepsilon\}$, where $\varepsilon \approx 0$, is a deleted #-neighborhood of $x_0$. We also say that it is a deleted $\varepsilon$-#-neighborhood of $x_0$.

**Theorem 12.1.** (a) The union of #-open sets is #-open:

(b) The #-intersection of #-closed sets is #-closed:

These statements apply to arbitrary collections, hyperfinite or hyperinfinite, of #-open and #-closed sets.

**Proof** (a) Let $L$ be a collection of #-open sets and $S = \bigcup \{G|G \in L\}$.

If $x_0 \in S$, then $x_0 \in G_0$ for some $G_0$ in $L$, and since $G_0$ is #-open, it contains some $\varepsilon$-#-neighborhood of $x_0$. Since $G_0 \subset S$, this $\varepsilon$-#-neighborhood is in $S$, which is consequently a #-neighborhood of $x_0$. Thus, $S$ is a #-neighborhood of each of its points, and therefore #-open, by definition.

(b) Let $F$ be a collection of #-closed sets and $T = \bigcap \{H|H \in F\}$. Then $T^c = \bigcup \{H^c|H \in F\}$ and, since each $H^c$ is #-open, $T^c$ is #-open, from (a). Therefore, $T$ is #-closed, by definition.

**Example 12.2.** If $-\infty^\# < a < b < \infty^\#$, the set $[a,b] = \{x|a \le x \le b\}$ is #-closed, since its complement is the union of the #-open sets $(-\infty^\#a)$ and $(b,\infty^\#)$. We say that $[a,b]$ is a #-closed interval. The set $[a,b) = \{x|a \le x < b\}$ is a half-#-closed or half-#-open interval if $-\infty^\# < a < b < \infty^\#$, as is $(a,b] = \{x|a < x \le b\}$ however, neither of these sets is #-open or #-closed. Semi-infinite #-closed intervals are sets of the form $[a, \infty^\#) = \{x|a \le x\}$ and $(-\infty^\#, a] = \{x|x \le a\}$, where $a$ is hyperfinite. They are #-closed sets, since their complements are the #-open intervals $(-\infty^\#, a)$ and $(a, \infty^\#)$, respectively.

**Definition 12.5.** Let $S$ be a subset of $\mathbb{R}_c^\# = (-\infty^\#, \infty^\#)$. Then

(a) $x_0$ is a #-limit point of $S$ if every deleted #-neighborhood of $x_0$ contains a point of $S$.

(b) $x_0$ is a boundary point of $S$ if every #-neighborhood of $x_0$ contains at least one point in $S$ and one not in $S$. The set of #-boundary points of $S$ is the #-boundary of $S$, denoted

by $\#-\partial S$. The #-closure of $S$, denoted by $\#-\bar{S}$, is $S \cup \#-\partial S$.

(c) $x_0$ is an #-isolated point of $S$ if $x_0 \in S$ and there is a #-neighborhood of $x_0$ that contains no other point of $S$.

(d) $x_0$ is #-exterior to $S$ if $x_0$ is in the #-interior of $S^c$. The collection of such points is the #-exterior of $S$.

**Theorem 12.2.** A set $S$ is #-closed if and only if no point of $S^c$ is a #-limit point of $S$.

**Proof.** Suppose that $S$ is #-closed and $x_0 \in S^c$. Since $S^c$ is #-open, there is a #-neighborhood of $x_0$ that is contained in $S^c$ and therefore contains no points of $S$.

Hence, $x_0$ cannot be a #-limit point of $S$. For the converse, if no point of $S^c$ is a #-limit point of $S$ then every point in $S^c$ must have a #-neighborhood contained in $S^c$. Therefore, $S^c$ is #-open and $S$ is #-closed.

**Corollary 12.1.** A set $S$ is #-closed if and only if it contains all its #-limit points.

If $S$ is #-closed and hyper bounded, then $\inf(S)$ and $\sup(S)$ are both in $S$.

**Proposition 12.1.** If $S$ is #-closed and hyper bounded, then $\inf(S)$ and $\sup(S)$ are both in $S$.
\#-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 12.4.** (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 w-\#-open neighborhood $N_x$ that contains no point of $S$ other than $x$. The collection $H = \{N_x|x \in S\}$ is an w-\#-open covering for $S$. Since $S$ is also hyper bounded, Theorem 12.3 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 non-Archimedean field *R_c^#.

via Cauchy completion of internal non-Archimedean field
\*R.

**Definition 13.1.** A hyper infinite sequence of hyperreal numbers from \*R is a function $a : \mathbb{N}^\# \to \*R$ from hypernatural numbers $\mathbb{N}^\#$ into the hyperreal numbers \*R.

We usually denote such a function by $n \mapsto a_n$, or by $a : n \to a_n$, so the terms in the sequence are written $\{a_1, a_2, a_3, \dots, a_n\}$. To refer to the whole hyper infinite sequence, we will write $\{a_n\}_{n=1}^{\infty}$, or $\{a_n\}_{n \in \mathbb{N}^{\#}}$, or for the sake of brevity simply $\{a_n\}$.

**Definition 13.2.** Let $\{a_n\}$ be a hyper infinite \*R-valued sequence mentioned above. Say that $\{a_n\}$ \#-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that, after $N$ (i.e. for all $n > N$), $|a_n| \le \varepsilon$. We denote this symbolically by $a_n \to_\# 0$.

We can also, at this point, define what it means for a hyper infinite \*R-valued sequence \#-tends to any given number $q \in \*R : \{a_n\}$ \#-tends to $q$ if the hyper infinite sequence $\{a_n - q\}$ \#-tends to 0 i.e., $a_n - q \to_\# 0$.

**Definition 13.3.** Let $\{a_n\}$ be a hyper infinite \*R-valued sequence. We call $\{a_n\}$ a Cauchy hyper infinite \*R-valued sequence if the difference between its terms \#-tends to 0. To be precise: given any hyperreal number such that $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 13.1.** If $\{a_n\}$ is a \#-convergent hyper infinite \*R-valued sequence (that is, $a_n \to_\# q$ for some hyperreal number $q \in \*R$), then $\{a_n\}$ is a Cauchy hyper infinite \*R-valued sequence.

**Proof.** We know that $a_n \to_\# q$. Here is a ubiquitous trick: instead of using $\varepsilon$ in the definition Definition 13.3, start with an arbitrary infinite small $\varepsilon > 0$, $\varepsilon \approx 0$ and then choose $N \in \mathbb{N}^* \setminus \mathbb{N}$ so that $|a_n - q| < \varepsilon/2$ when $n > N$. Then if $m, n > N$, we have $|a_n - a_m| = |(a_n - q) - (a_m - q)| \le |a_n - q| + |a_m - q| < \varepsilon/2 + \varepsilon/2 = \varepsilon$. This shows that $\{a_n\}_{n \in \mathbb{N}^*}$ is a Cauchy sequence.

**Theorem 13.2.** If $\{a_n\}$ is a Cauchy hyper infinite \*R-valued sequence, then it is bounded or hyper bounded; that is, there is some finite or hyperfinite $M \in \*R$ such that $|a_n| \le M$ for all $n \in \mathbb{N}^\#$.

**Proof.** Since $\{a_n\}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N \in \mathbb{N}^\#$ such that $|a_m - a_n| < 1$ whenever $m, n > N$. Thus, $|a_{N+1} - a_n| < 1$ for $n > N$. We can rewrite
## 12.2. #-Open Coverings

**Definition 12.6.** A collection $H$ of #-open sets of $\mathbb{R}_c^\#$ is an #-open covering of a set $S$ if every point in $S$ is contained in a set $H$ belonging to $H$; that is, if $S \subset \cup \{F | F \in H\}$.

**Definition 12.7.** A set $S \subset \mathbb{R}_c^\#$ is called #-compact (or hyper compact) if each of its #-open covers has a finite or hyperfinite subcover. .

**Theorem 12.3. (Generalized Heine–Borel Theorem)** If $H$ is an #-open covering of a #-closed and hyper bounded subset $S$ of the hyperreal line $\mathbb{R}_c^\#$ (or of the $\mathbb{R}_c^{\#n}, n \in \mathbb{N}^\#$)$
then $S$ has an #-open covering $\tilde{H}$ consisting of hyper finite many #-open sets belonging to $H$.

**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 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 12.4. (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 w-#-open neighborhood $N_x$ that contains no point of $S$ other than $x$. The collection $H = \{N_x | x \in S\}$ is an w-#-open covering for $S$. Since $S$ is also hyper bounded, Theorem 12.3 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 $S = \{x_k\}_{1 \le k \le n}, n \in \mathbb{N}^\#$.
this as $a_{N+1} - 1 < a_n < a_{N+1} + 1$. This means that $|a_n|$ is less than the maximum of $|a_{N+1} - 1|$ and $|a_{N+1} + 1|$. So, set $M$ equal to the maximum number in the following list: $\{|a_0|, |a_1|, \dots, |a_N|, |a_{N+1} - 1|, |a_{N+1} + 1|\}$. Then for any term $a_n$, if $n \le N$, then $|a_n|$ appears in the list and so $|a_n| \le M$; if $n > N$, then (as shown above) $|a_n|$ is less than at least one of the last two entries in the list, and so $|a_n| \le M$. Hence, $M \in \mathbb{R}$ is a bound for the sequence $\{a_n\}$.

**Definition 13.4.** Let $S$ be a set. A relation $x \sim y$ among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:
* Reflexivity: for any $s \in S$, $s \sim s$.
* Symmetry: for any $s, t \in S$, if $s \sim t$ then $t \sim s$.
* Transitivity: for any $s, t, r \in S$, if $s \sim t$ and $t \sim r$, then $s \sim r$.

**Theorem 13.3.** Let $S$ be a set, with an equivalence relation $(\cdot \sim \cdot)$ on pairs of elements. For $s \in S$, denote by $cl[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $cl[s] = cl[t]$ or $cl[s]$ and $cl[t]$ are disjoint. The hyperreal numbers $*\mathbb{R}^\#_c$ will be constructed as equivalence classes of Cauchy hyper infinite $*\mathbb{R}$-valued sequences. Let $F_{*\mathbb{R}}$ denote the set of all Cauchy hyper infinite $*\mathbb{R}$-valued sequences of hyperreal numbers. We define the equivalence relation on $F_{*\mathbb{R}}$.

**Definition 13.5.** Let $\{a_n\}$ and $\{b_n\}$ be in $F_{*\mathbb{R}}$. Say they are #-equivalent if $a_n - b_n \xrightarrow{\#} 0$ i.e., if and only if the hyper infinite $*\mathbb{R}$-valued sequence $a_n - b_n$ #-tends to 0.

**Theorem 13.4.** Definition 13.5 yields an equivalence relation on $F_{*\mathbb{R}}$.

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

**Reflexive:** $a_n - a_n = 0$, and the hyper infinite sequence all of whose terms are 0 clearly #-converges to 0. So $\{a_n\}$ is related to $\{a_n\}$.

**Symmetric:** Suppose $\{a_n\}$ is related to $\{b_n\}$, so $a_n - b_n \xrightarrow{\#} 0$.

But $b_n - a_n = -(a_n - b_n)$, and since only the absolute value $|a_n - b_n| = |b_n - a_n|$ comes into play in Definition 13.2, it follows that $b_n - a_n \xrightarrow{\#} 0$ as well. Hence, $\{b_n\}$ is related to $\{a_n\}$.

**Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 10.1. Suppose $\{a_n\}$ is related to $\{b_n\}$, and $\{b_n\}$ is related to $\{c_n\}$. This means that $a_n - b_n \xrightarrow{\#} 0$ and $b_n - c_n \xrightarrow{\#} 0$. To be fully precise, let us fix $\varepsilon > 0$, $\varepsilon \approx 0$; then there exists an $N \in \mathbb{N}^\#$ such that for all $n > N$, $|a_n - b_n| < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$, $|b_n - c_n| < \varepsilon/2$. Well, then, as long as $n$ is bigger than both $N$ and $M$, we have that $|a_n - c_n| = |(a_n - b_n) + (b_n - c_n)| \le |a_n - b_n| + |b_n - c_n| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|a_n - c_n| < \varepsilon$. This means that $a_n - c_n \xrightarrow{\#} 0$ i.e. $\{a_n\}$ is related to $\{c_n\}$.

