**Theorem 14.2.5.** Let S be a set, with an equivalence relation on pairs of elements.
For $s \in S$, denote by $[s]$ the set of all elements in S that are related to s. Then for
any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.

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

**Corollary 14.2.1.** If S is a set with an equivalence relation on pairs of elements, then the equivalence classes are non-empty disjoint sets whose union is all of S.

**Definition 14.2.23.** Let {$\Psi_{1,n}$}^*_{n=0} and {$\Psi_{2,n}$}^*_{n=0} be in $\widetilde{\Delta_{\omega}^{+10}}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_{\#} 0_{*\mathbb{R}^{\#}}$ as $n \to *\infty$, i.e. if the hyper infinite sequence {$|\Psi_{1,n} - \Psi_{2,n}|$}^*_{n=0} \#\text{-tends to } 0_{*\mathbb{R}^{\#}}$.

**Proposition 14.2.4.** Definition 4.2.23 yields an equivalence relation on

$$ \Xi = \overline{\{\Delta_{\omega}^{+10}, \{x_n\}_{n=0}^{\infty}, \{y_n\}_{n=0}^{\infty}\}}. $$

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

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

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

But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.2.20, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_{\#} 0_{*\mathbb{R}^\#}$ as well. Hence, {$\Psi_{2,n}$}^*_{n=0} is related to {$\Psi_{1,n}$}^*_{n=0}.

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

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

So, we really have equivalence relation, and so by Corollary 14.2.1, the set $\Xi$ is partitioned into disjoint subsets (equivalence classes).

**Definition 14.2.24.** The hyperreal numbers $^{*\mathbb{R}^\#}$ are the equivalence classes $[\{\Psi_{1,n}\}^*_{n=0}]$ of Cauchy hyper infinite sequences of, as per Definition 14.2.23.

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

**Definition 14.2.25.** Let $s, t \in ^{*\mathbb{R}^\#}$, so there are Cauchy hyper infinite sequences {$\Psi_n$}^*_{n=0} and {$\Phi_n$}^*_{n=0} with $s = [\{\Psi_n\}^*_{n=0}]$ and $t = [\{\Phi_n\}^*_{n=0}]$.

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

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

**Proposition 14.2.5.** The operations $+, \times$ in Definition 14.2.25 (a),(b) are well-defined.

**Proof.** Suppose that $[\{\Psi_n\}^*_{n=0}] = [\{\Psi_{1,n}\}^*_{n=0}]$ and $[\{\Phi_n\}^*_{n=0}] = [\{\Phi_{1,n}\}^*_{n=0}]$. Thus means that $\Psi_n - \Psi_{1,n} \to_{\#} 0_{*\mathbb{R}^\#}$ and $\Phi_n - \Phi_{1,n} \to_{\#} 0_{*\mathbb{R}^\#}$. Then
**Definition 14.2.20.** Let {$\Psi_n$}$_{n=0}^{*\infty}$ be in $\Delta_{\omega}^{+\downarrow 0}$, i.e. for all $n \in *\mathbb{N}$, $\Psi_n \in \Delta_{\omega}^{+\downarrow 0}$. Say {$\Psi_n$}$_{n=0}^{*\infty}$ #-tends to $0_{*\mathbb{R}_c^#}$ as $n \to *\infty$ iff for any given $\varepsilon > 0_{*\mathbb{R}_c^#}$, $\varepsilon \approx 0_{*\mathbb{R}_c^#}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.2.21.** Let {$\Psi_n$}$_{n=0}^{*\infty}$ be a hyper infinite sequence such that for all $n \in *\mathbb{N}$, $\Psi_n \in \Delta_{\omega}^{+\downarrow 0}$. We call {$\Psi_n$}$_{n=0}^{*\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{*\mathbb{R}_c^#}$. To be precise: given any $\varepsilon > 0_{*\mathbb{R}_c^#}$, $\varepsilon \approx 0_{*\mathbb{R}_c^#}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_m - \Psi_n| < \varepsilon$.

**Theorem 14.2.3.** Let {$\Psi_n$}$_{n=0}^{*\infty}$ be in $\widetilde{\Delta_{\omega}^{+\downarrow 0}}$. If {$\Psi_n$}$_{n=0}^{*\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_* \Phi$ as $n \to *\infty$ for some $\Phi \in \widetilde{\Delta_{\omega}^{+\downarrow 0}}$), then {$\Psi_n$}$_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence.

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

$$|\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \varepsilon/2 + \varepsilon/2 = \varepsilon.$$ 
This shows that {$\Psi_n$}$_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence.

**Theorem 14.2.4.** If {$\Psi_n$}$_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence, then it is bounded in $*\mathbb{R}_c^#$; that is, there is some number $M \in *\mathbb{R}_c^#$ such that $|\{\Psi_n\}_{n=0}^{*\infty}| \le M$ for all $n \in *\mathbb{N}$.

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

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

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

**Definition 14.2.22.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

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

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

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

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

**Theorem 14.2.5.** Let $S$ be a set, with an equivalence relation on pairs of elements. For $s \in S$, denote by $[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.

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

**Corollary 14.2.1.** If $S$ is a set with an equivalence relation on pairs of elements, then
($\Psi_n + \Phi_n$) - ($\Psi_{1,n} + \Phi_{1,n}$) = ($\Psi_n - \Psi_{1,n}$) + ($\Phi_n - \Phi_{1,n}$). Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to $0_{\mathbb{R}_c^\#}$, and so

$$[(\Psi_n + \Phi_n)] = [(\Psi_{1,n} + \Phi_{1,n})].$$

Multiplication is a little trickier; this is where we will use Theorem 14.2.4. We will also use another ubiquitous technique: adding $0_{\mathbb{R}_c^\#}$ in the form of $s - s$. Again, suppose that

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

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

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

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

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

Now, from Theorem 14.2.4, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in \mathbb{N}$. Taking some number $R$ (for example $R = M + L$) which is bigger than both, we have

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

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

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

**Theorem 14.2.6.** Given any hyperreal number $s \in \mathbb{R}_c^\#$, $s \ne 0_{\mathbb{R}_c^\#}$, there is a hyperreal number $t \in \mathbb{R}_c^\#$ such that $s \times t = 1_{\mathbb{R}_c^\#}$.

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

$$0_{\mathbb{R}_c^\#} = (0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, \dots). \qquad (14.2.12)$$

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

From this, we are to deduce the existence of a hyperreal number $t = [\{\Phi_n\}_{n=0}^{*\infty}]$ such that $s \times t = [\{\Psi_n \times \Phi_n\}_{n=0}^{*\infty}]$ is the same equivalence class as $1_{\mathbb{R}_c^\#} = [(1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, \dots)]$. Doing so is actually an easy

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

$$\Psi_n \ne 0_{\mathbb{R}_c^\#}.$$

That is,

**Lemma 14.2.1.** If $\{\Psi_n\}_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence which does not #-tends to $0_{\mathbb{R}_c^\#}$, then there is an $N \in \mathbb{N}$ such that, for $n > N$, $\Psi_n \ne 0_{\mathbb{R}_c^\#}$.

We will now use it to complete the proof of Theorem 14.2.6.

Let $N \in \mathbb{N}^\#$ be such that $\Psi_n \ne 0_{\mathbb{R}_c^\#}$ for $n > N$. Define hyper infinite sequence $\Phi_n$ of hyperreal numbers from $\mathbb{R}_c^\#$ as follows:
the equivalence classes are non-empty disjoint sets whose union is all of S.

**Definition 14.2.23.** Let {$\Psi_{1,n}$}$_{n=0}^{*\infty}$ and {$\Psi_{2,n}$}$_{n=0}^{*\infty}$ be in $\widetilde{\Delta}_{\omega}^{+|0|}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_* 0_{*\mathbb{R}^c}$ as $n \to *\infty$, i.e. if the hyper infinite sequence {$|\Psi_{1,n} - \Psi_{2,n}|$}$_{n=0}^{*\infty}$ #-tends to $0_{*\mathbb{R}^c}$.

**Proposition 14.2.4.** Definition 4.2.23 yields an equivalence relation on

$$ \Xi = \overline{\{\Delta_{\omega}^{+|0|}, \{x_n\}_{n=0}^{*\infty}, \{y_n\}_{n=0}^{*\infty}\}}. $$

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

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

* Symmetric: Suppose {$\Psi_{1,n}$}$_{n=0}^{*\infty}$ is related to {$\Psi_{2,n}$}$_{n=0}^{*\infty}$, so $\Psi_{1,n} - \Psi_{2,n} \to_* 0_{*\mathbb{R}^c}$. But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.2.20, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_* 0_{*\mathbb{R}^c}$ as well. Hence, {$\Psi_{2,n}$}$_{n=0}^{*\infty}$ is related to {$\Psi_{1,n}$}$_{n=0}^{*\infty}$.

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

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

So, we really have equivalence relation, and so by Corollary 14.2.1, the set $\Xi$ is partitioned into disjoint subsets (equivalence classes).

**Definition 14.2.24.** The hyperreal numbers $^{*\mathbb{R}}^c$ are the equivalence classes $[\{\Psi_{1,n}\}^{*\infty}_{n=0}]$ of Cauchy hyper infinite sequences of, as per Definition 14.2.23. That is, each such equivalence class is a hyperreal number in $^{*\mathbb{R}}^c$.

**Definition 14.2.25.** Let $s, t \in ^{*\mathbb{R}}^c$, so there are Cauchy hyper infinite sequences {$\Psi_n$}$_{n=0}^{*\infty}$ and {$\Phi_n$}$_{n=0}^{*\infty}$ with $s = [\{\Psi_n\}^{*\infty}_{n=0}]$ and $t = [\{\Phi_n\}^{*\infty}_{n=0}]$.

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

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

**Proposition 14.2.5.** The operations $+, \times$ in Definition 14.2.25 (a),(b) are well-defined.

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

Multiplication is a little trickier; this is where we will use Theorem 14.2.4. We will also use another ubiquitous technique: adding $0_{*\mathbb{R}^c}$ in the form of $s - s$. Again,
for $n \le N$, $\Phi_n = 0_{\cdot \mathbb{R}_c^\#}$, and for $n > N$, $\Phi_n = 1/\Psi_n$:

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

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

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

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

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

$$ 1_{\cdot \mathbb{R}_c^\#} = (1_{\cdot \mathbb{R}_c^\#}, 1_{\cdot \mathbb{R}_c^\#}, 1_{\cdot \mathbb{R}_c^\#}, 1_{\cdot \mathbb{R}_c^\#}, \dots), \qquad (14.2.13) $$

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

**Definition 14.2.26.** Let $s \in \widetilde{\cdot \mathbb{R}_c^\#}$. Say that $s$ is positive if $s \ne 0_{\cdot \mathbb{R}_c^\#}$, and if $s = [\{\Psi_n\}_{n=0}^{*\infty}]$ for some Cauchy hyper infinite sequence such that for some $N$, $\Psi_n > 0_{\cdot \mathbb{R}_c^\#}$ for all $n > N$. Given two hyperreal numbers $s, t \in \widetilde{\cdot \mathbb{R}_c^\#}$, say that $s > t$ if $s - t$ is positive.

**Theorem 14.2.7.** Let $s, t \in \widetilde{\cdot \mathbb{R}_c^\#}$ be hyperreal numbers such that $s > t$, and let $r \in \widetilde{\cdot \mathbb{R}_c^\#}$. Then $s + r > t + r$.

**Proof.** Let $s = [\{\Psi_n\}_{n=0}^{*\infty}]$, $t = [\{\Phi_n\}_{n=0}^{*\infty}]$, and $r = [\{\Theta_n\}_{n=0}^{*\infty}]$. Since $s > t$, i.e. $s - t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have $\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\cdot \mathbb{R}_c^\#}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\cdot \mathbb{R}_c^\#}$ as $n \to *\infty$, by the assumption that $s - t > 0_{\cdot \mathbb{R}_c^\#}$. Thus, by Definition 14.2.26, this means that:

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

**Definition 14.2.27.** There is canonical imbeding

$$ \cdot \mathbb{R}_c^\# \hookrightarrow \widetilde{\cdot \mathbb{R}_c^\#} \qquad (14.2.14) $$

defined by

$$ a \mapsto \tilde{a} \qquad (14.2.15) $$

where $\tilde{a}$ is hyper infinite sequence $\tilde{a} = (a, a, \dots) \in \cdot \mathbb{R}_c^\#$, $a \in \cdot \mathbb{R}_c^\# \cup \Delta_\omega^{+10}$.

**Notation 14.2.5.** $\tilde{a} = (a, a, \dots) \in \widetilde{\cdot \mathbb{R}_c^\#}$, $a \in \widetilde{\cdot \mathbb{R}_c^\#}$.

**Remark 14.2.11.** Let $a \in \widetilde{\cdot \mathbb{R}_c^\#}$. We will be identity $a \in \widetilde{\cdot \mathbb{R}_c^\#}$ with any $\{a_n\}_{n=0}^{*\infty} \subset \widetilde{\cdot \mathbb{R}_c^\#}$ such that $\#\text{-lim}_{n \to *\infty} a_n = a$ and we denote by $[[a]]$ the equivalence class corresponding to $a \in \widetilde{\cdot \mathbb{R}_c^\#}$.