**Definition 13.6.** The external hyperreal numbers $*\mathbb{R}^\#_c$ are the equivalence classes $cl[\{a_n\}]$ of Cauchy hyper infinite $*\mathbb{R}$-valued sequences of hyperreal numbers, as per Definition 13.5. That is, each such equivalence class is an external hyperreal number.

**Definition 13.7.** Given any hyperreal number $q \in *\mathbb{R}$, define a hyperreal number $q^\#$ to be the equivalence class of the hyper infinite $*\mathbb{R}$-valued sequence $q^\# = (q, q, q, q, \dots)$ consisting entirely of $q$. So we view $*\mathbb{R}$ as being inside $*\mathbb{R}^\#_c$ by thinking of each hyperreal number $q$ as its associated equivalence class $q^\#$. It is standard to abuse this notation, and simply refer to the equivalence class as $q$ as well.
## 13. External non-Archimedean field *$\mathbb{R}_c^#$.

via Cauchy completion of internal non-Archimedean field
*$\mathbb{R}$.

**Definition 13.1.** A hyper infinite sequence of hyperreal numbers from *$\mathbb{R}$ is a function $a : \mathbb{N}^# \to *\mathbb{R}$ from hypernatural numbers $\mathbb{N}^#$ into the hyperreal numbers *$\mathbb{R}$.
We usually denote such a function by $n \mapsto a_n$, or by $a : n \to a_n$, so the terms in the sequence are written $\{a_1, a_2, a_3, \dots, a_n\dots\}$. To refer to the whole hyper infinite sequence, we will write $\{a_n\}_{n=1}^\infty$, or $\{a_n\}_{n \in \mathbb{N}^#}$, or for the sake of brevity simply $\{a_n\}$.

**Definition 13.2.** Let $\{a_n\}$ be a hyper infinite *$\mathbb{R}$-valued sequence mentioned above. Say that $\{a_n\}$ #-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^# \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that, after $N$ (i.e. for all $n > N$), $|a_n| \le \varepsilon$. We denote this symbolically by $a_n \to_\# 0$.

We can also, at this point, define what it means for a hyper infinite *$\mathbb{R}$-valued
sequence #-tends to any given number $q \in *\mathbb{R} : \{a_n\}$ #-tends to $q$ if the hyper
infinite sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \to_\# 0$.

**Definition 13.3.** Let $\{a_n\}$ be a hyper infinite *$\mathbb{R}$-valued sequence. We call $\{a_n\}$ a Cauchy hyper infinite *$\mathbb{R}$-valued sequence if the difference between its terms #-tends to 0. To be precise: given any hyperreal number such that $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 13.1.** If $\{a_n\}$ is a #-convergent hyper infinite *$\mathbb{R}$-valued sequence (that is, $a_n \to_\# q$ for some hyperreal number $q \in *\mathbb{R}$), then $\{a_n\}$ is a Cauchy hyper infinite *$\mathbb{R}$-valued sequence.

**Proof.** We know that $a_n \to_\# q$. Here is a ubiquitous trick: instead of using $\varepsilon$ in the definition of *Definition 13.3*, start with an arbitrary infinite small $\varepsilon > 0$, $\varepsilon \approx 0$ and then choose $N \in \mathbb{N}^* \setminus \mathbb{N}$ so that $|a_n - q| < \varepsilon/2$ when $n > N$. Then if $m, n > N$, we have $|a_n - a_m| = |(a_n - q) - (a_m - q)| \le |a_n - q| + |a_m - q| < \varepsilon/2 + \varepsilon/2 = \varepsilon$. This shows that $\{a_n\}_{n \in \mathbb{N}^*}$ is a Cauchy sequence.

**Theorem 13.2.** If $\{a_n\}$ is a Cauchy hyper infinite *$\mathbb{R}$-valued sequence, then it is bounded or hyper bounded; that is, there is some finite or hyperfinite $M \in *\mathbb{R}$ such that $|a_n| \le M$ for all $n \in \mathbb{N}^*$.

**Proof.** Since $\{a_n\}$ is Cauchy, setting $\varepsilon = 1$ we know that there is some $N \in \mathbb{N}^*$ such that $|a_m - a_n| < 1$ whenever $m, n > N$. Thus, $|a_{N+1} - a_n| < 1$ for $n > N$. We can rewrite this as $a_{N+1} - 1 < a_n < a_{N+1} + 1$. This means that $|a_n|$ is less than the maximum of $|a_{N+1} - 1|$ and $|a_{N+1} + 1|$. So, set $M$ equal to the maximum number in the following list: $\{|a_0|, |a_1|, \dots, |a_N|, |a_{N+1} - 1|, |a_{N+1} + 1|\}$. Then for any term $a_n$, if $n \le N$, then $|a_n|$ appears in the list and so $|a_n| \le M$; if $n > N$, then (as shown above) $|a_n|$ is less than at least one of the last two entries in the list, and so $|a_n| \le M$. Hence, $M \in *\mathbb{R}$ is a bound for the sequence $\{a_n\}$.

**Definition 13.4.** Let $S$ be a set. A relation $x \sim y$ among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:
* Reflexivity: for any $s \in S$, $s \sim s$.
* Symmetry: for any $s, t \in S$, if $s \sim t$ then $t \sim s$.
* Transitivity: for any $s, t, r \in S$, if $s \sim t$ and $t \sim r$, then $s \sim r$.

**Theorem 13.3.** Let $S$ be a set, with an equivalence relation $(\cdot \sim \cdot)$ on pairs of elements.
**Definition 13.8.** Let $s, t \in *\mathbb{R}_c^\#$, so there are Cauchy hyper infinite $*\mathbb{R}$-valued sequences $\{a_n\}, \{b_n\}$ of hyperreal numbers with $s = \text{cl}[\{a_n\}]$ and $t = \text{cl}[\{b_n\}]$.

(a) Define $s+t$ to be the equivalence class of the sequence $\{a_n+b_n\}$.

(b) Define $s \times t$ to be the equivalence class of the sequence $\{a_n \times b_n\}$.

**Theorem 13.5.** The operations $+, \times$ in Definition 13.8 (a),(b) are well-defined.

**Proof.** Suppose that $\text{cl}[\{a_n\}] = \text{cl}[\{c_n\}]$ and $\text{cl}[\{b_n\}] = \text{cl}[\{d_n\}]$. Thus means that $a_n - c_n \xrightarrow{\#} 0$ and $b_n - d_n \xrightarrow{\#} 0$. Then $(a_n + b_n) - (c_n + d_n) = (a_n - c_n) + (b_n - d_n)$.

Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to 0, and so $\text{cl}[\{a_n + b_n\}] = \text{cl}[\{c_n + d_n\}]$.

Multiplication is a little trickier; this is where we will use Theorem 13.3. We will also use another ubiquitous technique: adding 0 in the form of $s-s$. Again, suppose that $\text{cl}[(a_n)] = \text{cl}[(c_n)]$ and $\text{cl}[(b_n)] = \text{cl}[(d_n)]$; we wish to show that

$\text{cl}[\{a_n \times b_n\}] = \text{cl}[\{c_n \times d_n\}]$, or, in other words, that $a_n \times b_n - c_n \cdot d_n \xrightarrow{\#} 0$. Well, we add and subtract one of the other cross terms, say

$$b_n \times c_n : a_n \times b_n - c_n \times d_n = a_n \times b_n + (b_n \times c_n - b_n \times c_n) - c_n \times d_n = \\ = (a_n \times b_n - b_n \times c_n) + (b_n \times c_n - c_n \times d_n) = b_n \times (a_n - c_n) + c_n \times (b_n - d_n).$$

Hence, we have $|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n|$. Now, from Theorem 13.2, there are numbers $M$ and $L$ such that $|b_n| \le M$ and $|c_n| \le L$ for all $n \in \mathbb{N}^\#$.

Taking some number $K$ which is bigger than both, we have

$$|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n| \le K(|a_n - c_n| + |b_n - d_n|).$$

Now, noting that both $a_n - c_n$ and $b_n - d_n$ tend to 0 and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2K$), we see that $a_n \times b_n - c_n \times d_n \xrightarrow{\#} 0$.

**Theorem 13.6.** Given any hyperreal number $s \in *\mathbb{R}_c^\#$, $s \ne 0$, there is a hyperreal number $t \in *\mathbb{R}_c^\#$ such that $s \times t = 1$.

**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 $\{0,0,0,0,\dots\}$. In other

words, $s = \text{cl}[\{a_n\}]$ where $a_n - 0$ does not $\#$-converge to 0. From this, we are to deduce

the existence of a hyperreal number $t = \text{cl}[\{b_n\}]$ such that $s \times t = \text{cl}[\{a_n \times b_n\}]$ is the same equivalence class as $\text{cl}[\{1,1,1,1,\dots\}]$. Doing so is actually an easy consequence

of the fact that nonzero hyperreal numbers have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{a_n\}$ does not tend to 0), there's no reason any number of the terms in $\{a_n\}$ can't equal 0. However, it turns out that eventually, $a_n \ne 0$.

That is:

**Lemma 13.1.** If $\{a_n\}$ is a Cauchy sequence which does not $\#$-tend to 0, then there is an $N \in \mathbb{N}^\#$ such that, for $n > N$, $a_n \ne 0$.

**Definition 13.9.** Let $s \in *\mathbb{R}_c^\#$. Say that $s$ is positive if $s \ne 0$, and if $s = \text{cl}[\{a_n\}]$ for some Cauchy sequence of hyperreal numbers such that for some $N \in \mathbb{N}^\#$, $a_n > 0$ for all $n > N$. Given two hyperreal numbers $s, t$, say that $s > t$ if $s - t$ is positive.

**Theorem 13.7.** Let $s, t \in *\mathbb{R}_c^\#$ be hyperreal numbers such that $s > t$, and let $r \in *\mathbb{R}_c^\#$. Then $s + r > t + r$.

**Proof.** Let $s = \text{cl}[\{a_n\}], t = \text{cl}[\{b_n\}],$ and $r = \text{cl}[\{c_n\}]$. Since $s > t$ i.e., $s - t > 0$, we
For $s \in S$, denote by $\mathbf{cl}[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $\mathbf{cl}[s] = \mathbf{cl}[t]$ or $\mathbf{cl}[s]$ and $\mathbf{cl}[t]$ are disjoint.

The hyperreal numbers $*\mathbb{R}^\#_c$ will be constructed as equivalence classes of Cauchy hyper infinite $*\mathbb{R}$-valued sequences. Let $\mathcal{F}_{*\mathbb{R}}$ denote the set of all Cauchy hyper infinite $*\mathbb{R}$-valued sequences of hyperreal numbers. We define the equivalence relation on $\mathcal{F}_{*\mathbb{R}}$.

**Definition 13.5.** Let $\{a_n\}$ and $\{b_n\}$ be in $\mathcal{F}_{*\mathbb{R}}$. Say they are #-equivalent if $a_n - b_n \xrightarrow{\#} 0$ i.e., if and only if the hyper infinite $*\mathbb{R}$-valued sequence $a_n - b_n$ #-tends to 0.

**Theorem 13.4.** *Definition 13.5 yields an equivalence relation on *$\mathcal{F}_{*\mathbb{R}}$*.*

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

**Reflexive:** $a_n - a_n = 0$, and the hyper infinite sequence all of whose terms are 0 clearly #-converges to 0. So $\{a_n\}$ is related to $\{a_n\}$.

**Symmetric:** Suppose $\{a_n\}$ is related to $\{b_n\}$, so $a_n - b_n \xrightarrow{\#} 0$.

But $b_n - a_n = -(a_n - b_n)$, and since only the absolute value $|a_n - b_n| = |b_n - a_n|$ comes into play in Definition 13.2, it follows that $b_n - a_n \xrightarrow{\#} 0$ as well. Hence, $\{b_n\}$ is related to $\{a_n\}$.

**Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 10.1. Suppose $\{a_n\}$ is related to $\{b_n\}$, and $\{b_n\}$ is related to $\{c_n\}$. This means that $a_n - b_n \xrightarrow{\#} 0$ and $b_n - c_n \xrightarrow{\#} 0$. To be fully precise, let us fix $\varepsilon > 0$, $\varepsilon \approx 0$; then there exists an $N \in \mathbb{N}^\#$ such that for all $n > N$, $|a_n - b_n| < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$, $|b_n - c_n| < \varepsilon/2$. Well, then, as long as $n$ is bigger than both $N$ and $M$, we have that $|a_n - c_n| = |(a_n - b_n) + (b_n - c_n)| \le |a_n - b_n| + |b_n - c_n| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So, choosing $L$ equal to the max of $N, M$, we see that given $\varepsilon > 0$ we can always choose $L$ so that for $n > L$, $|a_n - c_n| < \varepsilon$. This means that $a_n - c_n \xrightarrow{\#} 0$ i.e. $\{a_n\}$ is related to $\{c_n\}$.

**Definition 13.6.** The external hyperreal numbers $*\mathbb{R}^\#_c$ are the equivalence classes $\mathbf{cl}[\{a_n\}]$ of Cauchy hyper infinite $*\mathbb{R}$-valued sequences of hyperreal numbers, as per Definition 13.5. That is, each such equivalence class is an external hyperreal number.

**Definition 13.7.** Given any hyperreal number $q \in *\mathbb{R}$, define a hyperreal number $q^\#$ to be the equivalence class of the hyper infinite $*\mathbb{R}$-valued sequence $q^\# = (q, q, q, q, \dots)$ consisting entirely of $q$. So we view $*\mathbb{R}$ as being inside $*\mathbb{R}^\#_c$ by thinking of each hyperreal number $q$ as its associated equivalence class $q^\#$. It is standard to abuse this notation, and simply refer to the equivalence class as $q$ as well.

**Definition 13.8.** Let $s, t \in *\mathbb{R}^\#$, so there are Cauchy hyper infinite $*\mathbb{R}$-valued sequences $\{a_n\}, \{b_n\}$ of hyperreal numbers with $s = \mathbf{cl}[\{a_n\}]$ and $t = \mathbf{cl}[\{b_n\}]$.

(a) Define $s + t$ to be the equivalence class of the sequence $\{a_n + b_n\}$.

(b) Define $s \times t$ to be the equivalence class of the sequence $\{a_n \times b_n\}$.

**Theorem 13.5.** The operations $+, \times$ in Definition 13.8 (a),(b) are well-defined.

**Proof.** Suppose that $\mathbf{cl}[\{a_n\}] = \mathbf{cl}[\{c_n\}]$ and $\mathbf{cl}[\{b_n\}] = \mathbf{cl}[\{d_n\}]$. Thus means that $a_n - c_n \xrightarrow{\#} 0$ and $b_n - d_n \xrightarrow{\#} 0$. Then $(a_n + b_n) - (c_n + d_n) = (a_n - c_n) + (b_n - d_n)$.

Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to 0, and so $\mathbf{cl}[\{a_n + b_n\}] = \mathbf{cl}[\{c_n + d_n\}]$.

Multiplication is a little trickier; this is where we will use Theorem 13.3. We will also use another ubiquitous technique: adding 0 in the form of $s - s$. Again, suppose that $\mathbf{cl}[(a_n)] = \mathbf{cl}[(c_n)]$ and $\mathbf{cl}[(b_n)] = \mathbf{cl}[(d_n)]$; we wish to show that
know that there is an $N \in \mathbb{N}^\#$ such that, for $n > N$, $a_n - b_n > 0$. So $a_n > b_n$ for $n > N$. Now, adding $c_n$ to both sides of this inequality (as we know we can do for hyperreal numbers $*\mathbb{R}$), we have $a_n + c_n > b_n + c_n$ for $n > N$, or $(a_n + c_n) - (b_n + c_n) > 0$ for $n > N$. Note also that $(a_n + c_n) - (b_n + c_n) = a_n - b_n$ does not $\#\text{-converge to } 0$, by the assumption that $s - t > 0$. Thus, by Definition 13.8, this means that $s + r = \text{cl}[\{a_n + c_n\}] > \text{cl}[\{b_n + c_n\}] = t + r$.

**Theorem 13.8.** Let $s, t \in *\mathbb{R}_c^\#$, $s, t > 0$ be hyperreal numbers. Then there is $m \in \mathbb{N}^\#$ such that $m \times s > t$.

**Proof.** Let $s, t > 0$ be hyperreal numbers. We need to find a natural number $m$ so that $m \times s > t$. First, recall that, by $m$ in this context, we mean $\text{cl}[\{m, m, m, m, \dots\}]$. So, letting $s = \text{cl}[\{a_n\}]$ and $t = \text{cl}[\{b_n\}]$, what we need to show is that there exists $m$ with

$$ \begin{aligned} \text{cl}[\{m, m, m, m, \dots\}] \times \text{cl}[\{a_1, a_2, a_3, a_4, \dots\}] &= \\ \text{cl}[\{m \times a_1, m \times a_2, m \times a_3, m \times a_4, \dots\}] &> \\ &> \text{cl}[\{b_1, b_2, b_3, b_4, \dots\}]. \end{aligned} $$

Now, to say that $\text{cl}[\{m \times a_n\}] > \text{cl}[\{b_n\}]$, or $\text{cl}[\{m \times a_n - b_n\}]$ is positive, is, by Definition 13.9, just to say that there is $N \in \mathbb{N}^\#$ such that $m \times a_n - b_n > 0$ for all $n > N$, while $m \times a_n - b_n \not\to_\# 0$. To be precise, the first statement is: There exist $m, N \in \mathbb{N}^\#$ so that $m \times a_n > b_n$ for all $n > N$.

To produce a contradiction, we assume this is not the case; assume that (#) for every $m$ and $N$, there exists an $n > N$ so that $m \times a_n \le b_n$.

Now, since $\{b_n\}$ is a Cauchy sequence, by Theorem 13.2 it is hyperbounded - there is a hyperreal number $M \in *\mathbb{R}$ such that $b_n \le M$ for all $n \in \mathbb{N}^\#$. Now, by the properties for the hyperreal numbers $*\mathbb{R}$, given any hyperreal number such that $\varepsilon > 0, \varepsilon \approx 0$, there is an $m \in \mathbb{N}^\#$ such that $M/m < \varepsilon/2$. Fix such an $m$. Then if $m \times a_n \le b_n$, we have $a_n \le b_n/m \le M/m < \varepsilon/2$.

Now, $\{a_n\}$ is a Cauchy sequence, and so there exists $N$ so that for $n, k > N$, $|a_n - a_k| < \varepsilon/2$.

By Asumption (#), we also have an $n > N$ such that $m \times a_n \le b_n$, which means that $a_n < \varepsilon/2$. But then for every $k > N$, we have that $a_k - a_n < \varepsilon/2$, so $a_k < a_n + \varepsilon/2 < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Hence, $a_k < \varepsilon$ for all $k > N$. This proves that $a_k \to_\# 0$, which by Definition 13.9 contradicts the fact that $\text{cl}[\{a_n\}] = s > 0$.

Thus, there is indeed some $m \in N$ so that $m \times a_n - b_n > 0$ for all sufficiently infinite large $n \in \mathbb{N}^\#\mathbb{N}$. To conclude the proof, we must also show that $m \times a_n - b_n \not\to_\# 0$.

Actually, it is possible that $m \times a_n - b_n \to_\# 0$ (for example if $\{a_n\} = \{1, 1, 1, \dots\}$ and $\{b_n\} = \{m, m, m, \dots\}$). But that's okay: then we can simply choose a larger $m$. That is: let $m$ be a hypernatural number constructed as above, so that $m \times a_n - b_n > 0$ for all sufficiently large $n \in \mathbb{N}^\#\mathbb{N}$. If it happens to be true that $m \times a_n - b_n \not\to_\# 0$, then the proof is complete.

If, on the other hand, it turned out that $m \times a_n - b_n \to_\# 0$, then take instead the integer $m + 1$. Since $s = \text{cl}[\{a_n\}] > 0$, we have an $n > 0$ for all infinite large $n$, so $(m + 1) \times a_n - b_n = m \times a_n - b_n + a_n > a_n > 0$ for all infinite large $n$, so $m + 1$ works just as well as $m$ did in this regard; and since $m \times a_n - b_n \to_\# 0$, we have $(m + 1) \times a_n - b_n = (m \times a_n - b_n) + a_n \to_\# 0$ since $s = \text{cl}[\{a_n\}] > 0$ (so $a_n \to_\# 0$).

It will be handy to have one more Theorem about how the hyperreals $*\mathbb{R}$ and hyperreals $*\mathbb{R}_c^\#$ compare before we proceed. This theorem is known as the density
$\mathbf{cl}[\{a_n \times b_n\}] = \mathbf{cl}[\{c_n \times d_n\}]$, or, in other words, that $a_n \times b_n - c_n \cdot d_n \to_\# 0$. Well, we add and subtract one of the other cross terms, say

$$b_n \times c_n : a_n \times b_n - c_n \times d_n = a_n \times b_n + (b_n \times c_n - b_n \times c_n) - c_n \times d_n = \\ = (a_n \times b_n - b_n \times c_n) + (b_n \times c_n - c_n \times d_n) = b_n \times (a_n - c_n) + c_n \times (b_n - d_n).$$

Hence, we have $|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n|$. Now, from

Theorem 13.2, there are numbers $M$ and $L$ such that $|b_n| \le M$ and $|c_n| \le L$ for all $n \in \mathbb{N}^\#$. Taking some number $K$ which is bigger than both, we have

$$|a_n \times b_n - c_n \times d_n| \le |b_n| \times |a_n - c_n| + |c_n| \times |b_n - d_n| \le K(|a_n - c_n| + |b_n - d_n|).$$

Now, noting that both $a_n - c_n$ and $b_n - d_n$ tend to 0 and using the $\varepsilon/2$ trick (actually, this time we'll want to use $\varepsilon/2K$), we see that $a_n \times b_n - c_n \times d_n \to_\# 0$.

**Theorem 13.6.** Given any hyperreal number $s \in *\mathbb{R}_c^\#$, $s \ne 0$, there is a hyperreal number $t \in *\mathbb{R}_c^\#$ such that $s \times t = 1$.

**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 $\{0,0,0,0,\dots\}$. In other

words, $s = \mathbf{cl}[\{a_n\}]$ where $a_n - 0$ does not $\#$-converge to 0. From this, we are to deduce

the existence of a hyperreal number $t = \mathbf{cl}[\{b_n\}]$ such that $s \times t = \mathbf{cl}[\{a_n \times b_n\}]$ is the same equivalence class as $\mathbf{cl}[\{1,1,1,1,\dots\}]$. Doing so is actually an easy consequence

of the fact that nonzero hyperreal numbers have multiplicative inverses, but there is a subtle difficulty. Just because $s$ is nonzero (i.e. $\{a_n\}$ does not tend to 0), there's no reason any number of the terms in $\{a_n\}$ can't equal 0. However, it turns out that eventually, $a_n \ne 0$.

That is:

**Lemma 13.1.** If $\{a_n\}$ is a Cauchy sequence which does not $\#$-tend to 0, then there is an $N \in \mathbb{N}^\#$ such that, for $n > N$, $a_n \ne 0$.

**Definition 13.9.** Let $s \in *\mathbb{R}_c^\#$. Say that $s$ is positive if $s \ne 0$, and if $s = \mathbf{cl}[\{a_n\}]$ for some Cauchy sequence of hyperreal numbers such that for some $N \in \mathbb{N}^\#$, $a_n > 0$ for all $n > N$. Given two hyperreal numbers $s, t$, say that $s > t$ if $s - t$ is positive.

**Theorem 13.7.** Let $s, t \in *\mathbb{R}_c^\#$ be hyperreal numbers such that $s > t$, and let $r \in *\mathbb{R}_c^\#$. Then $s + r > t + r$.