**Definition 14.2.28.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\widetilde{\cdot \mathbb{R}_c^\#}$, $\{a_n\}_{n=0}^k \subset \widetilde{\cdot \mathbb{R}_c^\#}$. We define external hyper infinite sequence $\{\widetilde{a_n}\}_{n=0}^k \subset \widetilde{\cdot \mathbb{R}_c^\#}$ by
suppose that

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

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

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

Hence, we have $|\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n}| \le |\Phi_n| \times |\Psi_n - \Psi_{1,n}| + |\Psi_{1,n}| \times |\Phi_n - \Phi_{1,n}|$. Now, from Theorem 14.2.4, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in \mathbb{N}$. Taking some number $R$ (for example $R = M + L$) which is bigger than both, we have

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

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

$$\Psi_n \times \Phi_n - \Psi_{1,n} \times \Phi_{1,n} \to \mathcal{O}_{\mathbb{R}_c^\#}$$

**Theorem 14.2.6.** Given any hyperreal number $s \in \mathbb{R}_c^\#$, $s \ne \mathcal{O}_{\mathbb{R}_c^\#}$, there is a hyperreal number $t \in \mathbb{R}_c^\#$ such that $s \times t = 1_{\mathbb{R}_c^\#}$.

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

$$0_{\mathbb{R}_c^\#} = (0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, \dots). \qquad (14.2.12)$$

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

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

That is,

**Lemma 14.2.1.** If $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence which does not #-tends to $\mathcal{O}_{\mathbb{R}_c^\#}$, then there is an $N \in \mathbb{N}$ such that, for $n > N$, $\Psi_n \ne \mathcal{O}_{\mathbb{R}_c^\#}$.

We will now use it to complete the proof of Theorem 14.2.6.

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

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

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

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

Then $\Psi_n \times \Phi_n$ is equal to $\Psi_n \times \mathcal{O}_{\mathbb{R}_c^\#} = \mathcal{O}_{\mathbb{R}_c^\#}$ for $n \le N$, and equals
$$ \begin{aligned} \{A_n; k\}_{n=0}^{*\infty} &= \overbrace{\{a_n\}_{n=0}^k}^{\overbrace{\{a_n\}_{n=0}^k}^{*\overline{\mathbb{R}}_c^\#}} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, \widehat{a_k}) \in [[a_k]]. \end{aligned} \quad (14.2.16) $$

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

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

$$ \begin{aligned} \{A'_n; \infty\}_{n=0}^{*\infty} &= \overbrace{\{a_n\}_{n=0}^\infty}^{\overbrace{\{a_n\}_{n=0}^\infty}^{*\overline{\mathbb{R}}_c^\#}} = \\ &= (a_0, a_1, \dots, a_k, \dots, \overbrace{\{a_n\}_{n=0}^\infty}^{*\overline{\mathbb{R}}_c^\#}) \in [[[\{a_n\}_{n=0}^\infty]]]. \end{aligned} \quad (14.2.17) $$

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

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

$$ \begin{aligned} \{A_n; N\}_{n=0}^{*\infty} &= \overbrace{\{a_n\}_{n=0}^N}^{\overbrace{\{a_n\}_{n=0}^N}^{*\overline{\mathbb{R}}_c^\#}} = \\ &= (a_0, a_1, \dots, a_n, \dots, a_{N-1}, \widehat{a_N}) \in [[a_N]]. \end{aligned} \quad (14.2.18) $$

**Definition 14.2.29.** (i) Let $\{a_n\}_{n=0}^k, k \in N$ be finite sequence in $^{*\overline{\mathbb{R}}_c^\#}, \{a_n\}_{n=0}^N \subset ^{*\overline{\mathbb{R}}_c^\#}$.

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

$$ Ext-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k}^{\overbrace{\{c_n\}_{n=0}^k}^{*\overline{\mathbb{R}}_c^\#}} = (c_0, c_1, \dots, c_k, \dots, \widehat{c_k}) \in [[c_k]] \quad (14.2.19) $$

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

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

We define external countable sum $Ext-\sum_{n=0}^{n=\infty} a_n$ by

$$ \begin{aligned} Ext-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}_{n=0}^\infty}^{\overbrace{\{c_n\}_{n=0}^\infty}^{*\overline{\mathbb{R}}_c^\#}} = \\ &= (c_0, c_1, \dots, c_k, \dots, \overbrace{\{c_n\}_{n=0}^\infty, \{\overbrace{c_n\}_{n=0}^\infty\}}^{*\overline{\mathbb{R}}_c^\#}) \in [[[\{c_n\}_{n=0}^\infty]]] \end{aligned} \quad (14.2.20) $$

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

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

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

$$ Ext-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{N}}^{\overbrace{\{c_n\}_{n=0}^{N}}^{*\overline{\mathbb{R}}_c^\#}} = (c_0, c_1, \dots, c_k, \dots, c_N, \widehat{c_N}) \quad (14.2.21) $$

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

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

$$ Ext-\sum_{n=0}^{n=N} a_n = Ext-\sum_{n=0}^{n=\infty} a_n, \quad (14.2.22) $$

**Example 14.2.1.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *N, \alpha \in ^{*\overline{\mathbb{R}}_c^\#}$,

$r \in ^{*\overline{\mathbb{R}}_c^\#}$ be the first term and the ratio of the G.P respectively. Then for any
$$ \Psi_n \times \Phi_n = \Psi_n \times 1_{\cdot \mathbb{R}^\#} / \Psi_n = 1_{\cdot \mathbb{R}^\#} \text{ for } n > N $$

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

$$ 1_{\cdot \mathbb{R}^\#} = (1_{\cdot \mathbb{R}^\#}, 1_{\cdot \mathbb{R}^\#}, 1_{\cdot \mathbb{R}^\#}, 1_{\cdot \mathbb{R}^\#}, \dots), \qquad (14.2.13) $$

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

**Definition 14.2.26.** Let $s \in *\mathbb{R}_c^\#$. Say that $s$ is positive if $s \neq 0_{\cdot \mathbb{R}_c^\#}$, and if
$s = [\{\Psi_n\}_{n=0}^{*\infty}]$ for some Cauchy hyper infinite sequence such that for some $N$,
$\Psi_n > 0_{\cdot \mathbb{R}_c^\#}$ for all $n > N$. Given two hyperreal numbers $s, t \in *\mathbb{R}_c^\#$, say that $s > t$ if
$s - t$ is positive.

**Theorem 14.2.7.** 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 $s = [\{\Psi_n\}_{n=0}^{*\infty}]$, $t = [\{\Phi_n\}_{n=0}^{*\infty}]$, and $r = [\{\Theta_n\}_{n=0}^{*\infty}]$. Since $s > t$, i.e. $s - t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have $\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\cdot \mathbb{R}^\#}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\cdot \mathbb{R}^\#}$ as $n \to *\infty$, by the assumption that $s - t > 0_{\cdot \mathbb{R}_c^\#}$. Thus, by Definition 14.2.26, this means that:
$$ s + r = [\{\Psi_n + \Theta_n\}_{n=0}^{*\infty}] > [\{\Phi_n + \Theta_n\}_{n=0}^{*\infty}] = t + r. $$

**Definition 14.2.27.** There is canonical imbeding

$$ *\mathbb{R}_c^\# \hookrightarrow *\widetilde{\mathbb{R}}_c^\# \qquad (14.2.14) $$

defined by

$$ a \mapsto \tilde{a} \qquad (14.2.15) $$

where $\tilde{a}$ is hyper infinite sequence $\tilde{a} = (a, a, \dots) \in *\mathbb{R}_c^\#$, $a \in *\mathbb{R}_c^\# \cup \Delta_\omega^{+10}$.

**Notation 14.2.5.** $\hat{a} = (a, a, \dots) \in *\widetilde{\mathbb{R}}_c^\#$, $a \in *\widetilde{\mathbb{R}}_c^\#$.

**Remark 14.2.11.** Let $a \in *\widetilde{\mathbb{R}}_c^\#$. We will be identity $a \in *\widetilde{\mathbb{R}}_c^\#$ with any $\{a_n\}_{n=0}^{*\infty} \subset *\widetilde{\mathbb{R}}_c^\#$ such that $\#\text{-lim}_{n \to *\infty} a_n = a$ and we denote by $[[a]]$ the equivalence class corresponding to $a \in *\widetilde{\mathbb{R}}_c^\#$.

**Definition 14.2.28.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $*\widetilde{\mathbb{R}}_c^\#$, $\{a_n\}_{n=0}^k \subset *\widetilde{\mathbb{R}}_c^\#$.
We define external hyper infinite sequence $\{\overbrace{a_n}^k\}_{n=0} \subset *\widetilde{\mathbb{R}}_c^\#$ by

$$ \begin{aligned} \{A_n; k\}_{n=0}^{*\infty} &= \overbrace{\{a_n\}_{n=0}^k} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, \hat{a}_k) \in [[a_k]]. \end{aligned} \qquad (14.2.16) $$

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

We define hyper infinite sequence $\{A_n\}_{n=0}^{*\infty} = \overbrace{\{a_n\}_{n=0}^\infty} \subset *\widetilde{\mathbb{R}}_c^\#$ by
$N \in *\mathbb{N}$ by Proposition 14.2.6 and Definition 14.2.29 one obtains that

$$ \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#} - r^N}{1_{\widehat{\mathbb{R}}_c^\#} - r} = \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#}}{1_{\widehat{\mathbb{R}}_c^\#} - r} - \alpha \frac{1^N}{1_{\widehat{\mathbb{R}}_c^\#} - r}. \quad (14.2.23) $$

and

$$ \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#}}{1_{\widehat{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\widehat{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty}. \quad (14.2.24) $$

**Example 14.2.2.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}, \alpha \in *\mathbb{R}_c^\#, r \in *\mathbb{R}_c^\#,$
$r > 0, r \neq 1$. Note that

$$ \begin{aligned} \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#} - r^N}{1_{\widehat{\mathbb{R}}_c^\#} - r} &= \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \\ &= \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \\ &= \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#}}{1_{\widehat{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\widehat{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \end{aligned} \quad (14.2.25) $$

From (14.2.25) we obtain

$$ \begin{aligned} \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#} - r^N}{1_{\widehat{\mathbb{R}}_c^\#} - r} - \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#}}{1_{\widehat{\mathbb{R}}_c^\#} - r} + \alpha \left\{ \frac{r^n}{1_{\widehat{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} = \\ & \qquad \alpha \left\{ \frac{r^n}{1_{\widehat{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} - \alpha \frac{r^N}{1_{\widehat{\mathbb{R}}_c^\#} - r}. \end{aligned} \quad (14.2.26) $$

Assume that: (i) $r < 1_{\widehat{\mathbb{R}}_c^\#}$, then from (14.2.26) we obtain

$$ \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} > 0_{\widehat{\mathbb{R}}_c^\#}. \quad (14.2.27) $$

(ii) $r > 1_{\widehat{\mathbb{R}}_c^\#}$, then from (14.2.26) we obtain

$$ \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \alpha \left\{ \frac{r^n}{1_{\widehat{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} + \alpha \frac{r^N}{r - 1_{\widehat{\mathbb{R}}_c^\#}} > 0_{\widehat{\mathbb{R}}_c^\#}. \quad (14.2.28) $$

**Proposition 14.2.6.** (i) Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}$. Let $S_N$, $\alpha \in *\mathbb{R}_c^\#$, $r \in *\mathbb{R}_c^\#$ be the sum of $N$ terms, first term and the ratio of the G.P respectively. Then for any $N \in *\mathbb{N}$ the statement $\Phi_N$ holds

$$ \Phi_N \Leftrightarrow_s \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\widehat{\mathbb{R}}_c^\#} - r^N}{1_{\widehat{\mathbb{R}}_c^\#} - r}. \quad (14.2.29) $$

**Proof.** (i) Directly by hyperinfinite induction. Note that $\Phi_N \Rightarrow_s \Phi_{N+1}$:
$$
\begin{align}
\{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}}^{\sim} = \nonumber \\
&= (a_0, a_1, \dots, a_k, \dots, \overbrace{\{a_n\}_{n=0}^{\infty}}^{\sim}) \in [[\{a_n\}_{n=0}^{\infty}]]. \tag{14.2.17}
\end{align}
$$

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

$$
\begin{align}
\{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{N}}^{\sim} = \nonumber \\
&= (a_0, a_1, \dots, a_N, \dots, \overline{a_N}) \in [[a_N]]. \tag{14.2.18}
\end{align}
$$

**Definition 14.2.29.** (i) Let {$a_n$}$_{n=0}^k$, $k \in \mathbb{N}$ be finite sequence in $\widetilde{\mathbb{R}}_c^\# : \{a_n\}_{n=0}^N \subset \widetilde{\mathbb{R}}_c^\#$.
We define external finite sum $\mathrm{Ext}-\sum_{n=0}^{n=k} a_n$ by

$$
\mathrm{Ext}-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^{k}}^{\sim} = (c_0, c_1, \dots, c_k, \dots, \hat{c}_k) \in [[c_k]] \quad (14.2.19)
$$

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

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

$$
\begin{align}
\mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}_{n=0}^{\infty}}^{\sim} = \nonumber \\
&= \left( c_0, c_1, \dots, c_k, \dots, \overbrace{\{c_n\}_{n=0}^{\infty}}^{\sim}, \overbrace{\{c_n\}_{n=0}^{\infty}}^{\sim} \right) \in [\{c_n\}_{n=0}^{\infty}] \tag{14.2.20}
\end{align}
$$