**Proof.** Let $s = \mathbf{cl}[\{a_n\}], t = \mathbf{cl}[\{b_n\}],$ and $r = \mathbf{cl}[\{c_n\}]$. Since $s > t$ i.e., $s - t > 0$, we know that there is an $N \in \mathbb{N}^\#$ such that, for $n > N$, $a_n - b_n > 0$. So $a_n > b_n$ for $n > N$. Now, adding $c_n$ to both sides of this inequality (as we know we can do for hyperreal numbers $*\mathbb{R}$), we have $a_n + c_n > b_n + c_n$ for $n > N$, or

$(a_n + c_n) - (b_n + c_n) > 0$ for $n > N$. Note also that $(a_n + c_n) - (b_n + c_n) = a_n - b_n$ does not $\#$-converge to 0, by the assumption that $s - t > 0$. Thus, by Definition 13.8, this means that $s + r = \mathbf{cl}[\{a_n + c_n\}] > \mathbf{cl}[\{b_n + c_n\}] = t + r$.

**Theorem 13.8.** Let $s, t \in *\mathbb{R}_c^\#$, $s, t > 0$ be hyperreal numbers. Then there is $m \in \mathbb{N}^\#$ such that $m \times s > t$.

**Proof.** Let $s, t > 0$ be hyperreal numbers. We need to find a natural number $m$ so that $m \times s > t$. First, recall that, by $m$ in this context, we mean $\mathbf{cl}[\{m, m, m, m, \dots\}]$. So, letting $s = \mathbf{cl}[\{a_n\}]$ and $t = \mathbf{cl}[\{b_n\}]$, what we need to show is that there exists $m$ with
of $*\mathbb{R}$ in $*\mathbb{R}_c^#$, and it follows almost immediately from the construction of the $*\mathbb{R}_c^#$ from $*\mathbb{R}$.

**Theorem 13.9.** Given any hyperreal number $r \in *\mathbb{R}_c^#$, and any hyperreal number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hyperreal number $q \in *\mathbb{R}$ such that $|r-q|<\varepsilon$.

**Proof.** The hyperreal number $r$ is represented by a Cauchy $*\mathbb{R}$-valued sequence $\{a_n\}$. Since this sequence is Cauchy, given $\varepsilon > 0, \varepsilon \approx 0$, there is $N \in \mathbb{N}^#$ so that for all $m, n > N$,

$|a_n - a_m| < \varepsilon$. Picking some fixed $l > N$, we can take the hyperreal number $q$ given by $q = \text{cl}[\{a_l, a_{l+1}, a_{l+2}, \dots\}]$. Then we have $r-q = \text{cl}[\{a_n - a_l\}_{n \in \mathbb{N}^#}]$, and $q-r = \text{cl}[\{a_l - a_n\}_{n \in \mathbb{N}^#}]$.

Now, since $l > N$, we see that for $n > N, a_n - a_l < \varepsilon$ and $a_l - a_n < \varepsilon$, which means by Definition 13.9 that $r-q < \varepsilon$ and $q-r < \varepsilon$; hence, $|r-q| < \varepsilon$.

**Definition 13.10.** 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 13.1.** The order $\le$ given by Definition 10.9 obviously is $\le$-incomplete.

**Definition 13.11.** 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 or hyperbounded 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 exists $\alpha \in S$ 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 exists $\alpha \in S$ and $\beta \in L(S)$ such that $\alpha - \beta \le \varepsilon \approx 0$.

**Theorem 13.10.** (i) Any $\le$-admissible above subset $S \subset *\mathbb{R}_c^#$ has the least upper bound property. (ii) Any $\le$-admissible below subset $S \subset *\mathbb{R}_c^#$ has the greatest lower bound property.

**Proof.** Let $S \subset *\mathbb{R}_c^#$ be a nonempty subset, and let $M$ be an upper bound for $S$. We are going to construct two sequences of hyperreal numbers, $\{u_n\}$ and $\{l_n\}$. First, since $S$ is nonempty, there is some element $s_0 \in S$. Now, we go through the following hyperinductive procedure to produce numbers $u_0, u_1, u_2, \dots, u_n, \dots$ and $l_1, l_2, l_3, \dots, l_n, \dots$

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

(ii) Suppose that we have already defined $u_n$ and $l_n$. Consider the number $m_n = (u_n + l_n)/2$, 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 $l_{n+1} = l_n$.

**Remark 13.1.** Since $s < M$, it is easy to prove by hyper infinite induction that (i) $\{u_n\}$ is a non-increasing sequence: $u_{n+1} \le u_n, n \in \mathbb{N}^#$ and $\{l_n\}$ is a non-decreasing sequence $l_{n+1} \ge l_n, n \in \mathbb{N}^#$, (ii) $u_n$ is an upper bound for $S$ for all $n \in \mathbb{N}^#$ and $l_n$ is never an upper bound for $S$ for any $n \in \mathbb{N}^#$, (iii) $u_n - l_n = 2^{-n}(M-s)$. This gives us the following lemma.

**Lemma 13.2.** $\{u_n\}$ and $\{l_n\}$ are Cauchy $*\mathbb{R}$-valued sequences of hyperreal numbers.
$$ \begin{aligned} \mathrm{cl}[\{m, m, m, m, \dots\}] \times \mathrm{cl}[\{a_1, a_2, a_3, a_4, \dots\}] &= \\ & \mathrm{cl}[\{m \times a_1, m \times a_2, m \times a_3, m \times a_4, \dots\}] > \\ & > \mathrm{cl}[\{b_1, b_2, b_3, b_4, \dots\}]. \end{aligned} $$

Now, to say that $\mathrm{cl}[\{m \times a_n\}] > \mathrm{cl}[\{b_n\}]$, or $\mathrm{cl}[\{m \times a_n - b_n\}]$ is positive, is, by Definition 13.9, just to say that there is $N \in \mathbb{N}^\#$ such that $m \times a_n - b_n > 0$ for all $n > N$, while $m \times a_n - b_n \not\to_\# 0$. To be precise, the first statement is: There exist $m, N \in \mathbb{N}^\#$ so that $m \times a_n > b_n$ for all $n > N$.

To produce a contradiction, we assume this is not the case; assume that (#) for every $m$ and $N$, there exists an $n > N$ so that $m \times a_n \le b_n$.

Now, since $\{b_n\}$ is a Cauchy sequence, by Theorem 13.2 it is hyperbounded - there is a hyperreal number $M \in *\mathbb{R}$ such that $b_n \le M$ for all $n \in \mathbb{N}^\#$. Now, by the properties for the hyperreal numbers $*\mathbb{R}$, given any hyperreal number such that $\varepsilon > 0, \varepsilon \approx 0$, there is an $m \in \mathbb{N}^\#$ such that $M/m < \varepsilon/2$. Fix such an $m$. Then if $m \times a_n \le b_n$, we have $a_n \le b_n/m \le M/m < \varepsilon/2$.

Now, $\{a_n\}$ is a Cauchy sequence, and so there exists $N$ so that for $n, k > N, |a_n - a_k| < \varepsilon/2$.

By Asumption (#), we also have an $n > N$ such that $m \times a_n \le b_n$, which means that $a_n < \varepsilon/2$. But then for every $k > N$, we have that $a_k - a_n < \varepsilon/2$, so $a_k < a_n + \varepsilon/2 < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Hence, $a_k < \varepsilon$ for all $k > N$. This proves that $a_k \not\to_\# 0$, which by Definition 13.9 contradicts the fact that $\mathrm{cl}[\{a_n\}] = s > 0$.

Thus, there is indeed some $m \in N$ so that $m \times a_n - b_n > 0$ for all sufficiently infinite large $n \in \mathbb{N}^\# \setminus \mathbb{N}$. To conclude the proof, we must also show that $m \times a_n - b_n \not\to_\# 0$. Actually, it is possible that $m \times a_n - b_n \not\to_\# 0$ (for example if $\{a_n\} = \{1, 1, 1, \dots\}$ and $\{b_n\} = \{m, m, m, \dots\}$). But that's okay: then we can simply choose a larger $m$. That is: let $m$ be a hypernatural number constructed as above, so that $m \times a_n - b_n > 0$ for all sufficiently large $n \in \mathbb{N}^\# \setminus \mathbb{N}$. If it happens to be true that $m \times a_n - b_n \not\to_\# 0$, then the proof is complete.

If, on the other hand, it turned out that $m \times a_n - b_n \not\to_\# 0$, then take instead the integer $m+1$. Since $s = \mathrm{cl}[\{a_n\}] > 0$, we have a $n > 0$ for all infinite large $n$, so $(m+1) \times a_n - b_n = m \times a_n - b_n + a_n > a_n > 0$ for all infinite large $n$, so $m+1$ works just as well as $m$ did in this regard; and since $m \times a_n - b_n \to 0$, we have $(m+1) \times a_n - b_n = (m \times a_n - b_n) + a_n \not\to_\# 0$ since $s = \mathrm{cl}[\{a_n\}] > 0$ (so $a_n \not\to_\# 0$).

It will be handy to have one more Theorem about how the hyperreals $*\mathbb{R}$ and hyperreals $*\mathbb{R}_c^\#$ compare before we proceed. This theorem is known as the density of $*\mathbb{R}$ in $*\mathbb{R}_c^\#$, and it follows almost immediately from the construction of the $*\mathbb{R}_c^\#$ from $*\mathbb{R}$.

**Theorem 13.9.** Given any hyperreal number $r \in *\mathbb{R}_c^\#$, and any hyperreal number $\varepsilon > 0, \varepsilon \approx 0$, there is a hyperreal number $q \in *\mathbb{R}$ such that $|r - q| < \varepsilon$.

*Proof.* The hyperreal number $r$ is represented by a Cauchy $*\mathbb{R}$-valued sequence $\{a_n\}$. Since this sequence is Cauchy, given $\varepsilon > 0, \varepsilon \approx 0$, there is $N \in \mathbb{N}^\#$ so that for all $m, n > N$,

$|a_n - a_m| < \varepsilon$. Picking some fixed $l > N$, we can take the hyperreal number $q$ given by $q = \mathrm{cl}[\{a_l, a_l, a_l, \dots\}]$. Then we have $r - q = \mathrm{cl}[\{a_n - a_l\}_{n \in \mathbb{N}^\#}]$, and $q - r = \mathrm{cl}[\{a_l - a_n\}_{n \in \mathbb{N}^\#}]$.

Now, since $l > N$, we see that for $n > N, a_n - a_l < \varepsilon$ and $a_l - a_n < \varepsilon$, which means by
**Proof.** Note that each $l_n \le M$ for all $n \in \mathbb{N}^#$. Since $\{l_n\}$ is non-decreasing and $u_n - l_n = 2^{-n}(M-s)$, it follows directly that $\{l_n\}$ is Cauchy.

For $\{u_n\}$, we have $u_n \ge s_0$ for all $n \in \mathbb{N}^#$, and so $-u_n \le -s_0$.

Since $\{u_n\}$ is non-increasing, $\{-u_n\}$ is non-decreasing, and so as above, $\{-u_n\}$ is Cauchy. It is easy to verify that, therefore, $\{u_n\}$ is Cauchy.

The following Lemma shows that $\{u_n\}$ does #-tend to a hyperreal number $u \in *\mathbb{R}_c^#$.