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

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

$$
\mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{n=N}}^{\sim} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \quad (14.2.21)
$$

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

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

$$
\mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n, \qquad (14.2.22)
$$

**Example 14.2.1.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}, \alpha \in *\mathbb{R}_c^\#$,
$r \in *\mathbb{R}_c^\#$ be the first term and the ratio of the G.P respectively. Then for any
$N \in *\mathbb{N}$ by Proposition 14.2.6 and Definition 14.2.29 one obtains that

$$
\mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{\overbrace{1-r^N}^{*\mathbb{R}_c^\#}}{\overbrace{1-r}^{*\mathbb{R}_c^\#}} = \alpha \frac{\overbrace{1-r^N}^{*\mathbb{R}_c^\#}}{\overbrace{1-r}^{*\mathbb{R}_c^\#}} - \alpha \frac{\overbrace{r^N}^{*\mathbb{R}_c^\#}}{\overbrace{1-r}^{*\mathbb{R}_c^\#}}. \quad (14.2.23)
$$

and
$$
\begin{align}
S_{N+1} &= \operatorname{Ext}-\sum_{n=1}^{n=N} \alpha r^{n-1} = \operatorname{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} + \alpha r^N = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha r^N = \nonumber \\
&= \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha \frac{(1_{\mathbb{R}_c^\#} - r)r^N}{1_{\mathbb{R}_c^\#} - r} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N + r^N - r^{N+1}}{1_{\mathbb{R}_c^\#} - r} = \tag{14.2.30} \\
&= \alpha \frac{1_{\mathbb{R}_c^\#} - r^{N+1}}{1_{\mathbb{R}_c^\#} - r}. \nonumber
\end{align}
$$

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

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

$$
\tilde{\Phi}_N \iff_s \operatorname{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#} - r^N}{1_{\tilde{\mathbb{R}}_c^\#} - r}. \quad (14.2.31)
$$

Notice that (i)⇒(ii) by definitions.

**Definition 14.2.30.** Let {$a_n$}$_{n=0}^{\infty}$, $n \in *\mathbb{N}$ be external hyperinfinite sequence in *$\tilde{\mathbb{R}}_c^\#$* :
{$a_n$}$_{n=0}^{\infty} \subset *\tilde{\mathbb{R}}_c^\#$. We define external hyperinfinite sum $\operatorname{Ext}-\sum_{n=0}^{\infty} a_n$ by

$$
\operatorname{Ext}-\sum_{n=0}^{\infty} a_n = \#\operatorname{lim}_{N \to \infty} \left( \operatorname{Ext}-\sum_{n=0}^{N} a_n \right) \quad (14.2.32)
$$

if #-limit in (14.2.31) exists.

**Example 14.2.3.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{n-1}, n \in *\mathbb{N}^\#, \alpha \in *\tilde{\mathbb{R}}_c^\#, r \in *\tilde{\mathbb{R}}_c^\#$. From (14.2.27) we obtain

$$
\begin{align}
\operatorname{Ext}-\sum_{n=0}^{\infty} \alpha r^{n-1} &= \#\operatorname{lim}_{N \to \infty} \left( \operatorname{Ext}-\sum_{n=0}^{N} \alpha r^{n-1} \right) = \#\operatorname{lim}_{N \to \infty} \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#} - r^N}{1_{\tilde{\mathbb{R}}_c^\#} - r} = \tag{14.2.33} \\
&= \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} \nonumber
\end{align}
$$

since $\#\lim_{N \to \infty} r^N = 0_{\tilde{\mathbb{R}}_c^\#}$ if $|r| < 1$. From (14.2.33) and (14.2.25) we obtain

$$
\begin{align}
\alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} &= \operatorname{Ext}-\sum_{n=0}^{\infty} \alpha r^{n-1} = \operatorname{Ext}-\sum_{n=0}^{\infty} \alpha r^{n-1} + \operatorname{Ext}-\sum_{n \in *\mathbb{N}} \alpha r^{n-1} = \nonumber \\
&\quad \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} + \operatorname{Ext}-\sum_{n \in *\mathbb{N}} \alpha r^{n-1}. \tag{14.2.34}
\end{align}
$$

From (14.2.34) we obtain
$$ \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{1}{1_{\mathbb{R}_c^{\#}} - r} - \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^{\#}} - r} \right\}_{n=1}^{\infty} \quad (14.2.24) $$

**Example 14.2.2.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}$, $\alpha \in *\mathbb{R}_c^\#$, $r \in *\mathbb{R}_c^\#$, $r > 0, r \neq 1$. Note that

$$ \begin{aligned} \alpha \frac{1}{1_{\mathbb{R}_c^\#} - r} &= \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \\ &= \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \\ &= \alpha \frac{1}{1_{\mathbb{R}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \end{aligned} \quad (14.2.25) $$

From (14.2.25) we obtain

$$ \begin{aligned} \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{1}{1_{\mathbb{R}_c^\#} - r} - \alpha \frac{1}{1_{\mathbb{R}_c^\#} - r} + \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} = \\ & \qquad \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} - \alpha \frac{r^N}{1_{\mathbb{R}_c^\#} - r}. \end{aligned} \quad (14.2.26) $$

Assume that: (i) $r < 1_{\mathbb{R}_c^\#}$, then from (14.2.26) we obtain

$$ \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} > 0_{\mathbb{R}_c^\#} \quad (14.2.27) $$

(ii) $r > 1_{\mathbb{R}_c^\#}$, then from (14.2.26) we obtain

$$ \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \alpha \left\{ \frac{r^n}{1_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} + \alpha \frac{r^N}{r - 1_{\mathbb{R}_c^\#}} > 0_{\mathbb{R}_c^\#} \quad (14.2.28) $$

**Proposition 14.2.6.** (i) Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}$. Let $S_N$, $\alpha \in *\mathbb{R}_c^\#$, $r \in *\mathbb{R}_c^\#$ be the sum of $N$ terms, first term and the ratio of the G.P respectively. Then for any $N \in *\mathbb{N}$ the statement $\Phi_N$ holds

$$ \Phi_N \Leftrightarrow_s \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} \quad (14.2.29) $$

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

$$ \begin{aligned} S_{N+1} &= \mathrm{Ext}-\sum_{n=1}^{n=N} \alpha r^{n-1} = \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} + \alpha r^N = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha r^N = \\ &= \alpha \frac{1_{\mathbb{R}_c^\#} - r^N}{1_{\mathbb{R}_c^\#} - r} + \alpha \frac{(1_{\mathbb{R}_c^\#} - r) r^N}{1_{\mathbb{R}_c^\#} - r} = \alpha \frac{1_{\mathbb{R}_c^\#} - r^N + r^N - r^{N+1}}{1_{\mathbb{R}_c^\#} - r} = \\ &= \alpha \frac{1_{\mathbb{R}_c^\#} - r^{N+1}}{1_{\mathbb{R}_c^\#} - r}. \end{aligned} \quad (14.2.30) $$
$$
\begin{align}
\operatorname{Ext}\widehat{\sum}_{n \in \mathbb{N}} \alpha r^{n-1} &= \alpha \frac{\widehat{1_{\mathbb{R}_c^\#}}}{\widehat{1_{\mathbb{R}_c^\#}} - r} \alpha - \left( \frac{\widehat{1_{\mathbb{R}_c^\#}}}{\widehat{1_{\mathbb{R}_c^\#}} - r} - \alpha \left\{ \frac{r^n}{\widehat{1_{\mathbb{R}_c^\#}} - r} \right\}_{n=1}^\infty \right) = \tag{14.2.35} \\
&= \alpha \left\{ \frac{r^n}{\widehat{1_{\mathbb{R}_c^\#}} - r} \right\}_{n=1}^\infty > 0. \nonumber
\end{align}
$$

**Definition 14.2.31.** Let {$a_n$}$_{n=0}^\infty$ be *$\mathbb{R}_c^\#$- valued countable sequence
$a : \mathbb{N} \to *\mathbb{R}_c^\#$ such that:

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

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

We define a set $-\Xi_{\omega}^{\pm,0\vee=0}$ by $\{c_n\}_{n=0}^{\infty} \in -\Xi_{\omega}^{\pm,0\vee=0} \Leftrightarrow \{-c_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm,0\vee=0}$. Note that
$\Xi_{\omega}^{\pm,0\vee=0} = -\Xi_{\omega}^{\pm,0\vee=0}$.

**Definition 14.2.31.**

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

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

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

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

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

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

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

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

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

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

$$
a_n^{1*} = \begin{cases} a_n^{-1} & \text{if } a_n \neq 0_{*\mathbb{R}_c^\#} \\ 0_{*\mathbb{R}_c^\#} & \text{if } a_n = 0_{*\mathbb{R}_c^\#} \end{cases} \tag{14.2.36}
$$

Note that

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

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

$$
\alpha_n = \begin{cases} 1_{*\mathbb{R}_c^\#} & \text{if } a_n \neq 0_{*\mathbb{R}_c^\#} \\ 0_{*\mathbb{R}_c^\#} & \text{if } a_n = 0_{*\mathbb{R}_c^\#} \end{cases} \tag{14.2.37}
$$

**Definition 14.2.32.** We say that

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

**Definition 14.2.33.** (1) Let {$a_n$}$_{n=0}^\infty \in \Xi_\omega^{\pm,0\vee=0}$ and let {$A_n$}$_{n=0}^* = \overline{\{a_n\}_{n=0}^\infty}$ be a hyper infinite sequence
Thus $S_{N+1} = \alpha \frac{1_{\mathbb{R}_c^\#} + r^{N+1}}{1_{\mathbb{R}_c^\#} - r}$ and therefore $\Phi_{N+1}$ holds.

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

$$ \tilde{\Phi}_N \Leftrightarrow_s \mathrm{Ext} - \sum_{n=1}^{n=N-1} ar^{n-1} = \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#} - r^N}{1_{\tilde{\mathbb{R}}_c^\#} - r}. \quad (14.2.31) $$

Notice that (i)⇒(ii) by definitions.

**Definition 14.2.30.** Let $\{a_n\}_{n=0}^{*\infty}$, $n \in *\mathbb{N}$ be external hyperinfinite sequence in $*\tilde{\mathbb{R}}_c^\#$:
$\{a_n\}_{n=0}^{*\infty} \subset *\tilde{\mathbb{R}}_c^\#$. We define external hyperinfinite sum $\mathrm{Ext} - \sum_{n=0}^{*\infty} a_n$ by

$$ \mathrm{Ext} - \sum_{n=0}^{*\infty} a_n = \#\lim_{N \to *\infty} \left( \mathrm{Ext} - \sum_{n=0}^{N} a_n \right) \quad (14.2.32) $$

if $\#\lim_{N \to *\infty} r^N$ exists.

**Example 14.2.3.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{n-1}$, $n \in *\mathbb{N}^\#$, $\alpha \in *\tilde{\mathbb{R}}_c^\#$, $r \in *\tilde{\mathbb{R}}_c^\#$. From (14.2.27) we obtain

$$ \begin{aligned} \mathrm{Ext} - \sum_{n=0}^{*\infty} ar^{n-1} &= \#\lim_{N \to *\infty} \left( \mathrm{Ext} - \sum_{n=0}^{N} ar^{n-1} \right) = \#\lim_{N \to *\infty} \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#} - r^N}{1_{\tilde{\mathbb{R}}_c^\#} - r} = \\ &= \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} \end{aligned} \quad (14.2.33) $$

since $\#\lim_{N \to *\infty} r^N = 0_{\tilde{\mathbb{R}}_c^\#}$ if $|r| < 1$. From (14.2.33) and (14.2.25) we obtain

$$ \begin{aligned} \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} &= \mathrm{Ext} - \sum_{n=0}^{*\infty} ar^{n-1} = \mathrm{Ext} - \sum_{n=0}^{*\infty} ar^{n-1} + \mathrm{Ext} - \sum_{n \in *\mathbb{N}^\#} ar^{n-1} = \\ &\quad \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{*\infty} + \mathrm{Ext} - \sum_{n \in *\mathbb{N}^\#} ar^{n-1}. \end{aligned} \quad (14.2.34) $$

From (14.2.34) we obtain

$$ \begin{aligned} \mathrm{Ext} - \sum_{n \in *\mathbb{N}^\#} ar^{n-1} &= \alpha \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} \alpha - \left( \frac{1_{\tilde{\mathbb{R}}_c^\#}}{1_{\tilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{*\infty} \right) = \\ &= \alpha \left\{ \frac{r^n}{1_{\tilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{*\infty} > 0. \end{aligned} \quad (14.2.35) $$

**Definition 14.2.31.** Let $\{a_n\}_{n=0}^\infty$ be $*\tilde{\mathbb{R}}_c^\#$-valued countable sequence $a: \mathbb{N} \to *\tilde{\mathbb{R}}_c^\#$$ such that:

(i) there is $M \in *\mathbb{N}$ such that for all $n > M$, $a_n \neq 0_{*\tilde{\mathbb{R}}_c^\#}$, we denote a set of the all these sequences by $\Xi_{\tilde{\omega}}^{\pm, \pm 0}$. We define a set $-\Xi_{\tilde{\omega}}^{\pm, \pm 0}$ by $\{c_n\}_{n=0}^\infty \in -\Xi_{\tilde{\omega}}^{\pm, \pm 0} \Leftrightarrow \{-c_n\}_{n=0}^\infty \in \Xi_{\tilde{\omega}}^{\pm, \pm 0}$. Note that $\Xi_{\tilde{\omega}}^{\pm, \pm 0} = -\Xi_{\tilde{\omega}}^{\pm, \pm 0}$.
$$ \{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots) \quad (14.2.38) $$

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

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

$$ \{x_n + y_n A_n\}_{n=0}^{\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} = (x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, x_k + y_k a_k, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots), \quad (14.2.39) $$

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

**Definition 14.2.34.** Let $\{A_n\}_{n=0}^{\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ and $\{B_n\}_{n=0}^{\infty} = \overline{\{b_n\}_{n=0}^{\infty}}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \neq 0}$. Then we define:

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

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

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

**Definition 14.2.35.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \neq 0}$, i.e. for all $n \in *\mathbb{N}$, $\Psi_n \in \Xi_{\omega}^{\pm, \neq 0, \neq 0}$.

Say $\{\Psi_n\}_{n=0}^{\infty}$ #-tends to $0_{*\mathbb{R}^c}$ as $n \to *\infty$ iff for any given $\varepsilon > 0_{*\mathbb{R}^c}$, $\varepsilon \approx 0_{*\mathbb{R}^c}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.2.36.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be a hyper infinite sequence such that for all $n \in *\mathbb{N}$, $\Psi_n \in \tilde{\Xi}_{\omega}^{\pm, \neq 0, \neq 0}$. We call $\{\Psi_n\}_{n=0}^{\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{*\mathbb{R}^c}$. To be precise: given any $\varepsilon > 0_{*\mathbb{R}^c}$, $\varepsilon \approx 0_{*\mathbb{R}^c}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.2.8.** Let $\{\Psi_n\}_{n=0}^{\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \neq 0}$. If $\{\Psi_n\}_{n=0}^{\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_* \Phi$ as $n \to *\infty$ for some $\Phi \in \tilde{\Xi}_{\omega}^{\pm, \neq 0, \neq 0}$), then $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

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

$$ |\Psi_n - \Psi_m| = |(\Psi_n - \Phi) - (\Psi_m - \Phi)| \le |\Psi_n - \Phi| + |\Psi_m - \Phi| < \varepsilon/2 + \varepsilon/2 = \varepsilon. $$
This shows that $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence.

**Theorem 14.2.9.** If $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence, then it is bounded in $*\mathbb{R}^c$; that is, there is some number $M \in *\mathbb{R}^c$ such that $|\{\Psi_n\}_{n=0}^{\infty}| \le M$ for all $n \in *\mathbb{N}$.

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

Then for any term $\Psi_n$, if $n \le N$, then $|\Psi_n|$ appears in the list and so $|\Psi_n| \le M$; if $n > N$,
(ii) there is countable subsequence {$a_{n_k}$}$_{k=m}^{\infty}$ $\subset$ {$a_n$}$_{n=0}^{\infty}$ such that $a_{n_k} = 0_{*\mathbb{R}_c^\#}$, $k \ge m$ and $a_n \ne 0_{*\mathbb{R}_c^\#}$ iff $a_n \notin \{a_{n_k}\}_{k=m}^{\infty}$, we denote a set of the all these countable sequences by $\Xi_{\omega}^{\pm, \ne 0, \vee=0}$.

We define a set $-\Xi_{\omega}^{\pm, \ne 0, \vee=0}$ by $\{c_n\}_{n=0}^{\infty} \in -\Xi_{\omega}^{\pm, \ne 0, \vee=0} \iff \{-c_n\}_{n=0}^{\infty} \in \Xi_{\omega}^{\pm, \ne 0, \vee=0}$. Note that
$\Xi_{\omega}^{\pm, \ne 0, \vee=0} = -\Xi_{\omega}^{\pm, \ne 0, \vee=0}$.

**Definition 14.2.31.**

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

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

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

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

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

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

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

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

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

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

$$
a_n^{1*} = \begin{cases} a_n^{-1} & \text{if } a_n \neq 0_{*\mathbb{R}_c^\#} \\ 0_{*\mathbb{R}_c^\#} & \text{if } a_n = 0_{*\mathbb{R}_c^\#} \end{cases} \tag{14.2.36}
$$

Note that

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

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

$$
\alpha_n = \begin{cases} 1_{\mathbb{R}_c^\#} & \text{if } a_n \neq 0_{\mathbb{R}_c^\#} \\ 0_{\mathbb{R}_c^\#} & \text{if } a_n = 0_{\mathbb{R}_c^\#} \end{cases} \tag{14.2.37}
$$

**Definition 14.2.32.** We say that

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

**Definition 14.2.33. (1)** Let {$a_n$}$_{n=0}^{\infty}$ $\in \Xi_{\omega}^{\pm, \ne 0, \vee=0}$ and let {$A_n$}$_{n=0}^{*\infty}$ $=$ $\overline{\{a_n\}_{n=0}^{\infty}}$ be a hyper infinite sequence

$$
\{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}} = (a_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \dots, \{a_n\}_{n=0}^{\infty}, \dots) \quad (14.2.38)
$$

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

(2) Let {x_n + y_n a_n} ∞_{n=0} ∈ Ξ_{ω}^{±, ≠ 0, ∨=0} and let

$$
\{x_n + y_n A_n\}_{n=0}^{*\infty} = \overline{\{x_n + y_n a_n\}_{n=0}^{\infty}} = \\
(x_0 + y_0 a_0, x_1 + y_1 a_1, \dots, x_k + y_k a_k, \dots, \{x_n + y_n a_n\}_{n=0}^{\infty}, \dots), \tag{14.2.39}
$$

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

**Definition 14.2.34.** Let $\{A_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty}}$ and $\{B_n\}_{n=0}^{*\infty} = \overline{\{b_n\}_{n=0}^{\infty}}$ be in $\tilde{\Xi}_{\omega}^{\pm, \ne 0, \vee=0}$.
Then we define:
then (as shown above) $|\Psi_n|$ is less than at least one of the last two entries in the list,
and so $|\Psi_n|\le M$. Hence, $M$ is a bound for the sequence.

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

**Definition 14.2.37.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

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

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

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

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

**Theorem 14.2.10.** Let $S$ be a set, with an equivalence relation on pairs of elements.
For $s \in S$, denote by $[s]$ the set of all elements in $S$ that are related to $s$. Then for
any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.

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

**Corollary 14.2.2.** If $S$ is a set with an equivalence relation on pairs of elements, then
the equivalence classes are non-empty disjoint sets whose union is all of $S$.

**Definition 14.2.38.** Let $\{\Psi_{1,n}\}_{n=0}^{*\infty}$ and $\{\Psi_{2,n}\}_{n=0}^{*\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_{\#} 0_{*\mathbb{R}_c^{\#}}$ as $n \to *\infty$, i.e. if the hyper infinite sequence $\{|\Psi_{1,n} - \Psi_{2,n}|\}_{n=0}^{*\infty}$ #-tends to $0_{*\mathbb{R}_c^{\#}}$.

**Proposition 14.2.4.** Definition 4.2.38 yields an equivalence relation on $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$.

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

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

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

But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.2.35, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_{\#} 0_{*\mathbb{R}_c^{\#}}$ as well. Hence, $\{\Psi_{2,n}\}_{n=0}^{*\infty}$ is related to $\{\Psi_{1,n}\}_{n=0}^{*\infty}$.

• **Transitive:** Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 14.2.4.
Suppose $\{\Psi_{1,n}\}_{n=0}^{*\infty}$ is related to $\{\Psi_{2,n}\}_{n=0}^{*\infty}$, and $\{\Psi_{2,n}\}_{n=0}^{*\infty}$ is related to $\{\Psi_{3,n}\}_{n=0}^{*\infty}$.
This means that $\Psi_{1,n} - \Psi_{2,n} \to_{\#} 0_{*\mathbb{R}_c^{\#}}$ and $\Psi_{2,n} - \Psi_{3,n} \to_{\#} 0_{*\mathbb{R}_c^{\#}}$.

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

So, we really have equivalence relation, and so by Corollary 14.2.2, the set $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$
is partitioned into disjoint subsets (equivalence classes).

**Definition 14.2.39.** (1) The hyperreal numbers $*\mathbb{R}_c^{\#}$ are the equivalence classes
$$ (i) \{A_n\}_{n=0}^{*\infty} + \{B_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty} + \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n + b_n\}_{n=0}^{\infty}} = \{A_n + B_n\}_{n=0}^{*\infty} \in \Xi_{\omega}^{\pm, \neq 0, \nu=0} $$

$$ (ii) \{A_n\}_{n=0}^{*\infty} - \{B_n\}_{n=0}^{*\infty} = \overline{\{a_n\}_{n=0}^{\infty} - \{b_n\}_{n=0}^{\infty}} \triangleq \overline{\{a_n - b_n\}_{n=0}^{\infty}} = \{A_n - B_n\}_{n=0}^{*\infty} \in \Xi_{\omega}^{\pm, \neq 0, \nu=0} $$

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

**Definition 14.2.35.** Let $\{\Psi_n\}_{n=0}^{*\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$, i.e. for all $n \in *\mathbb{N}$, $\Psi_n \in \Xi_{\omega}^{\pm, \neq 0, \nu=0}$. Say $\{\Psi_n\}_{n=0}^{*\infty}$ #-tends to $0_{*\mathbb{R}^c}$ as $n \to *\infty$ iff for any given $\varepsilon > 0_{*\mathbb{R}^c}$, $\varepsilon \approx 0_{*\mathbb{R}^c}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $n > N$, $|\Psi_n| < \varepsilon$.

**Definition 14.2.36.** Let $\{\Psi_n\}_{n=0}^{*\infty}$ be a hyper infinite sequence such that for all $n \in *\mathbb{N}$, $\Psi_n \in \tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$. We call $\{\Psi_n\}_{n=0}^{*\infty}$ a Cauchy hyper infinite sequence if the difference between its terms #-tends to $0_{*\mathbb{R}^c}$. To be precise: given any $\varepsilon > 0_{*\mathbb{R}^c}$, $\varepsilon \approx 0_{*\mathbb{R}^c}$ there is a hypernatural number $N \in *\mathbb{N}\mathbb{N}$, $N = N(\varepsilon)$ such that for any $m, n > N$, $|\Psi_n - \Psi_m| < \varepsilon$.

**Theorem 14.2.8.** Let $\{\Psi_n\}_{n=0}^{*\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$. If $\{\Psi_n\}_{n=0}^{*\infty}$ is a #-convergent hyper infinite sequence (that is, $\Psi_n \to_* \Phi$ as $n \to *\infty$ for some $\Phi \in \tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$), then $\{\Psi_n\}_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence.

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

**Theorem 14.2.9.** If $\{\Psi_n\}_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence, then it is bounded in $*\mathbb{R}^c$; that is, there is some number $M \in *\mathbb{R}^c$ such that $|\{\Psi_n\}_{n=0}^{*\infty}| \le M$ for all $n \in *\mathbb{N}$.

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

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

**Definition 14.2.37.** Let $S$ be a set of objects. A relation among pairs of elements of $S$ is said to be an equivalence relation if the following three properties hold:

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

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

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

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

**Theorem 14.2.10.** Let $S$ be a set, with an equivalence relation on pairs of elements.
$[ \{ \Psi_{1,n} \}_{n=0}^{*\infty} ]$ of Cauchy hyper infinite sequences of, as per Definition 14.2.38 and
(2) the all gyperreals $*\mathbb{R}_c^\# \subset \widetilde{\mathbb{R}}_c^\# $ by the canonical imbedding $ *\mathbb{R}_c^\# \hookrightarrow \widetilde{\mathbb{R}}_c^\#$
(14.1.42)-(14.1.43).

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

**Definition 14.2.40.** Let $s, t \in \widetilde{\mathbb{R}}_c^\#$, so there are Cauchy hyper infinite sequences $\{\Psi_n\}_{n=0}^{*\infty}$ and $\{\Phi_n\}_{n=0}^{*\infty}$ with $s = [\{\Psi_n\}_{n=0}^{*\infty}]$ and $t = [\{\Phi_n\}_{n=0}^{*\infty}]$.

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

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

**Proposition 14.2.5.** The operations $+, \times$ in Definition 14.2.25 (a), (b) are well-defined.

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

Multiplication is a little trickier; this is where we will use Theorem 14.2.10. We will
also use another ubiquitous technique: adding $0_{*\mathbb{R}_c^\#}$ in the form of $s-s$. Again,
suppose that

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

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

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

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

Now, from Theorem 14.2.9, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in *\mathbb{N}$. Taking some number $R$ (for example $R = M+L$) which is bigger than both, we have

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

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

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

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

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

$$
0_{*\mathbb{R}_c^\#} = (0_{*\mathbb{R}_c^\#}, 0_{*\mathbb{R}_c^\#}, 0_{*\mathbb{R}_c^\#}, 0_{*\mathbb{R}_c^\#}, \dots). \quad (14.2.40)
$$

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

From this, we are to deduce the existence of a hyperreal number $t = [\{\Phi_n\}_{n=0}^{*\infty}]$
For $s \in S$, denote by $[s]$ the set of all elements in $S$ that are related to $s$. Then for any $s, t \in S$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint.