**Lemma 13.3.** There is a hyperreal number $u \in *\mathbb{R}_c^#$ such that $u_n \to_* u$.

**Proof.** Fix a term $u_n$ in the sequence $\{u_n\}$. By Theorem 13.9, there is a hyperreal number $q_n \in *\mathbb{R}$, $n \in \mathbb{N}^#$ such that $|u_n - q_n| < 1/n$. Consider the sequence $\{q_1, q_2, q_3, \dots, q_n, \dots\}$ of hyperreal numbers. We will show this sequence is Cauchy. Fix $\varepsilon > 0$, $\varepsilon \approx 0$. By the Theorem 13.8, we can choose $N \in \mathbb{N}^#$ so that $1/N < \varepsilon/3$. We know, since $\{u_n\}$ is Cauchy, that there is an $M \in \mathbb{N}^#$ such that for $n, m > M$, $|u_n - u_m| < \varepsilon/3$. Then, so long as $n, m > \max\{N, M\}$, we have

$$|q_n - q_m| = |(q_n - u_n) + (u_n - u_m) + (u_m - q_m)| \leq \\ \leq |q_n - u_n| + |u_n - u_m| + |u_m - q_m| < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon.$$

Thus, $\{q_n\}$ is a Cauchy sequence of internal hyperreal numbers, and so it represents the external hyperreal number $u = \text{cl}[\{q_n\}]$. We must show that $u_n - u \to_* 0$, but this is practically built into the definition of $u$. To be precise, letting $q_n^*$ be the hyperreal number

$\text{cl}[\{q_n, q_n, q_n, \dots\}]$, we see immediately that $q_n^* - u \to_* 0$ (this is precisely equivalent to the statement that $\{q_n\}$ is Cauchy). But $u_n - q_n^* < 1/n$ by construction; it is easily verify that the assertion that if a sequence $q_n^* \to_* u$ and $u_n - q_n^* \to_* 0$, then $u_n \to_* u$. So $\{u_n\}$, a non-increasing sequence of upper bounds for $S$, tends to a hyperreal number $u$. As you've guessed, $u$ is the least upper bound of our set $S$. To prove this, we need one more lemma.

**Lemma 13.4.** $l_n \to_* u$.

**Proof.** First, note in the first case above, we have that

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

In the second case, we also have

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

Now, this means that $u_1 - l_1 = \frac{1}{2}(M-s)$, and so $u_2 - l_2 = \frac{1}{2}(u_1 - l_1) = \frac{1}{2^2}(L-s)$, and in general by hyperinfinite induction, $u_n - l_n = 2^{-n}(M-s)$. Since $M > s$ so $M - s > 0$, and since $2^{-n} < 1/n$, by the Theorem 13.8, we have for any $\varepsilon > 0$ that $2^{-n}(M-s) < \varepsilon$ for all sufficiently large $n \in \mathbb{N}^#$. Thus, $u_n - l_n = 2^{-n}(M-s) < \varepsilon$ as well, and so $u_n - l_n \to_* 0$. Again, it is easily verify that, since $u_n \to_* u$, we have $l_n \to_* u$ as well.

**Remark 13.2.** Note that assumption in Theorem 13.10 that $S$ is $\le$-admissible above subset of $\mathbb{R}_c^#$ is necessarily, othervice Theorem 13.10 is not holds.

**Theorem 13.11. (Generalized Nested Intervals Theorem)**

Let $\{I_n\}_{n \in \mathbb{N}^#} = \{[a_n, b_n]\}_{n \in \mathbb{N}^#}$, $[a_n, b_n] \subset \mathbb{R}_c^#$ be a hyper infinite sequence of closed intervals satisfying each of the following conditions:
Definition 13.9 that $r-q < \varepsilon$ and $q-r < \varepsilon$; hence, $|r-q| < \varepsilon$.

**Definition 13.10.** 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 13.1.** The order $\le$ given by Definition 10.9 obviously is $\le$-incomplete.

**Definition 13.11.** 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 or hyperbounded 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 13.10.** (i) Any $\le$-admissible above subset $S \subsetneq *\mathbb{R}_c^\#$$ has the least upper bound property.(ii) Any $\le$-admissible below subset $S \subsetneq *\mathbb{R}_c^\#$$ has the greatest lower bound property.

**Proof.** Let $S \subsetneq *\mathbb{R}_c^\#$ be a nonempty subset, and let $M$ be an upper bound for $S$. We are going to construct two sequences of hyperreal numbers, $\{u_n\}$ and $\{l_n\}$. First, since $S$ is nonempty, there is some element $s_0 \in S$. Now, we go through the following hyperinductive procedure to produce numbers $u_0, u_1, u_2, \dots, u_n, \dots$ and $l_1, l_2, l_3, \dots, l_n, \dots$

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

(ii) Suppose that we have already defined $u_n$ and $l_n$. Consider the number $m_n = (u_n + l_n)/2$, 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 $l_{n+1} = l_n$.

**Remark 13.1.** Since $s < M$, it is easy to prove by hyper infinite induction that (i) $\{u_n\}$ is a non-increasing sequence: $u_{n+1} \le u_n, n \in \mathbb{N}^\#$ and $\{l_n\}$ is a non-decreasing sequence $l_{n+1} \ge l_n, n \in \mathbb{N}^\#$, (ii) $u_n$ is an upper bound for $S$ for all $n \in \mathbb{N}^\#$ and $l_n$ is never an upper bound for $S$ for any $n \in \mathbb{N}^\#$, (iii) $u_n - l_n = 2^{-n}(M-s)$. This gives us the following lemma.

**Lemma 13.2.** $\{u_n\}$ and $\{l_n\}$ are Cauchy $*\mathbb{R}$-valued sequences of hyperreal numbers.

**Proof.** Note that each $l_n \le M$ for all $n \in \mathbb{N}^\#$. Since $\{l_n\}$ is non-decreasing and $u_n - l_n = 2^{-n}(M-s)$, it follows directly that $\{l_n\}$ is Cauchy.

For $\{u_n\}$, we have $u_n \ge s_0$ for all $n \in \mathbb{N}^\#$, and so $-u_n \le -s_0$.

Since $\{u_n\}$ is non-increasing, $\{-u_n\}$ is non-decreasing, and so as above, $\{-u_n\}$ is Cauchy. It is easy to verify that, therefore, $\{u_n\}$ is Cauchy.

The following Lemma shows that $\{u_n\}$ does $\#$-tend to a hyperreal number $u \in *\mathbb{R}_c^\#$.

**Lemma 13.3.** There is a hyperreal number $u \in *\mathbb{R}_c^\#$ such that $u_n \to_s u$.

**Proof.** Fix a term $u_n$ in the sequence $\{u_n\}$. By Theorem 13.9, there is a hyperreal number $q_n \in *\mathbb{R}, n \in \mathbb{N}^\#$ such that $|u_n - q_n| < 1/n$. Consider the sequence $\{q_1, q_2, q_3, \dots, q_n, \dots\}$ of hyperreal numbers. We will show this sequence is Cauchy. Fix $\varepsilon > 0, \varepsilon \approx 0$. By the Theorem 13.8, we can choose $N \in \mathbb{N}^\#$ so that $1/N < \varepsilon/3$. We know, since $\{u_n\}$ is Cauchy, that there is an $M \in \mathbb{N}^\#$ such that for $n, m > M$,
(i) $I_1 \supseteq I_2 \supseteq I_3 \supseteq \dots \supseteq I_n \supseteq \dots,$

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

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

**Proof.** Note that: (a) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is hyperbounded above by $b_1$ and (b) the set $B = \{b_n|n \in \mathbb{N}^\#\}$ is $\le$-admissible above subset of $\mathbb{R}_c^\#$.

By Theorem 13.10 there exists $\sup A$. Let $\xi = \sup A$.

Since $I_n$ are nested, for any positive hyperintegers $m$ and $n$ we have $a_m \le a_{m+n} \le b_{m+n} \le b_n$, so that $\xi \le b_n$ for each $n \in \mathbb{N}^\#$. Since we obviously have $a_n \le \xi$ for each $n \in \mathbb{N}^\#$, we have $a_n \le \xi \le b_n$ for all $n \in \mathbb{N}^\#$, which implies $\xi \in \cap_{n=1}^\infty I_n$. Finally, if $\xi, \eta \in \cap_{n=1}^\infty I_n$, with $\xi \le \eta$, then we get $0 \le \eta - \xi \le b_n - a_n$, for all $n \in \mathbb{N}^\#$, so that $0 \le \eta - \xi \le \inf_{n \in \mathbb{N}^\#} |b_n - a_n| = 0$.

**Theorem 13.12. (Generalized Squeeze Theorem)**

Let $\{a_n\}$, $\{c_n\}$ be two hyper infinite sequences #-converging to $L$, and $\{b_n\}$ a hyper infinite sequence. If $\forall n \ge K, K \in \mathbb{N}^\#$ we have $a_n \le b_n \le c_n$, then $\{b_n\}$ also #-converges to $L$.

**Proof.** Choose an $\varepsilon > 0, \varepsilon \approx 0$. By definition of the #-limit, there is an $N_1 \in \mathbb{N}^\#$ such that for all $n > N_1$ we have $|a_n - L| < \varepsilon$, in other words $L - \varepsilon < a_n < L + \varepsilon$. Similarly, there is an $N_2 \in \mathbb{N}^\#$ such that for all $n > N_2$ we have $L - \varepsilon < c_n < L + \varepsilon$. Denote $N = \max(N_1, N_2, K)$. Then for $n > N, L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon$, in other words $|b_n - L| < \varepsilon$. Since $\varepsilon > 0, \varepsilon \approx 0$ was arbitrary, by definition of the #-limit this says that $\#\text{-lim}_{n \to \infty}^\# b_n = L$.

**Theorem 13.13. (Corollary of the Generalized Squeeze Theorem).**

If $\#\text{-lim}_{n \to \infty}^\# |a_n| = 0$ then $\#\text{-lim}_{n \to \infty}^\# a_n = 0$.

**Proof.** We know that $-|a_n| \le a_n \le |a_n|$. We want to apply the Generalized Squeeze Theorem. We are given that $\#\text{-lim}_{n \to \infty}^\# |a_n| = 0$. This also implies that $\#\text{-lim}_{n \to \infty}^\# (-|a_n|) = 0$. So by the Generalized Squeeze Theorem, $\#\text{-lim}_{n \to \infty}^\# a_n = 0$.

**Theorem 13.14. (Generalized Bolzano-Weierstrass Theorem)**

Every hyperbounded hyper infinite $\mathbb{R}_c^\#$-valued sequence has a #-convergent hyper infinite subsequence.

**Proof.** Let $\{w_n\}_{n \in \mathbb{N}^\#}$ be a hyperbounded hyper infinite sequence. Then, there exists an interval $[a_1, b_1]$ such that $a_1 \le w_n \le b_1$ for all $n \in \mathbb{N}^\#$.

Either $[a_1, \frac{a_1+b_1}{2}]$ or $[\frac{a_1+b_1}{2}, b_1]$ contains hyper infinitely many terms of $\{w_n\}$. That is, there exists hyper infinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[a_1, \frac{a_1+b_1}{2}]$ or there exists hyper infinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[\frac{a_1+b_1}{2}, b_1]$. If $[a_1, \frac{a_1+b_1}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_2, b_2] = [a_1, \frac{a_1+b_1}{2}]$. Otherwise, let $[a_2, b_2] = [\frac{a_1+b_1}{2}, b_1]$.

Either $[a_2, \frac{a_2+b_2}{2}]$ or $[\frac{a_2+b_2}{2}, b_2]$ contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^\#}$. If $[a_2, \frac{a_2+b_2}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_3, b_3] = [a_2, \frac{a_2+b_2}{2}]$. Otherwise, let $[a_3, b_3] = [\frac{a_2+b_2}{2}, b_2]$. By hyper infinite induction, we can continue this construction and obtain hyper infinite sequence of intervals $\{[a_n, b_n]\}_{n \in \mathbb{N}^\#}$ such that:

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

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

(iii) for each $n \in \mathbb{N}^\#$, $b_{n+1} - a_{n+1} = \frac{1}{2}(b_n - a_n)$.
$|u_n - u_m| < \varepsilon/3$. Then, so long as $n, m > \max\{N, M\}$, we have

$$ \begin{aligned} |q_n - q_m| &= |(q_n - u_n) + (u_n - u_m) + (u_m - q_m)| \le \\ &\le |q_n - u_n| + |u_n - u_m| + |u_m - q_m| < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon. \end{aligned} $$

Thus, $\{q_n\}$ is a Cauchy sequence of internal hyperreal numbers, and so it represents the external hyperreal number $u = \text{cl}[\{q_n\}]$. We must show that $u_n - u \to_\# 0$, but this is practically built into the definition of $u$. To be precise, letting $q_n^* \to$ the hyperreal number

$\text{cl}[\{q_n, q_n, q_n, \dots\}]$, we see immediately that $q_n^* - u \to_\# 0$ (this is precisely equivalent to the statement that $\{q_n\}$ is Cauchy). But $u_n - q_n^* < 1/n$ by construction; it is easily verify that the assertion that if a sequence $q_n^* \to_\# u$ and $u_n - q_n^* \to_\# 0$, then $u_n \to_\# u$. So $\{u_n\}$, a non-increasing sequence of upper bounds for $S$, tends to a hyperreal

number $u$. As you've guessed, $u$ is the least upper bound of our set $S$. To prove this, we

need one more lemma.