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

**Corollary 14.2.2.** If $S$ is a set with an equivalence relation on pairs of elements, then the equivalence classes are non-empty disjoint sets whose union is all of $S$.

**Definition 14.2.38.** Let $\{\Psi_{1,n}\}_{n=0}^{*\infty}$ and $\{\Psi_{2,n}\}_{n=0}^{*\infty}$ be in $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$. Say they are equivalent (i.e. related) if $|\Psi_{1,n} - \Psi_{2,n}| \to_{\#} 0_{*\mathbb{R}_c^{\#}}$ as $n \to *\infty$, i.e. if the hyper infinite sequence $\{|\Psi_{1,n} - \Psi_{2,n}|\}_{n=0}^{*\infty}$ $\#$-tends to $0_{*\mathbb{R}_c^{\#}}$.

**Proposition 14.2.4.** Definition 4.2.38 yields an equivalence relation on $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$.

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

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

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

But $\Psi_{2,n} - \Psi_{1,n} = -(\Psi_{1,n} - \Psi_{2,n})$, and since only the absolute value $|\Psi_{1,n} - \Psi_{2,n}| = |\Psi_{2,n} - \Psi_{1,n}|$ comes into play in Definition 14.2.35, it follows that $\Psi_{2,n} - \Psi_{1,n} \to_{\#} 0_{*\mathbb{R}_c^{\#}}$ as well. Hence, $\{\Psi_{2,n}\}_{n=0}^{*\infty}$ is related to $\{\Psi_{1,n}\}_{n=0}^{*\infty}$.

* Transitive: Here we will use the $\varepsilon/2$ trick we applied to prove Theorem 14.2.4.

Suppose $\{\Psi_{1,n}\}_{n=0}^{*\infty}$ is related to $\{\Psi_{2,n}\}_{n=0}^{*\infty}$, and $\{\Psi_{2,n}\}_{n=0}^{*\infty}$ is related to $\{\Psi_{3,n}\}_{n=0}^{*\infty}$.
This means that $\Psi_{1,n} - \Psi_{2,n} \to_{\#} 0_{*\mathbb{R}_c^{\#}}$ and $\Psi_{2,n} - \Psi_{3,n} \to_{\#} 0_{*\mathbb{R}_c^{\#}}$.

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

So, we really have equivalence relation, and so by Corollary 14.2.2, the set $\tilde{\Xi}_{\omega}^{\pm, \neq 0, \nu=0}$ is partitioned into disjoint subsets (equivalence classes).

**Definition 14.2.39.** (1) The hyperreal numbers $^{\tilde{\mathbb{R}}}^{\#}$ are the equivalence classes $[\{\Psi_{1,n}\}_{n=0}^{*\infty}]$ of Cauchy hyper infinite sequences of, as per Definition 14.2.38 and (2) the all gyperreals $^{\tilde{\mathbb{R}}}^{\#} \subset ^{\tilde{\mathbb{R}}}^{\#}$ by the canonical imbedding $^{\tilde{\mathbb{R}}}^{\#} \hookrightarrow ^{\tilde{\mathbb{R}}}^{\#}$ (14.1.42)-(14.1.43).

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

**Definition 14.2.40.** Let $s, t \in ^{\tilde{\mathbb{R}}}^{\#}$, so there are Cauchy hyper infinite sequences $\{\Psi_n\}_{n=0}^{*\infty}$ and $\{\Phi_n\}_{n=0}^{*\infty}$ with $s = [\{\Psi_n\}_{n=0}^{*\infty}]$ and $t = [\{\Phi_n\}_{n=0}^{*\infty}]$.

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

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

**Proposition 14.2.5.** The operations $+, \times$ in Definition 14.2.25 (a),(b) are well-defined.

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

That is,

**Lemma 14.2.2.** If $\{\Psi_n\}_{n=0}^{\infty}$ is a Cauchy hyper infinite sequence which does not #-tends to $0_{\mathbb{R}_c^\#}$, then there is an $N \in \mathbb{N}$ such that, for $n > N$, $\Psi_n \neq 0_{\mathbb{R}_c^\#}$. We will now use it to complete the proof of Theorem 14.2.11.

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

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

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

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

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

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

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

$$ 1_{\widehat{\mathbb{R}}_c^\#} = (1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, \dots), \qquad (14.2.41) $$

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

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

**Theorem 14.2.7.** Let $s, t \in \widehat{\mathbb{R}}_c^\#$ be hyperreal numbers such that $s > t$, and let $r \in \widehat{\mathbb{R}}_c^\#$. Then $s + r > t + r$.

**Proof.** Let $s = [\{\Psi_n\}_{n=0}^{\infty}]$, $t = [\{\Phi_n\}_{n=0}^{\infty}]$, and $r = [\{\Theta_n\}_{n=0}^{\infty}]$. Since $s > t$, i.e. $s - t > 0$, we know that there is an $N$ such that, for $n > N$, $\Psi_n - \Phi_n > 0$. So $\Psi_n > \Phi_n$ for $n > N$. Now, adding $\Theta_n$ to both sides of this inequality, we have $\Psi_n + \Theta_n > \Phi_n + \Theta_n$ for $n > N$, or $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) > 0_{\mathbb{R}_c^\#}$ for $n > N$. Note also that $(\Psi_n + \Theta_n) - (\Phi_n + \Theta_n) = \Psi_n - \Phi_n$ does not #-converge to $0_{\mathbb{R}_c^\#}$ as $n \to \infty$, by the assumption that $s - t > 0_{\mathbb{R}_c^\#}$. Thus, by Definition 14.2.41, this means that: $s + r = [\{\Psi_n + \Theta_n\}_{n=0}^{\infty}] > [\{\Phi_n + \Theta_n\}_{n=0}^{\infty}] = t + r$.
($\Psi_n + \Phi_n$) - ($\Psi_{1,n} + \Phi_{1,n}$) = ($\Psi_n - \Psi_{1,n}$) + ($\Phi_n - \Phi_{1,n}$). Now, using the familiar $\varepsilon/2$ trick, you can construct a proof that this tends to $0_{\mathbb{R}^\#}$, and so

$$[(\Psi_n + \Phi_n)] = [(\Psi_{1,n} + \Phi_{1,n})].$$

Multiplication is a little trickier; this is where we will use Theorem 14.2.10. We will also use another ubiquitous technique: adding $0_{\mathbb{R}^\#}$ in the form of $s - s$. Again, suppose that

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

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

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

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

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

Now, from Theorem 14.2.9, there are numbers $M$ and $L$ such that $|\Phi_n| \le M$ and $|\Psi_{1,n}| \le L$ for all $n \in *\mathbb{N}$. Taking some number $R$ (for example $R = M + L$) which is bigger than both, we have

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

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

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

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

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

$$0_{*\mathbb{R}^\#} = (0_{*\mathbb{R}^\#}, 0_{*\mathbb{R}^\#}, 0_{*\mathbb{R}^\#}, 0_{*\mathbb{R}^\#}, \dots). \qquad (14.2.40)$$

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

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

$$\Psi_n \ne 0_{*\mathbb{R}^\#}.$$

That is,

**Lemma 14.2.2.** If $\{\Psi_n\}_{n=0}^{*\infty}$ is a Cauchy hyper infinite sequence which does not #-tends to $0_{\mathbb{R}^\#}$, then there is an $N \in *\mathbb{N}$ such that, for $n > N$, $\Psi_n \ne 0_{\mathbb{R}^\#}$.

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

Let $N \in *\mathbb{N}^\#$ be such that $\Psi_n \ne 0_{\mathbb{R}^\#}$ for $n > N$. Define hyper infinite sequence $\Phi_n$ of hyperreal numbers from $\tilde{\mathbb{R}}^\#$ as follows:
**Definition 14.2.42.** There is canonical imbeding

$$ *\mathbb{R}_c^\# \hookrightarrow *\widetilde{\mathbb{R}}_c^\# \qquad (14.2.42) $$

defined by

$$ a \mapsto \tilde{a} \qquad (14.2.43) $$

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

**Notation 14.2.5.** $\hat{a} = (a, a, \dots) \in *\widetilde{\mathbb{R}}_c^\#$, $a \in *\widetilde{\mathbb{R}}_c^\#$.

**Definition 14.2.43.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $*\widetilde{\mathbb{R}}_c^\#$, $\{a_n\}_{n=0}^k \subset *\widetilde{\mathbb{R}}_c^\#$.
We define external hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^k} \subset *\widetilde{\mathbb{R}}_c^\#$ by

$$ \begin{aligned} \{A_n; k\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^k} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, k, \hat{a}_k). \end{aligned} \qquad (14.2.44) $$

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

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

$$ \begin{aligned} \{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}} = \\ &= (a_0, a_1, \dots, a_k, \dots \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}_{n=0}^{\infty}}). \end{aligned} \qquad (14.2.45) $$

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

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

$$ \begin{aligned} \{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^N} = \\ &= (a_0, a_1, \dots, a_m, \dots, a_{N-1}, a_N, \hat{a}_N) \in [[a_N]]. \end{aligned} \qquad (14.2.46) $$

**Definition 14.2.44.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $*\widetilde{\mathbb{R}}_c^\#$, $\{a_n\}_{n=0}^N \subset *\widetilde{\mathbb{R}}_c^\#$.

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

$$ Ext-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k} = (c_0, c_1, \dots, c_m, \dots, c_k, \hat{c}_k) \in [[c_k]] \qquad (14.2.47) $$

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

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

$$ \begin{aligned} Ext-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}_{n=0}^\infty} = \\ &= (c_0, c_1, \dots, c_k, \dots \{c_n\}_{n=0}^\infty, \overbrace{\{c_n\}_{n=0}^\infty} \in [[\overbrace{\{c_n\}_{n=0}^\infty}]] \end{aligned} \qquad (14.2.48) $$

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

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

We define external hyperfinite sum $Ext-\sum_{n=0}^{n=N} a_n$ by
for $n \le N$, $\Phi_n = 0_{\mathbb{R}_c^\#}$, and for $n > N$, $\Phi_n = 1_{\mathbb{R}_c^\#}/\Psi_n$:
$$ \{\Phi_n\}_{n=0}^{\infty} = (0_{\mathbb{R}_c^\#}, 0_{\mathbb{R}_c^\#}, \dots, 0_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}/\Psi_{N+1}, 1_{\mathbb{R}_c^\#}/\Psi_{N+2}, \dots). $$

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

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

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

$$ 1_{\mathbb{R}_c^\#} = (1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, 1_{\mathbb{R}_c^\#}, \dots), \qquad (14.2.41) $$

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

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

**Theorem 14.2.7.** Let $s, t \in \widetilde{\mathbb{R}_c^\#}$ be hyperreal numbers such that $s > t$, and let $r \in \widetilde{\mathbb{R}_c^\#}$. Then $s+r > t+r$.

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

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

**Definition 14.2.42.** There is canonical imbeding

$$ \ast_{\mathbb{R}_c^\#} \to \widetilde{\ast_{\mathbb{R}_c^\#}} \qquad (14.2.42) $$

defined by

$$ a \mapsto \tilde{a} \qquad (14.2.43) $$

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

**Notation 14.2.5.** $\hat{a} = (a, a, \dots) \in \widetilde{\ast_{\mathbb{R}_c^\#}}, a \in \widetilde{\ast_{\mathbb{R}_c^\#}}$.

**Definition 14.2.43.** (i) Let $\{a_n\}_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $\widetilde{\ast_{\mathbb{R}_c^\#}}, \{a_n\}_{n=0}^k \subset \widetilde{\ast_{\mathbb{R}_c^\#}}$.

We define external hyper infinite sequence $\widetilde{\{a_n\}_{n=0}^k} \subset \widetilde{\ast_{\mathbb{R}_c^\#}}$ by
$$ \mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{n=N}} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \in [[c_N]] \quad (14.2.49) $$

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

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

$$ \mathrm{Ext}-\sum_{n=0}^{n=N} a_n = \mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n. \quad (14.2.50) $$

**Example 14.2.3.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}, \alpha \in \widehat{\mathbb{R}}_c^\#$, $r \in \widehat{\mathbb{R}}_c^\#$ be the first term and the ratio of the G.P respectively. Then for any $N \in *\mathbb{N}$ by Proposition 14.2.6 and Definition 14.2.44 one obtains that

$$ \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#} - r^N}{\widehat{1}_{\mathbb{R}_c^\#} - r} = \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#}}{\widehat{1}_{\mathbb{R}_c^\#} - r} - \alpha \frac{\widehat{r}^N}{\widehat{1}_{\mathbb{R}_c^\#} - r}. \quad (14.2.51) $$

and

$$ \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#}}{\widehat{1}_{\mathbb{R}_c^\#} - r} - \alpha \left\{ \frac{\widehat{r^n}}{\widehat{1}_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty}. \quad (14.2.52) $$

**Example 14.2.4.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}, \alpha \in \widehat{\mathbb{R}}_c^\#, r \in \widehat{\mathbb{R}}_c^\#$, $r < 0$.
Note that

$$ \begin{aligned} \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#} - r^N}{\widehat{1}_{\mathbb{R}_c^\#} - r} &= \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \\ &= \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \\ &= \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#}}{\widehat{1}_{\mathbb{R}_c^\#} - r} - \alpha \left\{ \frac{\widehat{r^n}}{\widehat{1}_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \end{aligned} \quad (14.2.53) $$

From (14.2.53) we obtain

$$ \begin{aligned} \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#} - r^N}{\widehat{1}_{\mathbb{R}_c^\#} - r} - \alpha \frac{\widehat{1}_{\mathbb{R}_c^\#}}{\widehat{1}_{\mathbb{R}_c^\#} - r} + \alpha \left\{ \frac{\widehat{r^n}}{\widehat{1}_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} = \\ &\qquad \alpha \left\{ \frac{\left(\widehat{-1}_{\mathbb{R}_c^\#}\right)^n |r|^n}{\widehat{1}_{\mathbb{R}_c^\#} - r} \right\}_{n=1}^{\infty} - \alpha \frac{\widehat{r^N}}{\widehat{1}_{\mathbb{R}_c^\#} - r}. \end{aligned} \quad (14.2.54) $$

Assume that: (i) $r < 0$, $|r| < 1$ then from (14.2.54) we obtain

$$ \mathrm{Ext}-\sum_{(n \in *\mathbb{N}) \wedge (n \le N-1)} \alpha (-1)_{\mathbb{R}_c^\#}^{n-1} |r|^{n-1} \neq 0. \quad (14.2.55) $$

## 15.1. Basic analysis on external non-Archimedean field $\mathbb{R}_c^\#$.
$$
\begin{align}
\{A_n; k\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^k}^{\infty} = \tag{14.2.44} \\
&= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, k, \hat{a}_k). \nonumber
\end{align}
$$

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

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

$$
\begin{align}
\{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}}^{\infty} = \tag{14.2.45} \\
&= \left( a_0, a_1, \dots, a_k, \dots \middle| \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}_{n=0}^{\infty}}^{*\mathbb{R}_c^\#} \right). \nonumber
\end{align}
$$

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

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

$$
\begin{align}
\{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{N}}^{\infty} = \tag{14.2.46} \\
&= (a_0, a_1, \dots, a_m, \dots, a_{N-1}, a_N, \hat{a}_N) \in [[a_N]]. \nonumber
\end{align}
$$

**Definition 14.2.44.** (i) Let {$a_n$}$_{n=0}^k, k \in \mathbb{N}$ be finite sequence in $^{*}\overline{\mathbb{R}}_c^\#$, {$a_n$}$_{n=0}^N \subset ^{*}\overline{\mathbb{R}}_c^\#$.

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

$$
Ext-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k} = (c_0, c_1, \dots, c_m, \dots, c_k, \hat{c}_k) \in [[c_k]] \quad (14.2.47)
$$

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

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

countable sum $Ext-\sum_{n=0}^{n=\infty} a_n$ by

$$
\begin{align}
\mathrm{Ext}-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}_{n=0}^{\infty}}^{n=\infty} = \tag{14.2.48} \\
&= \left( c_0, c_1, \dots, c_k, \dots \middle| \{c_n\}_{n=0}^{\infty}, \overbrace{\{c_n\}_{n=0}^{\infty}}^{n=\infty} \right) \in \left[ \left[ \overbrace{\{c_n\}_{n=0}^{\infty}}^{n=\infty} \right] \right] \nonumber
\end{align}
$$

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

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

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

$$
Ext-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{n=N}}^{n=N} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \in [[c_N]] \quad (14.2.49)
$$

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

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

$$
Ext-\sum_{n=0}^{n=N} a_n = Ext-\sum_{n=0}^{n=\infty} a_n. \qquad (14.2.50)
$$

**Example 14.2.3.** Consider the G.P: $\alpha, \alpha r, \alpha r^2, \dots, \alpha r^{N-1}, N \in *\mathbb{N}, \alpha \in ^{*}\overline{\mathbb{R}}_c^\#$,
$r \in ^{*}\overline{\mathbb{R}}_c^\#$ be the first term and the ratio of the G.P respectively. Then for any
15.1. The #-limit of a function $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$

**Definition 15.1.** The $(\varepsilon, \delta)$ definition of the #-limit of a function $f : D \to \mathbb{R}_c^\#$ is as follows: Let $f$ be a $\mathbb{R}_c^\#$-valued function defined on a subset $D \subset \mathbb{R}_c^\#$ of the Cauchy hyperreal numbers. Let $c$ be a limit point of $D$ and let $L$ be a hyperreal number. We say that

$$ \#\lim_{x \to c} f(x) = L \qquad (15.1) $$

if for every $\varepsilon \approx 0, \varepsilon > 0$ there exists a $\delta \approx 0, \delta > 0$ such that, for all $x \in D$, if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$, symbolically:

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon (\varepsilon \approx 0 \land \varepsilon > 0) \exists \delta (\delta \approx 0 \land \delta > 0) \forall x \in D, 0 < |x-c| < \delta \Rightarrow |f(x)-L| < \varepsilon. \qquad (15.2) $$

**Definition 15.2.** The function $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is #-continuous (or micro continuous) at some point $c$ of its domain if the #-limit of $f(x)$, as $x$ #-approaches $c$ through the domain of $f$, exists and is equal to $f(c)$:

$$ \#\lim_{x \to c} f(x) = f(c). \qquad (15.3) $$

**Theorem 15.1.** If $\#\lim_{x \to c} x_0 f(x)$ exists; then it is unique that is; if $\#\lim_{x \to c} x_0 f(x) = L_1$ and $\#\lim_{x \to c} x_0 f(x) = L_2$, then $L_1 = L_2$.

**Theorem 15.2.** If $\#\lim_{x \to c} x_0 f_1(x) = L_1$ and $\#\lim_{x \to c} x_0 f_2(x) = L_2$ then

$$ \begin{aligned} \#\lim_{x \to c} x_0 [f_1(x) \pm f_2(x)] &= L_1 \pm L_2, \\ \#\lim_{x \to c} x_0 [f_1(x) \times f_2(x)] &= L_1 \times L_2, \\ \#\lim_{x \to c} x_0 \frac{f_1(x)}{f_2(x)} &= \frac{L_1}{L_2}, L_2 \neq 0. \end{aligned} \qquad (15.4) $$

**Definition 15.3.** (a) We say that $f(x)$ #-approaches the left-hand #-limit $L$ as $x$ #-approaches $x_0$ from the left, and write $\#\lim_{x \to x_0^-} f(x) = L$, if $f(x)$ is defined on some #-open interval $(a, x_0)$ and, for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - L| < \varepsilon$ if $x_0 - \delta < x < x_0$.

(b) We say that $f(x)$ #-approaches the right-hand #-limit $L$ as $x$ #-approaches $x_0$ from the right, and write $\#\lim_{x \to x_0^+} f(x) = L$, if $f(x)$ is defined on some open interval $(x_0, b)$ and,

for each $\varepsilon > 0$, there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - L| < \varepsilon$, $\varepsilon > 0$, $\varepsilon \approx 0$ if $x_0 < x < x_0 + \delta$.

Left- and right-hand #-limits are also called one-sided #-limits. We will often simplify the notation by writing $\#\lim_{x \to c} x_0^- f(x) = f(x_0^-)$ and $\#\lim_{x \to c} x_0^+ f(x) = f(x_0^+)$.

**Theorem 15.3.** A function $f$ has a #-limit at $x_0$ if and only if it has left- and right-hand #-limits at $x_0$; and they are equal. More specifically; $\#\lim_{x \to c} x_0 f(x) = L$ if and only if $f(x_0^+) = f(x_0^-) = L$.

**Definition 15.4.** We say that $f(x)$ approaches the #-limit $L$ as $x$ approaches $\infty^\#$, and write $\#\lim_{x \to c} \infty^\# f(x) = L$, if $f$ is defined on an interval $(a, \infty^\#)$ and, for each $\varepsilon > 0, \varepsilon \approx 0$, there is a number $\beta$ such that $|f(x) - L| < \varepsilon$ if $x > \beta$.

**Definition 15.5.** We say that $f(x)$ approaches $\infty^\#$ as $x$ approaches $x_0$ from the left, and write
$N \in *\mathbb{N}$ by Proposition 14.2.6 and Definition 14.2.44 one obtains that

$$ \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#} - r^N}{1_{\widetilde{\mathbb{R}}_c^\#} - r} = \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#}}{1_{\widetilde{\mathbb{R}}_c^\#} - r} - \alpha \frac{r^N}{1_{\widetilde{\mathbb{R}}_c^\#} - r}. \quad (14.2.51) $$

and

$$ \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} = \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#}}{1_{\widetilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\widetilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty}. \quad (14.2.52) $$

**Example 14.2.4.** Consider the G.P: $\alpha, ar, ar^2, \dots, ar^{N-1}, N \in *\mathbb{N}, \alpha \in *\widetilde{\mathbb{R}}_c^\#, r \in *\widetilde{\mathbb{R}}_c^\#,$
$r < 0_{\widetilde{\mathbb{R}}_c^\#}, |r| < 1$. Note that

$$ \begin{aligned} \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#} - r^N}{1_{\widetilde{\mathbb{R}}_c^\#} - r} &= \mathrm{Ext}-\sum_{n=1}^{n=N-1} \alpha r^{n-1} = \\ &= \mathrm{Ext}-\sum_{n=1}^{\infty} \alpha r^{n-1} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} = \\ &= \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#}}{1_{\widetilde{\mathbb{R}}_c^\#} - r} - \alpha \left\{ \frac{r^n}{1_{\widetilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} + \mathrm{Ext}-\sum_{(n \in *\mathbb{N}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1}. \end{aligned} \quad (14.2.53) $$

From (14.2.53) we obtain

$$ \begin{aligned} \mathrm{Ext}-\sum_{(n \in *\mathbb{N}\mathbb{N}) \wedge (n \le N-1)} \alpha r^{n-1} &= \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#} - r^N}{1_{\widetilde{\mathbb{R}}_c^\#} - r} - \alpha \frac{1_{\widetilde{\mathbb{R}}_c^\#}}{1_{\widetilde{\mathbb{R}}_c^\#} - r} + \alpha \left\{ \frac{r^n}{1_{\widetilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} = \\ & \qquad \alpha \left\{ \frac{(-1_{\widetilde{\mathbb{R}}_c^\#})^n |r|^n}{1_{\widetilde{\mathbb{R}}_c^\#} - r} \right\}_{n=1}^{\infty} - \alpha \frac{r^N}{1_{\widetilde{\mathbb{R}}_c^\#} - r}. \end{aligned} \quad (14.2.54) $$

Assume that: (i) $r < 0_{\widetilde{\mathbb{R}}_c^\#}, |r| < 1$ then from (14.2.54) we obtain

$$ \mathrm{Ext}-\sum_{(n \in *\mathbb{N}\mathbb{N}) \wedge (n \le N-1)} \alpha (-1_{\widetilde{\mathbb{R}}_c^\#})^{n-1} |r|^{n-1} \neq 0_{\widetilde{\mathbb{R}}_c^\#}. \quad (14.2.55) $$

## 15.1. Basic analysis on external non-Archimedean field $\mathbb{R}_c^\#$.

### 15.1. The #-limit of a function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$

**Definition 15.1.** The $(\varepsilon, \delta)$ definition of the #-limit of a function $f: D \to \mathbb{R}_c^\#$ is as follows: Let $f$ be a $\mathbb{R}_c^\#$-valued function defined on a subset $D \subset \mathbb{R}_c^\#$ of the Cauchy hyperreal numbers. Let $c$ be a limit point of $D$ and let $L$ be a hyperreal number. We say that

$$ \#\text{-lim}_{x \to c} f(x) = L \quad (15.1) $$

if for every $\varepsilon \approx 0, \varepsilon > 0$ there exists a $\delta \approx 0, \delta > 0$ such that, for all $x \in D$, if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$, symbolically:

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon (\varepsilon \approx 0 \wedge \varepsilon > 0) \exists \delta (\delta \approx 0 \wedge \delta > 0) \forall x \in D, 0 < |x - c| < \delta \Rightarrow |f(x) - L| < \varepsilon. \quad (15.2) $$
$$ \#\lim_{x \to \# x_0} f(x) = \infty^{\#} \text{ or } f(x_0 - ) = \infty^{\#} \qquad (15.5) $$

if $f$ is defined on an interval $(a, x_0)$ and, for each hyperreal number $M$, there is a $\delta \approx 0, \delta > 0$ such that $f(x) > M$ if $x_0 - \delta < x < x_0$.

Similarly we define: $\#\lim_{x \to x_0} f(x) = -\infty^{\#}$, $\#\lim_{x \to \# x_0} f(x) = -\infty^{\#}$, $\#\lim_{x \to \# x_0+} f(x) = +\infty^{\#}$, $\#\lim_{x \to \# x_0-} f(x) = +\infty^{\#}$.

**Example 15.1.** (i) $\#\lim_{x \to \# x_0} x_0 - x^{-1} = -\infty^{\#}$, (ii) $\#\lim_{x \to \# x_0} x_0 + x^{-1} = +\infty^{\#}$, (iii) $\#\lim_{x \to \# -\infty} x^2 = \#\lim_{x \to \# +\infty} x^2 = \infty^{\#}$.

**Remark 15.1.** Throughout this paper, $\#\lim_{x \to \# x_0} f(x)$ exists" will mean that $\#\lim_{x \to \# x_0} f(x) = L$, where $L$ is finite or hyperfinite.