**Lemma 13.4.** $l_n \to_\# u$.

**Proof.** First, note in the first case above, we have that

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

In the second case, we also have

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

Now, this means that $u_1 - l_1 = \frac{1}{2}(M-s)$, and so $u_2 - l_2 = \frac{1}{2}(u_1 - l_1) = \frac{1}{2^2}(L-s)$, and in general by hyperinfinite induction, $u_n - l_n = 2^{-n}(M-s)$. Since $M > s$ so $M-s > 0$, and since $2^{-n} < 1/n$, by the Theorem 13.8, we have for any $\varepsilon > 0$ that $2^{-n}(M-s) < \varepsilon$ for all sufficiently large $n \in \mathbb{N}^\#$. Thus, $u_n - l_n = 2^{-n}(M-s) < \varepsilon$ as well, and so $u_n - l_n \to_\# 0$. Again, it is easily verify that, since $u_n \to_\# u$, we have $l_n \to_\# u$ as well.

**Remark 13.2.** Note that assumption in Theorem 13.10 that $S$ is $\leq$-admissible above subset of $\mathbb{R}_c^\#$ is necessarily, otherwise Theorem 13.10 is not holds.

**Theorem 13.11.** (Generalized Nested Intervals Theorem)

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

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

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

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

**Proof.** Note that: (a) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is hyperbouded above by $b_1$ and (b) the set $A = \{a_n|n \in \mathbb{N}^\#\}$ is $\leq$-admissible above subset of $\mathbb{R}_c^\#$.

By Theorem 13.10 there exists $\sup A$. Let $\xi = \sup A$.

Since $I_n$ are nested, for any positive hyperintegers $m$ and $n$ we have $a_m \le a_{m+n} \le b_{m+n} \le b_n$, so that $\xi \le b_n$ for each $n \in \mathbb{N}^\#$. Since we obviously have $a_n \le \xi$ for each $n \in \mathbb{N}^\#$, we have $a_n \le \xi \le b_n$ for all $n \in \mathbb{N}^\#$, which implies $\xi \in \cap_{n=1}^\infty I_n$. Finally, if $\xi, \eta \in \cap_{n=1}^\infty I_n$, with $\xi \le \eta$, then we get $0 \le \eta - \xi \le b_n - a_n$, for all $n \in \mathbb{N}^\#$, so that $0 \le \eta - \xi \le \inf_{n \in \mathbb{N}^\#} |b_n - a_n| = 0$.
Then generalized nested intervals theorem implies that the intersection of all of the intervals $[a_n, b_n]$ is a single point $w$. We will now construct a hyper infinite subsequence of $\{w_n\}_{n \in \mathbb{N}^\#}$ which will #-converge to $w$.

Since $[a_1, b_1]$ contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^\#}$, there exists $k_1 \in \mathbb{N}^\#$ such that $w_{k_1}$ is in $[a_1, b_1]$. Since $[a_2, b_2]$ contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^\#}$, there exists $k_2 \in \mathbb{N}^\#$, $k_2 > k_1$, such that $w_{k_2}$ is in $[a_2, b_2]$. Since $[a_3, b_3]$ contains hyper infinitely many terms of $\{w_n\}_{n \in \mathbb{N}^\#}$, there exists $k_3 \in \mathbb{N}^\#$, $k_3 > k_2$, such that $w_{k_3}$ is in $[a_3, b_3]$. Continuing this process by hyper infinite induction, we obtain hyper infinite sequence $\{w_{k_n}\}_{n \in \mathbb{N}^\#}$ such that $w_{k_n} \in [a_n, b_n]$ for each $n \in \mathbb{N}^\#$. The sequence $\{w_{k_n}\}_{n \in \mathbb{N}^\#}$ is a subsequence of $\{w_n\}_{n \in \mathbb{N}^\#}$ since $k_{n+1} > k_n$ for each $n \in \mathbb{N}^\#$. Since $a_n \to_\# w$, and $a_n \le w_n \le b_n$ for each $n \in \mathbb{N}^\#$, the squeeze theorem implies that $w_{k_n} \to_\# w$.

**Definition 13.12.** Let $\{a_n\}$ be a hyperreal sequence i.e., $a_n \in *\mathbb{R}_c^\#$, $n \in \mathbb{N}^\#$. Say that $\{a_n\}$ #-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^\#\mathbb{N}$, $N = N(\varepsilon)$ such that, for all $n > N$, $|a_n| \le \varepsilon$. We often denote this symbolically by $a_n \to_\# 0$. We can also, at this point, define what it means for a hyperreal sequence #-tends to a given number $q \in *\mathbb{R}_c^\# : \{a_n\}$ #-tends to $q$ if the hyperreal sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \to_\# 0$.

**Definition 13.13.** Let $\{a_n\}$, $n \in \mathbb{N}^\#$ be a hyperreal sequence. We call $\{a_n\}$ a Cauchy hyperreal sequence if the difference between its terms #-tends to 0. To be precise: given any hyperreal number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 13.15.** If $\{a_n\}$ is a #-convergent hyperreal sequence (that is, $a_n \to_\# b$ for some hyperreal number $b \in \mathbb{R}_c^\#$), then $\{a_n\}$ is a Cauchy hyperreal sequence.

**Theorem 13.16.** If $\{a_n\}$ is a Cauchy hyperreal sequence, then it is hyper bounded; that is, there is some $M \in \mathbb{R}_c^\#$ such that $|a_n| \le M$ for all $n \in \mathbb{N}^\#$.

**Theorem 13.17.** Any Cauchy hyperreal sequence $\{a_n\}$ has a #-limit in $*\mathbb{R}_c^\#$ i.e., there exists $b \in *\mathbb{R}_c^\#$ such that $a_n \to_\# b$.

**Proof.** By Definition 13.13 given $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $n, n' > N$,

$$|a_n - a_{n'}| < \varepsilon. \qquad (13.1)$$

From (13.1) for any $n, n' > N$ we get

$$a_{n'} - \varepsilon < a_n < a_n + \varepsilon. \qquad (13.2)$$

The generalized Bolzano-Weierstrass theorem implies there is a #-convergent hyper infinite subsequence $\{a_{n_k}\} \subset \{a_n\}$ such that $a_{n_k} \to_\# b$ for some hyperreal number $b \in *\mathbb{R}_c^\#$. Let us show that the sequence $\{a_n\}$ also #-convergent to this $b \in *\mathbb{R}_c^\#$.

We can choose $k \in \mathbb{N}^\#$ so large that $n_k > N$ and

$$|a_{n_k} - b| < \varepsilon. \qquad (13.3)$$

We choose now in (13.1) $n' = n_k$ and therefore

$$|a_n - a_{n_k}| < \varepsilon. \qquad (13.4)$$

From (13.3) and (13.4) for any $n > N$ we get

$$(a_{n_k} - b) + (a_n - a_{n_k}) = |a_n - b| < 2\varepsilon. \qquad (13.5)$$

Thus $a_n \to_\# b$ as well.
**Theorem 13.12. (Generalized Squeeze Theorem)**

Let {$a_n$}, {$c_n$} be two hyper infinite sequences #-converging to $L$, and {$b_n$} a hyper infinite sequence. If $\forall n \ge K, K \in \mathbb{N}^\#$ we have $a_n \le b_n \le c_n$, then {$b_n$} also #-converges to $L$.

**Proof.** Choose an $\varepsilon > 0, \varepsilon \approx 0$. By definition of the #-limit, there is an $N_1 \in \mathbb{N}^\#$ such that for all $n > N_1$ we have $|a_n - L| < \varepsilon$, in other words $L - \varepsilon < a_n < L + \varepsilon$. Similarly, there is an $N_2 \in \mathbb{N}^\#$ such that for all $n > N_2$ we have $L - \varepsilon < c_n < L + \varepsilon$. Denote $N = \max(N_1, N_2, K)$. Then for $n > N, L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon$, in other words $|b_n - L| < \varepsilon$. Since $\varepsilon > 0, \varepsilon \approx 0$ was arbitrary, by definition of the #-limit this says that $\#\text{-lim}_{n \to \infty}^# b_n = L$.

**Theorem 13.13. (Corollary of the Generalized Squeeze Theorem).**

If $\#\text{-lim}_{n \to \infty}^# |a_n| = 0$ then $\#\text{-lim}_{n \to \infty}^# a_n = 0$.

**Proof.** We know that $-|a_n| \le a_n \le |a_n|$. We want to apply the Generalized Squeeze Theorem. We are given that $\#\text{-lim}_{n \to \infty}^# |a_n| = 0$. This also implies that $\#\text{-lim}_{n \to \infty}^# (-|a_n|) = 0$. So by the Generalized Squeeze Theorem, $\#\text{-lim}_{n \to \infty}^# a_n = 0$.

**Theorem 13.14. (Generalized Bolzano-Weierstrass Theorem)**

Every hyperbounded hyper infinite $\mathbb{R}^\#$-valued sequence has a #-convergent hyper infinite subsequence.

**Proof.** Let {$w_n$}$_{n \in \mathbb{N}^\#$} be a hyperbounded hyper infinite sequence. Then, there exists an interval $[a_1, b_1]$ such that $a_1 \le w_n \le b_1$ for all $n \in \mathbb{N}^\#$.

Either $[a_1, \frac{a_1+b_1}{2}]$ or $[\frac{a_1+b_1}{2}, b_1]$ contains hyper infinitely many terms of $\{w_n\}$. That is, there exists hyper infinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[a_1, \frac{a_1+b_1}{2}]$ or there exists hyper infinitely many $n$ in $\mathbb{N}^\#$ such that $a_n$ is in $[\frac{a_1+b_1}{2}, b_1]$. If $[a_1, \frac{a_1+b_1}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_2, b_2] = [a_1, \frac{a_1+b_1}{2}]$. Otherwise, let $[a_2, b_2] = [\frac{a_1+b_1}{2}, b_1]$.

Either $[a_2, \frac{a_2+b_2}{2}]$ or $[\frac{a_2+b_2}{2}, b_2]$ contains hyper infinitely many terms of $\{w_n\}$$_{n \in \mathbb{N}^\#}$. If $[a_2, \frac{a_2+b_2}{2}]$ contains hyper infinitely many terms of $\{w_n\}$, let $[a_3, b_3] = [a_2, \frac{a_2+b_2}{2}]$. Otherwise, let $[a_3, b_3] = [\frac{a_2+b_2}{2}, b_2]$. By hyper infinite induction, we can continue this construction and obtain hyper infinite sequence of intervals $\{[a_n, b_n]\}$$_{n \in \mathbb{N}^\#}$ such that:
(i) for each $n \in \mathbb{N}^\#$, $[a_n, b_n]$ contains hyper infinitely many terms of $\{w_n\}$$_{n \in \mathbb{N}^\#}$,
(ii) for each $n \in \mathbb{N}^\#$, $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$ and
(iii) for each $n \in \mathbb{N}^\#$, $b_{n+1} - a_{n+1} = \frac{1}{2}(b_n - a_n)$.

Then generalized nested intervals theorem implies that the intersection of all of the intervals $[a_n, b_n]$ is a single point $w$. We will now construct a hyper infinite subsequence of $\{w_n\}$$_{n \in \mathbb{N}^\#}$ which will #-converge to $w$.