To leave open the possibility that $L = \pm\infty^{\#}$, we will say that $\#\lim_{x \to x_0} f(x)$ exists in the extended hyperreals.

This convention also applies to one-sided limits and limits as $x$ approaches $\pm\infty^{\#}$.

## 15.2. Monotonic Functions $f: \mathbb{R}^{\#}_c \to \mathbb{R}^{\#}_c$.

**Definition 17.6.** A function $f: \mathbb{R}^{\#}_c \to \mathbb{R}^{\#}_c$ is nondecreasing on an interval $I \subset \mathbb{R}^{\#}_c$ if

$$ f(x_1) \le f(x_2) \qquad (15.6) $$

whenever $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$, or nonincreasing on $I$ if

$$ f(x_1) \ge f(x_2) \qquad (15.7) $$

whenever $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$.

In either case, $f$ is on $I$. If $\le$ can be replaced by $<$ in (15.6), $f$ is increasing on $I$. If $\ge$ can be replaced by $>$ in (15.7), $f$ is decreasing on $I$. In either of these two cases, $f$ is strictly monotonic on $I$.

**Theorem 15.4.** Suppose that $f(x)$ is monotonic on $(a, b)$ and define $\alpha = \inf_{a<x<b} f(x)$ and $\beta = \sup_{a<x<b} f(x)$. Suppose that $\exists \alpha$ and $\exists \beta$, then:

(a) If $f$ is nondecreasing, then $f(a +) = \alpha$ and $f(b -) = \beta$.

(b) If $f$ is nonincreasing; then $f(a +) = \beta$ and $f(b -) = \alpha$.

Here $a += -\infty^{\#}$ if $a = -\infty^{\#}$ and $b += \infty^{\#}$ if $b = \infty^{\#}$.

(c) If $a < x_0 < b$, then $f(x_0 +)$ and $f(x_0 -)$ exist and are finite or hyperfinite; moreover, $f(x_0 +) \le f(x_0) \le f(x_0 -)$ if $f$ is nondecreasing, and $f(x_0 +) \ge f(x_0) \ge f(x_0 -)$ if $f$ is nonincreasing:

**Proof** (a) We first show that $f(a +) = \alpha$. If $M > \alpha$, there is an $x_0$ in $(a, b)$ such that $f(x_0) < M$. Since $f$ is nondecreasing, $f(x) < M$ if $a < x < x_0$. Therefore, if $\alpha = -\infty^{\#}$, then $f(a +) = -\infty^{\#}$. If $\alpha > -\infty^{\#}$, let $M = \alpha + \varepsilon$, where $\varepsilon \approx 0, \varepsilon > 0$. Then $\alpha \le f(x) < \alpha + \varepsilon$, so (i) $|f(x) - \alpha| < \varepsilon$ if $a < x < x_0$.

If $a = -\infty^{\#}$, this implies that $f(-\infty^{\#}) = \alpha$. If $a > -\infty^{\#}$, let $\delta = x_0 - a$. Then (i) is equivalent to $|f(x) - \alpha| < \varepsilon$ if $a < x < a + \delta$, which implies that $f(a +) = \alpha$. We now show that $f(b +) = \beta$. If $M < \beta$, there is an $x_0$ in $(a, b)$ such that $f(x_0) > M$. Since $f$ is nondecreasing, $f(x) > M$ if $x_0 < x < b$. Therefore, if $\beta = \infty^{\#}$, then $f(b -) = \infty^{\#}$. If $\beta < \infty^{\#}$, let $M = \beta - \varepsilon$, where $\varepsilon \approx 0$. Then $\beta - \varepsilon < f(x) \le \beta$, so (ii) $|f(x) - \beta| < \varepsilon$ if $x_0 < x < b$.

If $b = \infty^{\#}$, this implies that $f(\infty^{\#}) = \beta$. If $b < \infty^{\#}$, let $\delta = b - x_0$. Then (ii) is equivalent to $f(x) < b - \delta < x < b$, which implies that $f(b -) = \beta$.

(b) The proof is similar to the proof of (a).

(c) Suppose that $f(x)$ is nondecreasing. Applying (a) to $f(x)$ on $(a, x_0)$ and $(x_0, b)$ separately shows that $f(x_0 -) = \sup_{a<x<x_0} f(x)$ and $f(x_0 +) = \inf_{x_0<x<b} f(x)$.
**Definition 15.2.** The function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is \#-continuous (or micro continuous) at some point $c$ of its domain if the \#-limit of $f(x)$, as $x$ \#-approaches $c$ through the domain of $f$, exists and is equal to $f(c)$:

$$ \#\lim_{x \to \#} c f(x) = f(c). \qquad (15.3) $$

**Theorem 15.1.** If \#-lim$_{x \to \#} x_0 f(x)$ exists; then it is unique that is; if \#-lim$_{x \to \#} x_0 f(x) = L_1$ and \#-lim$_{x \to \#} x_0 f(x) = L_2$, then $L_1 = L_2$.

**Theorem 15.2.** If \#-lim$_{x \to \#} x_0 f_1(x) = L_1$ and \#-lim$_{x \to \#} x_0 f_2(x) = L_2$ then

$$ \begin{align} \#\lim_{x \to \#} x_0 [f_1(x) \pm f_2(x)] &= L_1 \pm L_2, \notag \\ \#\lim_{x \to \#} x_0 [f_1(x) \times f_2(x)] &= L_1 \times L_2, \tag{15.4} \\ \#\lim_{x \to \#} x_0 \frac{f_1(x)}{f_2(x)} &= \frac{L_1}{L_2}, \quad L_2 \neq 0. \notag \end{align} $$

**Definition 15.3.** (a) We say that $f(x)$ \#-approaches the left-hand \#-limit $L$ as $x$ \#-approaches $x_0$ from the left, and write \#-lim$_{x \to x_0^-} f(x) = L$, if $f(x)$ is defined on some \#-open interval $(a, x_0)$ and, for each $\varepsilon > 0$, $\varepsilon \approx 0$ there is a $\delta > 0$, $\delta \approx 0$ such that $|f(x) - L| < \varepsilon$ if $x_0 - \delta < x < x_0$.

(b) We say that $f(x)$ \#-approaches the right-hand \#-limit $L$ as $x$ \#-approaches $x_0$ from the right, and write \#-lim$_{x \to \#} x_0^+$, $f(x) = L$, if $f(x)$ is defined on some open interval $(x_0, b)$ and, for each $\varepsilon > 0$, there is a $\delta > 0$, $\delta \approx 0$ such that $|f(x) - L| < \varepsilon$, $\varepsilon > 0$, $\varepsilon \approx 0$ if $x_0 < x < x_0 + \delta$. Left- and right-hand \#-limits are also called one-sided \#-limits. We will often simplify the notation by writing \#-lim$_{x \to \#} x_0^- f(x) = f(x_0^-)$ and \#-lim$_{x \to \#} x_0^+ f(x) = f(x_0^+)$.

**Theorem 15.3.** A function $f$ has a \#-limit at $x_0$ if and only if it has left- and right-hand \#-limits at $x_0$; and they are equal. More specifically; \#-lim$_{x \to \#} x_0^+$, $f(x) = L$ if and only if $f(x_0^+) = f(x_0^-) = L$.

**Definition 15.4.** We say that $f(x)$ approaches the \#-limit $L$ as $x$ approaches $\infty^\#$, and write \#-lim$_{x \to \#} \infty^\#$, $f(x) = L$, if $f$ is defined on an interval $(a, \infty^\#)$ and, for each $\varepsilon > 0$, $\varepsilon \approx 0$, there is a number $\beta$ such that $|f(x) = L| < \varepsilon$ if $x > \beta$.

**Definition 15.5.** We say that $f(x)$ approaches $\infty^\#$ as $x$ approaches $x_0$ from the left, and write

$$ \#\lim_{x \to \#} x_0^- f(x) = \infty^\# \text{ or } f(x_0^-) = \infty^\# \qquad (15.5) $$

if $f$ is defined on an interval $(a, x_0)$ and, for each hyperreal number $M$, there is a $\delta \approx 0$, $\delta > 0$ such that $f(x) > M$ if $x_0 - \delta < x < x_0$.

Similarly we define: \#-lim$_{x \to x_0^-} f(x) = -\infty^\#$, \#-lim$_{x \to \#} x_0^+$, $f(x) = -\infty^\#$, \#-lim$_{x \to \#} x_0^+$, $f(x) = \infty^\#$.

**Example 15.1.** (i) \#-lim$_{x \to \#} x_0^- x^{-1} = -\infty^\#$, (ii) \#-lim$_{x \to \#} x_0^+ x^{-1} = +\infty^\#$, (iii) \#-lim$_{x \to \#} -\infty^\#$, $x^2 = \#-\lim_{x \to \#} \infty^\#$, $x^2 = \infty^\#$.

**Remark 15.1.** Throughout this paper, \#-lim$_{x \to \#} x_0^+$, $f(x)$ exists” will mean that \#-lim$_{x \to \#} x_0^+$, $f(x) = L$, where $L$ is finite or hyperfinite.

To leave open the possibility that $L = \pm\infty^\#$, we will say that \#-lim$_{x \to x_0^-} f(x)$ exists in the extended hyperreals.

This convention also applies to one-sided limits and limits as $x$ approaches $\pm\infty^\#$.
However, if $x_1 < x_0 < x_2$, then $f(x_1) \le f(x_0) \le f(x_2)$ and hence, $f(x_0 -) \le f(x_0) \le f(x_0 +)$.

## 15.3. #-Limits Inferior and Superior

**Definition 15.7.** We say that: (i) $f$ is bounded on a set $S \subseteq \mathbb{R}^c$ if there is a constant $M \in \mathbb{R}, M < \infty$ such that $f(x) \le M$ for all $x \in S$, (ii) $f$ is hyperbounded on a set $S \subseteq \mathbb{R}^c$ if $f$ is not bounded on a set $S$ and there is a constant $M \in \mathbb{R}^c/\mathbb{R}, M < \infty$ such that $f(x) \le M$ for all $x \in S$.

**Definition 15.8.** Suppose that $f$ is bounded or hyperbounded on $[a, x_0)$, where $x_0$ may be finite or hyperfinite or $\infty$. For $a \le x < x_0$, define (i) $S_f(x; x_0) = \sup_{x \le f < x_0} f(t)$ and (ii) $I_f(x; x_0) = \inf_{x \le f < x_0} f(t)$.

Then the left #-limit superior of $f(x)$ at $x_0$ is defined to be

$$ \#-\overline{\lim}_{x \to x_0} f(x) = \#-\lim_{x \to x_0} S_f(x; x_0) \quad (15.8) $$

and the left limit inferior of $f(x)$ at $x_0$ is defined to be

$$ \#-\underline{\lim}_{x \to x_0} f(x) = \#-\lim_{x \to x_0} S_f(x; x_0). \quad (15.9) $$

If $x_0 = \infty$, we define $x_0 = -\infty$.

**Theorem 15.5.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then $\beta = \#-\overline{\lim}_{x \to x_0} f(x)$ exists and is the unique hyperreal number with the following properties:

(a) If $\varepsilon > 0, \varepsilon \approx 0$, there is an $a_1$ in $[a, x_0)$ such that

(i) $f(x) < \beta + \varepsilon$ if $a_1 \le x < x_0$

(b) If $\varepsilon > 0, \varepsilon \approx 0$ and $a_1$ is in $[a, x_0)$, then

$f(\bar{x}) > \beta - \varepsilon$ for some $\bar{x} \in [a, x_0)$.

**Proof.** Since $f(x)$ is bounded or hyperbounded on $[a, x_0)$, $S_f(x; x_0)$ is nonincreasing and bounded or hyperbounded on $[a, x_0)$. By applying Theorem 17.4(b) to $S_f(x; x_0)$, we conclude that $\beta$ exists finite or hyperfinite.

Therefore, if $\varepsilon > 0, \varepsilon \approx 0$, there is an $\bar{a}$ in $[a, x_0)$ such that

(ii) $\beta - \varepsilon/2 < S_f(x; x_0) < \beta + \varepsilon/2$ if $\bar{a} \le x < x_0$.

Since $S_f(x; x_0)$ is an upper bound of $\{f(t) | x \le t < x_0\}$, $f(x) < S_f(x; x_0)$. Therefore, the second inequality in (ii) implies the inequality (i) with $a_1 = \bar{a}$. This proves (a). To prove (b), let $a_1$ be given and define $x_1 = \max\{a_1, \bar{a}\}$. Then the first inequality in (ii) implies that (iii) $S_f(x; x_0) > \beta - \varepsilon/2$. Since $S_f(x; x_0)$ is the supremum of

$\{f(t) | x_1 \le t < x_0\}$, there is an $\bar{x}$ in $[x_1, x_0)$ such that $f(\bar{x}) > S_f(x; x_0) - \varepsilon/2$. This and (iii) imply that $f(\bar{x}) > \beta - \varepsilon/2$. Since $\bar{x}$ is in $[a_1, x_0)$, this proves (b).