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

**Definition 13.12.** Let {$a_n$} be a hyperreal sequence i.e., $a_n \in \mathbb{R}^\#$, $n \in \mathbb{N}^\#$. Say that
**Remark 13.3.** Note that there exist canonical natural embedings

$$ \mathbb{R} \hookrightarrow *\mathbb{R} \hookrightarrow *\mathbb{R}_c^\# \qquad (13.6) $$

## 13.1. The Extended Hyperreal Number System *$\hat{\mathbb{R}}_c^\#$*

**Definition 13.14.** (a) A set $S \subset \mathbb{N}^\#$ is hyperfinite if $card(S) = card(\{x|0 \le x \le n\})$, $n \in \mathbb{N}^\# \setminus \mathbb{N}$. (b) A set $S \subset \mathbb{N}^\#$ is hyper infinite if $card(S) = card(\mathbb{N}^\#)$.

**Notation 13.2.** If $F$ is an arbitrary collection of subsets of $*\mathbb{R}_c^\#$, then $\cup\{S|S \in F\}$ is the set of all elements that are members of at least one of the sets in $F$, and $\cap\{S|S \in F\}$ is the set of all elements that are members of every set in $F$. The union and intersection of finitely or hyperfinitely many sets $S_k, 0 \le k \le n \in \mathbb{N}^\#$ are also written as $\cup_{k=0}^n S_k$ and $\cap_{k=0}^n S_k$. The union and intersection of an hyperinfinite sequence $S_k, k \in \mathbb{N}^\#$ of sets are written as $\cup_{k=0}^\infty S$ or $\cup_{n \in \mathbb{N}^\#} S$ and $\cap_{k=0}^\infty S$ or $\cap_{n \in \mathbb{N}^\#} S$ correspondingly. A nonempty set $S$ of hyperreal numbers $*\mathbb{R}_c^\#$ is unbounded above if it has no hyperfinite

upper bound, or unbounded below if it has no hyperfinite lower bound. It is convenient to adjoin to the hyperreal number system two points, $+\infty^\#$ (which we also write more simply as $\infty^\#$) and $-\infty^\#$, and to define the order relationships between them and any hyperreal number $x \in *\mathbb{R}_c^\#$ by $-\infty^\# < x < \infty^\#$.

We call $-\infty^\#$ and $\infty^\#$ points at hyperinfinity. If $S$ is a nonempty set of hyperreals, we write $supS = \infty^\#$ to indicate that $S$ is unbounded above, and $infS = -\infty^\#$ to indicate that $S$ is unbounded below.

## 13.2. \#-Open and \#-Closed Sets on $*\hat{\mathbb{R}}_c^\#$$^\#$

**Definition 13.15.** If $a$ and $b$ are in the extended hyperreals and $a < b$, then the open interval $(a,b)$ is defined by $(a,b) \triangleq \{x|a < x < b\}$. :

The open intervals $(a, +\infty^\#)$ and $(-\infty^\# , b)$ are semi-hyperinfinite if $a$ and $b$ are finite or hyperfinite, and $(-\infty^\# , \infty^\#)$ is the entire hyperreal line.

If $-\infty^\# < a < b < \infty^\#$, the set $[a,b] \triangleq \{x|a \le x \le b\}$ is \#-closed, since its complement is the union of the \#-open sets $(-\infty^\# , a)$ and $(b, \infty^\#)$. We say that $[a,b]$ is a \#-closed interval. Semi-hyper infinite \#-closed intervals are sets of the form $[a, \infty) = \{x|a \le x\}$ and $(-\infty^\# , a] = \{x|x \le a\}$, where $a$ is finite or hyperfinite. They are \#-closed sets, since their complements are the \#-open intervals $(-\infty^\# , a)$ and $(a, \infty^\#)$, respectively.

**Definition 13.16.** If $x_0 \in \mathbb{R}_c^\#$ is a hyperreal number and $\varepsilon > 0, \varepsilon \approx 0$ then the open interval $(x_0 - \varepsilon, x_0 + \varepsilon)$ is an \#-neighborhood of $x_0$. If a set $S \subset *\mathbb{R}_c^\#$ contains an \#-neighborhood of $x_0$, then $S$ is a \#-neighborhood of $x_0$, and $x_0$ is an \#-interior point of $S$.

The set of \#-interior points of $S$ is the \#-interior of $S$, denoted by \#-Int($S$).

(i) If every point of $S$ is an \#-interior point (that is, $S = \#-\text{Int}(S)$), then $S$ is \#-open.

(ii) A set $S$ is \#-closed if $S^c = *\mathbb{R}_c^\# \setminus S$ is \#-open.

**Example 13.1.** An open interval $(a,b)$ is an \#-open set, because if $x_0 \in (a,b)$ and $\varepsilon \le \min\{x_0 - a; b - x_0\}$, then $(x_0 - \varepsilon, x_0 + \varepsilon) \subset (a,b)$

**Remark 13.4.** The entire hyperline $*\hat{\mathbb{R}}_c^\# = (-\infty^\# , \infty^\#)$ is \#-open, and therefore $\emptyset$ is \#-closed.

However, $\emptyset$ is also \#-open, for to deny this is to say that $\emptyset$ contains a point that is not
$\{a_n\}$ #-tends to 0 if, given any $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N \in \mathbb{N}^\# \setminus \mathbb{N}$, $N = N(\varepsilon)$ such that, for all $n > N$, $|a_n| \le \varepsilon$. We often denote this symbolically by $a_n \to_\# 0$. We can also, at this point, define what it means for a hyperreal sequence #-tends to a given number $q \in *\mathbb{R}_c^\# : \{a_n\}$ #-tends to $q$ if the hyperreal sequence $\{a_n - q\}$ #-tends to 0 i.e., $a_n - q \to_\# 0$.

**Definition 13.13.** Let $\{a_n\}, n \in \mathbb{N}^\#$ be a hyperreal sequence. We call $\{a_n\}$ a Cauchy hyperreal sequence if the difference between its terms #-tends to 0. To be precise: given any hyperreal number $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $m, n > N$, $|a_n - a_m| < \varepsilon$.

**Theorem 13.15.** If $\{a_n\}$ is a #-convergent hyperreal sequence (that is, $a_n \to_\# b$ for some hyperreal number $b \in *\mathbb{R}_c^\#$), then $\{a_n\}$ is a Cauchy hyperreal sequence.

**Theorem 13.16.** If $\{a_n\}$ is a Cauchy hyperreal sequence, then it is hyper bounded; that is, there is some $M \in *\mathbb{R}_c^\#$ such that $|a_n| \le M$ for all $n \in \mathbb{N}^\#$.

**Theorem 13.17.** Any Cauchy hyperreal sequence $\{a_n\}$ has a #-limit in $*\mathbb{R}_c^\#$ i.e., there exists $b \in *\mathbb{R}_c^\#$ such that $a_n \to_\# b$.

**Proof.** By Definition 13.13 given $\varepsilon > 0$, $\varepsilon \approx 0$, there is a hypernatural number $N = N(\varepsilon)$ such that for any $n, n' > N$,

$$|a_n - a_{n'}| < \varepsilon. \qquad (13.1)$$

From (13.1) for any $n, n' > N$ we get

$$a_{n'} - \varepsilon < a_n < a_n + \varepsilon. \qquad (13.2)$$

The generalized Bolzano-Weierstrass theorem implies there is a #-convergent hyper infinite subsequence $\{a_{n_k}\} \subset \{a_n\}$ such that $a_{n_k} \to_\# b$ for some hyperreal number $b \in *\mathbb{R}_c^\#$. Let us show that the sequence $\{a_n\}$ also #-convergent to this $b \in *\mathbb{R}_c^\#$.

We can choose $k \in \mathbb{N}^\#$ so large that $n_k > N$ and

$$|a_{n_k} - b| < \varepsilon. \qquad (13.3)$$

We choose now in (13.1) $n' = n_k$ and therefore

$$|a_n - a_{n_k}| < \varepsilon. \qquad (13.4)$$

From (13.3) and (13.4) for any $n > N$ we get

$$(a_{n_k} - b) + (a_n - a_{n_k}) = |a_n - b| < 2\varepsilon. \qquad (13.5)$$

Thus $a_n \to_\# b$ as well.

**Remark 13.3.** Note that there exist canonical natural embedings

$$\mathbb{R} \hookrightarrow *\mathbb{R} \hookrightarrow *\mathbb{R}_c^\#. \qquad (13.6)$$

## 13.1. The Extended Hyperreal Number System *$\mathbb{R}_c^\#$*

**Definition 13.14.** (a) A set $S \subset \mathbb{N}^\#$ is hyperfinite if $\mathrm{card}(S) = \mathrm{card}(\{x | 0 \le x \le n\})$, $n \in \mathbb{N}^\# \setminus \mathbb{N}$. (b) A set $S \subset \mathbb{N}^\#$ is hyper infinite if $\mathrm{card}(S) = \mathrm{card}(\mathbb{N}^\#)$.

**Notation 13.2.** If $F$ is an arbitrary collection of subsets of $*\mathbb{R}_c^\#$, then $\cup\{S | S \in F\}$ is the set of all elements that are members of at least one of the sets in $F$, and $\cap\{S | S \in F\}$ is the set of all elements that are members of every set in $F$. The union and intersection of finitely or hyperfinitely many sets $S_k, 0 \le k \le n \in \mathbb{N}^\#$ are also written as $\cup_{k=0}^n S_k$ and $\cap_{k=0}^n S_k$. The union and intersection of an hyperinfinite sequence $S_k, k \in \mathbb{N}^\#$ of sets are written as $\cup_{k=0}^\infty S_k$ or $\cup_{n \in \mathbb{N}^\#} S$ and $\cap_{k=0}^\infty S_k$ or $\cap_{n \in \mathbb{N}^\#} S$ correspondingly.
an #-interior point, which is absurd because $\emptyset$ contains no points. Since $\emptyset$ is #-open, $*\mathbb{R}_c^\#$ is #-closed. Thus, $*\mathbb{R}_c^\#$ and $\emptyset$ are both #-open and #-closed.

**Remark 13.5.** They are not the only subsets of $*\mathbb{R}_c^\#$ with this property.

**Definition 13.17.** A deleted #-neighborhood of a point $x_0$ is a set that contains every point

of some #-neighborhood of $x_0$ except for $x_0$ itself. For example, $S = \{x|0 < |x - x_0| < \varepsilon\}$, where $\varepsilon \approx 0$, is a deleted #-neighborhood of $x_0$. We also say that it is a deleted $\varepsilon$-#-neighborhood of $x_0$.

**Theorem 13.18.** (a) The union of #-open sets is #-open:

(b) The #-intersection of #-closed sets is #-closed:

These statements apply to arbitrary collections, hyperfinite or hyperinfinite, of #-open and #-closed sets.

**Proof** (a) Let $L$ be a collection of #-open sets and $S = \bigcup \{G|G \in L\}$.

If $x_0 \in S$, then $x_0 \in G_0$ for some $G_0$ in $L$, and since $G_0$ is #-open, it contains some $\varepsilon$-#-neighborhood of $x_0$. Since $G_0 \subset S$, this $\varepsilon$-#-neighborhood is in $S$, which is consequently a #-neighborhood of $x_0$. Thus, $S$ is a #-neighborhood of each of its points, and therefore #-open, by definition.

(b) Let $F$ be a collection of #-closed sets and $T = \bigcap \{H|H \in F\}$. Then $T^c = \bigcup \{H^c|H \in F\}$ and, since each $H^c$ is #-open, $T^c$ is #-open, from (a). Therefore, $T$ is #-closed, by definition.

**Example 13.2.** If $-\infty^\# < a < b < \infty^\#$, the set $[a, b] = \{x|a \le x \le b\}$ is #-closed, since its complement is the union of the #-open sets $(-\infty^\#a)$ and $(b, \infty^\#)$. We say that $[a, b]$ is a #-closed interval. The set $[a, b) = \{x|a \le x < b\}$ is a half-#-closed or half-#-open interval if $-\infty^\# < a < b < \infty^\#$, as is $(a, b] = \{x|a < x \le b\}$ however, neither of these sets is #-open or #-closed. Semi-infinite #-closed intervals are sets of the form $[a, \infty^\#) = \{x|a \le x\}$ and $(-\infty^\#, a] = \{x|x \le a\}$, where $a$ is hyperfinite. They are #-closed sets, since their complements are the #-open intervals $(-\infty^\#, a)$ and

$(a, \infty^\#)$, respectively.

**Definition 13.18.** Let $S$ be a subset of $\mathbb{R}_c^\# = (-\infty^\#, \infty^\#)$. Then

(a) $x_0$ is a #-limit point of $S$ if every deleted #-neighborhood of $x_0$ contains a point of $S$.

(b) $x_0$ is a boundary point of $S$ if every #-neighborhood of $x_0$ contains at least one point in $S$ and one not in $S$. The set of #-boundary points of $S$ is the #-boundary of $S$, denoted

$$ \text{by } \#\partial S. \text{ The } \#\text{-closure of } S, \text{ denoted by } \#-\bar{S}, \text{ is } S \cup \#\partial S. $$

(c) $x_0$ is an #-isolated point of $S$ if $x_0 \in S$ and there is a #-neighborhood of $x_0$ that contains

no other point of $S$.

(d) $x_0$ is #-exterior to $S$ if $x_0$ is in the #-interior of $S^c$. The collection of such points is the #-exterior of $S$.

**Theorem 13.19.** A set $S$ is #-closed if and only if no point of $S^c$ is a #-limit point of $S$.

**Proof.** Suppose that $S$ is #-closed and $x_0 \in S^c$. Since $S^c$ is #-open, there is a #-neighborhood of $x_0$ that is contained in $S^c$ and therefore contains no points of $S$.

Hence, $x_0$ cannot be a #-limit point of $S$. For the converse, if no point of $S^c$ is a #-limit point of $S$ then every point in $S^c$ must have a #-neighborhood contained in $S^c$. Therefore, $S^c$ is #-open and $S$ is #-closed.

**Corollary 13.1.** A set $S$ is #-closed if and only if it contains all its #-limit points.
A nonempty set $S$ of hyperreal numbers $*\mathbb{R}_c^\#$ is unbounded above if it has no hyperfinite

upper bound, or unbounded below if it has no hyperfinite lower bound. It is convenient to adjoin to the hyperreal number system two points, $+\infty^\#$ (which we also write more simply as $\infty^\#)$) and $-\infty^\#$, and to define the order relationships between them and any hyperreal number $x \in *\mathbb{R}_c^\#$ by $-\infty^\# < x < \infty^\#$.

We call $-\infty^\#$ and $\infty^\#$ points at hyperinfinity. If $S$ is a nonempty set of hyperreals, we write $\sup S = \infty^\#$ to indicate that $S$ is unbounded above, and $\inf S = -\infty^\#$ to indicate that $S$ is unbounded below.

## 13.2. #-Open and #-Closed Sets on $*\mathbb{R}_c^\#$.

**Definition 13.15.** If $a$ and $b$ are in the extended hyperreals and $a < b$, then the open interval $(a,b)$ is defined by $(a,b) \triangleq \{x|a < x < b\}$. :

The open intervals $(a, +\infty^\#)$ and $(-\infty^\#, b)$ are semi-hyper infinite if $a$ and $b$ are finite or hyperfinite, and $(-\infty^\#, \infty^\#)$ is the entire hyperreal line.

If $-\infty^\# < a < b < \infty^\#$, the set $[a,b] \triangleq \{x|a \le x \le b\}$ is #-closed, since its complement is the union of the #-open sets $(-\infty^\#, a)$ and $(b, \infty^\#)$. We say that $[a,b]$ is a #-closed interval. Semi-hyper infinite #-closed intervals are sets of the form $[a, \infty) = \{x|a \le x\}$ and $(-\infty^\#, a] = \{x|x \le a\}$, where $a$ is finite or hyperfinite. They are #-closed sets, since their complements are the #-open intervals $(-\infty^\#, a)$ and $(a, \infty^\#)$, respectively.

**Definition 13.16.** If $x_0 \in \mathbb{R}_c^\#$ is a hyperreal number and $\varepsilon > 0, \varepsilon \approx 0$ then the open interval $(x_0 - \varepsilon, x_0 + \varepsilon)$ is an #-neighborhood of $x_0$. If a set $S \subset *\mathbb{R}_c^\#$ contains an #-neighborhood of $x_0$, then $S$ is a #-neighborhood of $x_0$, and $x_0$ is an #-interior point of $S$.

The set of #-interior points of $S$ is the #-interior of $S$, denoted by $\#\mathrm{Int}(S)$.

(i) If every point of $S$ is an #-interior point (that is, $S = \#\mathrm{Int}(S)$), then $S$ is #-open.

(ii) A set $S$ is #-closed if $S^c = *\mathbb{R}_c^\# \setminus S$ is #-open.

**Example 13.1.** An open interval $(a,b)$ is an #-open set, because if $x_0 \in (a,b)$ and $\varepsilon \le \min\{x_0 - a; b - x_0\}$, then $(x_0 - \varepsilon, x_0 + \varepsilon) \subset (a,b)$

**Remark 13.4.** The entire hyperline $*\hat{\mathbb{R}}_c^\# = (-\infty^\#, \infty^\#)$ is #-open, and therefore $\emptyset$ is #-closed.

However, $\emptyset$ is also #-open, for to deny this is to say that $\emptyset$ contains a point that is not an #-interior point, which is absurd because $\emptyset$ contains no points. Since $\emptyset$ is #-open, $*\hat{\mathbb{R}}_c^\#$ is #-closed. Thus, $*\hat{\mathbb{R}}_c^\#$ and $\emptyset$ are both #-open and #-closed.

**Remark 13.5.** They are not the only subsets of $*\hat{\mathbb{R}}_c^\#$ with this property.

**Definition 13.17.** A deleted #-neighborhood of a point $x_0$ is a set that contains every point

of some #-neighborhood of $x_0$ except for $x_0$ itself. For example, $S = \{x|0 < |x - x_0| < \varepsilon\}$, where $\varepsilon \approx 0$, is a deleted #-neighborhood of $x_0$. We also say that it is a deleted $\varepsilon$-#-neighborhood of $x_0$.

**Theorem 13.18.** (a) The union of #-open sets is #-open:

(b) The #-intersection of #-closed sets is #-closed:

These statements apply to arbitrary collections, hyperfinite or hyperinfinite, of #-open and #-closed sets.

**Proof** (a) Let $L$ be a collection of #-open sets and $S = \cup \{G|G \in L\}$.
If S is #-closed and hyper bounded, then inf(S) and sup(S) are both in S.

**Proposition 13.1.** If S is #-closed and hyper bounded, then inf(S) and sup(S) are both in S.

## 13.3. #-Open Coverings

**Definition 13.19.** A collection $H$ of #-open sets of $\mathbb{R}_c^\#$ is an #-open covering of a set $S$ if every point in $S$ is contained in a set $H$ belonging to $H$; that is, if $S \subset \cup\{F|F \in H\}$.

**Definition 13.20.** A set $S \subset \mathbb{R}_c^\#$ is called #-compact (or hyper compact) if each of its #-open covers has a hyperfinite subcover. .

**Theorem 13.20. (Generalized Heine–Borel Theorem)** If $H$ is an #-open covering of a #-closed and hyper bounded subset $S$ of the hyperreal line $*\mathbb{R}_c^\#$ (or of the $*\mathbb{R}_c^{\#n}$, $n \in \mathbb{N}^\#$) then $S$ has an #-open covering $\tilde{H}$ consisting of hyper finite many #-open sets belonging to $H$.

**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 an 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 $S = \{x_k\}_{1\le k \le n}, n \in \mathbb{N}^\#$.
If $x_0 \in S$, then $x_0 \in G_0$ for some $G_0$ in $L$, and since $G_0$ is #-open, it contains some $\varepsilon$-#-neighborhood of $x_0$. Since $G_0 \subset S$, this $\varepsilon$-#-neighborhood is in $S$, which is consequently a #-neighborhood of $x_0$. Thus, $S$ is a #-neighborhood of each of its points, and therefore #-open, by definition.

(b) Let $F$ be a collection of #-closed sets and $T = \cap \{H | H \in F\}$. Then $T^c = \cup \{H^c | H \in F\}$ and, since each $H^c$ is #-open, $T^c$ is #-open, from (a). Therefore, $T$ is #-closed, by definition.

**Example 13.2.** If $-\infty^# < a < b < \infty^#$, the set $[a,b] = \{x|a \le x \le b\}$ is #-closed, since its complement is the union of the #-open sets $(-\infty^#a)$ and $(b, \infty^#)$. We say that $[a,b]$ is a #-closed interval. The set $[a,b) = \{x|a \le x < b\}$ is a half-#-closed or half-#-open interval if $-\infty^# < a < b < \infty^#$, as is $(a,b] = \{x|a < x \le b\}$ however, neither of these sets is #-open or #-closed. Semi-infinite #-closed intervals are sets of the form $[a, \infty^#) = \{x|a \le x\}$ and $(-\infty^#, a] = \{x|x \le a\}$, where $a$ is hyperfinite. They are #-closed sets, since their complements are the #-open intervals $(-\infty^#, a)$ and $(a, \infty^#)$, respectively.

**Definition 13.18.** Let $S$ be a subset of $\mathbb{R}_c^# = (-\infty^#, \infty^#)$. Then

(a) $x_0$ is a #-limit point of $S$ if every deleted #-neighborhood of $x_0$ contains a point of $S$.

(b) $x_0$ is a boundary point of $S$ if every #-neighborhood of $x_0$ contains at least one point in $S$ and one not in $S$. The set of #-boundary points of $S$ is the #-boundary of $S$, denoted

by $\#-\partial S$. The #-closure of $S$, denoted by $\#-\bar{S}$, is $S \cup \#-\partial S$.

(c) $x_0$ is an #-isolated point of $S$ if $x_0 \in S$ and there is a #-neighborhood of $x_0$ that contains

no other point of $S$.

(d) $x_0$ is #-exterior to $S$ if $x_0$ is in the #-interior of $S^c$. The collection of such points is the #-exterior of $S$.

**Theorem 13.19.** A set $S$ is #-closed if and only if no point of $S^c$ is a #-limit point of $S$.

**Proof.** Suppose that $S$ is #-closed and $x_0 \in S^c$. Since $S^c$ is #-open, there is a #-neighborhood of $x_0$ that is contained in $S^c$ and therefore contains no points of $S$.

Hence, $x_0$ cannot be a #-limit point of $S$. For the converse, if no point of $S^c$ is a #-limit point of $S$ then every point in $S^c$ must have a #-neighborhood contained in $S^c$. Therefore, $S^c$ is #-open and $S$ is #-closed.

**Corollary 13.1.** A set $S$ is #-closed if and only if it contains all its #-limit points.

If $S$ is #-closed and hyper bounded, then $\inf(S)$ and $\sup(S)$ are both in $S$.

**Proposition 13.1.** If $S$ is #-closed and hyper bounded, then $\inf(S)$ and $\sup(S)$ are both in $S$.

## 13.3. #-Open Coverings

**Definition 13.19.** A collection $H$ of #-open sets of $\mathbb{R}_c^#$ is an #-open covering of a set $S$ if every point in $S$ is contained in a set $H$ belonging to $H$; that is, if $S \subset \cup \{F | F \in H\}$.

**Definition 13.20.** A set $S \subset \mathbb{R}_c^#$ is called #-compact (or hyper compact) if each of its #-open covers has a hyperfinite subcover.

**Theorem 13.20. (Generalized Heine-Borel Theorem)** If $H$ is an #-open covering of a #-closed and hyper bounded subset $S$ of the hyperreal line $*\mathbb{R}_c^#$ (or of the $*\mathbb{R}_c^{\#n}$, $n \in \mathbb{N}^#$) then $S$ has an #-open covering $\tilde{H}$ consisting of hyper finite many #-open sets belonging to $H$.
## 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 $\mathbb{R}^\#$, $x \times (y + z) = (x \times y) + (x \times z)$.

(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. $(\mathbb{R}^\#, \leq)$ 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

$\mathbb{Q}^\# \subset \mathbb{R}^\#$ 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