Now we show that there cannot be more than one hyperreal number with properties (a) and (b). Suppose that $\beta_1 < \beta_2$ and $\beta_2$ has property (b); thus, if $\varepsilon \approx 0, \varepsilon > 0$ and $a_1$ is in $[a, x_0)$ there is an $\bar{x}$ in $[a_1, x_0)$ such that $f(\bar{x}) > \beta_2 - \varepsilon$. Letting $\varepsilon = \beta_2 - \beta_1$, we see that there is an $\bar{x}$ in $[a_1, b)$ such that $f(\bar{x}) > \beta_2 - (\beta_2 - \beta_1) = \beta_1$ so $\beta_1$ cannot have property (a). Therefore, there cannot be more than one hyperreal number that satisfies both (a) and (b).

**Theorem 15.6.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then $\alpha = \lim_{x \to x_0} f(x)$ exists and there is the unique hyperreal number with the following properties:

(a) If $\varepsilon \approx 0, \varepsilon > 0$ there is an $a_1$ in $[a, x_0)$ such that $f(x) > \alpha - \varepsilon$ if $a_1 \le x < x_0$.

(b) If $\varepsilon \approx 0, \varepsilon > 0$ and $a_1$ is in $[a, x_0)$, then
## 15.2. Monotonic Functions $f: \mathbb{R}^\#_c \to \mathbb{R}^\#_c$.

**Definition 17.6.** A function $f: \mathbb{R}^\#_c \to \mathbb{R}^\#_c$ is nondecreasing on an interval $I \subset \mathbb{R}^\#_c$ if

$$f(x_1) \le f(x_2) \qquad (15.6)$$

whenever $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$, or nonincreasing on $I$ if

$$f(x_1) \ge f(x_2) \qquad (15.7)$$

whenever $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$.

In either case, $f$ is on $I$. If $\le$ can be replaced by $<$ in (15.6), $f$ is increasing on $I$. If $\ge$ can be replaced by $>$ in (15.7), $f$ is decreasing on $I$. In either of these two cases, $f$ is strictly monotonic on $I$.

**Theorem 15.4.** Suppose that $f(x)$ is monotonic on $(a, b)$ and define

$\alpha = \inf_{a<x<b} f(x)$ and $\beta = \sup_{<x<b} f(x)$. Suppose that $\exists \alpha$ and $\exists \beta$, then:

(a) If $f$ is nondecreasing, then $f(a^+) = \alpha$ and $f(b^-) = \beta$.

(b) If $f$ is nonincreasing; then $f(a^+) = \beta$ and $f(b^-) = \alpha$.

Here $a^+ = -\infty^\#$ if $a = -\infty^\#$ and $b^+ = \infty^\#$ if $b = \infty^\#$.

(c) If $a < x_0 < b$, then $f(x_0^+) = \alpha$ and $f(x_0^-) = \beta$ exist and are finite or hyperfinite; moreover, $f(x_0^+) \le f(x_0^-) \le f(x_0^-)$ if $f$ is nondecreasing, and $f(x_0^+) \ge f(x_0^-) \ge f(x_0^-)$ if $f$ is nonincreasing:

**Proof (a)** We first show that $f(a^+) = \alpha$. If $M > \alpha$, there is an $x_0$ in $(a, b)$ such that $f(x_0^-) < M$. Since $f$ is nondecreasing, $f(x) < M$ if $a < x < x_0$. Therefore, if $\alpha = -\infty^\#$, then $f(a^+) = -\infty^\#$. If $\alpha > -\infty^\#$, let $M = \alpha + \varepsilon$, where $\varepsilon \approx 0, \varepsilon > 0$.

Then $\alpha \le f(x) < \alpha + \varepsilon$, so (i) $|f(x) - \alpha| < \varepsilon$ if $a < x < x_0$.

If $\alpha = -\infty^\#$, this implies that $f(-\infty^\#) = \alpha$. If $\alpha > -\infty^\#$, let $\delta = x_0 - \alpha$. Then (i) is equivalent to $|f(x) - \alpha| < \varepsilon$ if $a < x < a + \delta$, which implies that $f(a^+) = \alpha$.

We now show that $f(b^+) = \beta$. If $M < \beta$, there is an $x_0$ in $(a, b)$ such that $f(x_0) > M$.

Since $f(x)$ is nondecreasing, $f(x) > M$ if $x_0 < x < b$. Therefore, if $\beta = \infty^\#$, then $f(b^-) = \infty^\#$. If $\beta < \infty^\#$, let $M = \beta - \varepsilon$, where $\varepsilon \approx \varepsilon > 0$. Then $\beta - \varepsilon < f(x) \le \beta$, so (ii) $|f(x) - \beta| < \varepsilon$ if $x_0 < x < b$.

If $b = \infty^\#$, this implies that $f(\infty^\#) = \beta$. If $b < \infty^\#$, let $\delta = b - x_0$. Then (ii) is equivalent to $f(x) < f(b - \delta) < x < b$, which implies that $f(b^-) = \beta$.

(b) The proof is similar to the proof of (a).

(c) Suppose that $f(x)$ is nondecreasing. Applying (a) to $f(x)$ on $(a, x_0)$ and $(x_0, b)$ separately shows that $f(x_0^-) = \sup_{a<x<x_0} f(x)$ and $f(x_0^+) = \inf_{x_0<x<b} f(x)$.

However, if $x_1 < x_0 < x_2$, then $f(x_1) \le f(x_0) \le f(x_2)$ and hence, $f(x_0^-) \le f(x_0) \le f(x_0^+)$.

## 15.3. #-Limits Inferior and Superior

**Definition 15.7.** We say that: (i) $f$ is bounded on a set $S \subset \mathbb{R}^\#_c$ if there is a constant $M \in \mathbb{R}, M < \infty$ such that $f(x) \le M$ for all $x \in S$, (ii) $f$ is hyperbounded on a set $S \subset \mathbb{R}^\#_c$ if $f$ is not bounded on a set $S$ and there is a constant $M \in \mathbb{R}^\#_c/\mathbb{R}, M < \infty^\#$ such that $f(x) \le M$ for all $x \in S$.

**Definition 15.8.** Suppose that $f$ is bounded or hyperbounded on $[a, x_0)$, where $x_0$ may be finite or hyperfinite or $\infty^\#$. For $a \le x < x_0$, define (i) $S_f(x; x_0) = \sup_{x \le t < x_0} f(t)$ and

(ii) $I_f(x; x_0) = \inf_{x \le t < x_0} f(t)$.

Then the left #-limit superior of $f(x)$ at $x_0$ is defined to be
$f(\bar{x}) < \alpha + \varepsilon$ for some $\bar{x} \in [a, x_0)$.

**Theorem 15.7.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then

(i) $\#\underline{\lim}_{x \to x_0} f(x) \le \#\overline{\lim}_{x \to x_0} f(x)$;

(ii) $\#\underline{\lim}_{x \to x_0} f(-x) = \#\overline{\lim}_{x \to x_0} f(x)$;

(iii) $\#\overline{\lim}_{x \to x_0} f(-x) = \#\underline{\lim}_{x \to x_0} f(x)$;

(iv) $\#\underline{\lim}_{x \to x_0} f(x) = \#\overline{\lim}_{x \to x_0} f(x)$ if and only if $\#\lim_{x \to x_0} f(x)$ exists, in which case
$\#\lim_{x \to x_0} f(x) = \#\underline{\lim}_{x \to x_0} f(x) = \#\overline{\lim}_{x \to x_0} f(x)$

**Theorem 15.8.** Suppose that $f(x)$ and $g(x)$ are bounded or hyperbounded on $[a, x_0)$.

Then: (i) $\#\overline{\lim}_{x \to x_0} (f+g)(x) \le \#\underline{\lim}_{x \to x_0} f(x) + \#\overline{\lim}_{x \to x_0} g(x)$;

(ii) $\#\underline{\lim}_{x \to x_0} (f+g)(x) \ge \#\overline{\lim}_{x \to x_0} f(x) + \#\underline{\lim}_{x \to x_0} g(x)$.

**Theorem 15.9.** The $\alpha = \lim_{x \to x_0} f(x)$ exists i.e., $\alpha$ is finite or hyperfinite if and only if for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that $|f(x_1) - f(x_2)| < \varepsilon$ if $x_0 - \delta < x_1, x_2 < x_0$.

**Theorem 15.10.** (i) Suppose that $f(x)$ is bounded or hyperbounded on an interval $(x_0, b]$, then $\#\underline{\lim}_{x \to x_0} f(x) = \#\overline{\lim}_{x \to x_0} f(x)$ if and only if $\#\lim_{x \to x_0} f(x)$ exists, in which case $\#\underline{\lim}_{x \to x_0} f(x) = \#\overline{\lim}_{x \to x_0} f(x) = \#\overline{\lim}_{x \to x_0} f(x)$.

(ii) Suppose that $f(x)$ is bounded or hyperbounded on an open interval containing $x_0$, then $\#\lim_{x \to x_0} f(x)$ exists if and only if
$\#\overline{\lim}_{x \to x_0} f(x) = \#\underline{\lim}_{x \to x_0} f(x) = \#\overline{\lim}_{x \to x_0} f(x) = \#\underline{\lim}_{x \to x_0} f(x)$.

## 15.4. The #-continuity of a function $f: \mathbb{R}^\#_c \to \mathbb{R}^\#_c$.

**Definition 15.9.** (i) We say that a function $f: \mathbb{R}^\#_c \to \mathbb{R}^\#_c$ is #-continuous at $x_0$ if $f$ is defined on an open interval $(a, b)$ containing $x_0$ and $\lim_{x \to x_0} f(x_0) = x_0$.

(ii) We say that $f$ is #-continuous from the left at $x_0$ if $f$ is defined on an open interval $(a, x_0)$ and $f(x_0^-) = f(x_0)$.

(iii) We say that $f$ is #-continuous from the right at $x_0$ if $f$ is defined on an open interval $(x_0, b)$ and $f(x_0^+) = f(x_0)$.

**Theorem 15.11.** (i) A function $f$ is #-continuous at $x_0$ if and only if $f$ is defined on an open interval $(a, b)$ containing $x_0$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that

$$|f(x) - f(x_0)| < \varepsilon \qquad (15.10)$$

whenever $|x - x_0| < \delta$.

(ii) A function $f$ is #-continuous from the right at $x_0$ if and only if $f$ is defined on an interval $[x_0, b)$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that (17.10) holds whenever $x_0 \le x < x_0 + \delta$.

(iii) A function $f$ is #-continuous from the left at $x_0$ if and only if $f$ is defined on an interval $(a, x_0]$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that (15.10) holds whenever $x_0 - \delta < x \le x_0$.

Note that from Definition 15.9 and Theorem 15.8, $f$ is #-continuous at $x_0$ if and only if $f(x_0^+) = f(x_0^-) = f(x_0)$ or, equivalently, if and only if it is #-continuous from the right and left at $x_0$.

**Definition 15.10.** A function $f: \mathbb{R}^\#_c \to \mathbb{R}^\#_c$ is #-continuous on an open interval $(a, b)$ if it is

#-continuous at every point in $(a, b)$. If, in addition,
$$ \#\overline{\lim}_{x \to \# x_0} f(x) = \#\lim_{x \to \# x_0} S_f(x; x_0) \quad (15.8) $$

and the left limit inferior of $f(x)$ at $x_0$ is defined to be

$$ \#\overline{\lim}_{x \to \# x_0} f(x) = \#\lim_{x \to \# x_0} I_f(x; x_0). \quad (15.9) $$

If $x_0 = \infty^{\#}$, we define $x_0 = -\infty^{\#}$.

**Theorem 15.5.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then $\beta = \#\overline{\lim}_{x \to \# x_0} f(x)$ exists and is the unique hyperreal number with the following properties:

(a) If $\varepsilon > 0, \varepsilon \approx 0$, there is an $a_1$ in $[a, x_0)$ such that

(i) $f(x) < \beta + \varepsilon$ if $a_1 \le x < x_0$

(b) If $\varepsilon > 0, \varepsilon \approx 0$ and $a_1$ is in $[a, x_0)$, then

$f(\bar{x}) > \beta - \varepsilon$ for some $\bar{x} \in [a, x_0)$.

**Proof.** Since $f(x)$ is bounded or hyperbounded on $[a, x_0)$, $S_f(x; x_0)$ is nonincreasing and bounded or hyperbounded on $[a, x_0)$. By applying Theorem 17.4(b) to $S_f(x; x_0)$, we conclude that $\beta$ exists finite or hyperfinite.

Therefore, if $\varepsilon > 0, \varepsilon \approx 0$, there is an $\bar{a}$ in $[a, x_0)$ such that

(ii) $\beta - \varepsilon/2 < S_f(x; x_0) < \beta + \varepsilon/2$ if $\bar{a} \le x < x_0$.

Since $S_f(x; x_0)$ is an upper bound of $\{f(t)|x \le t < x_0\}$, $f(x) < S_f(x; x_0)$. Therefore, the second inequality in (ii) implies the inequality (i) with $a_1 = \bar{a}$. This proves (a).

To prove (b), let $a_1$ be given and define $x_1 = \max\{a_1, \bar{a}\}$. Then the first inequality in (ii) implies that (iii) $S_f(x; x_0) > \beta - \varepsilon/2$. Since $S_f(x; x_0)$ is the supremum of

$\{f(t)|x_1 \le t < x_0\}$, there is an $\bar{x}$ in $[x_1, x_0)$ such that

$f(\bar{x}) > S_f(x; x_0) - \varepsilon/2$. This and (iii) imply that $f(\bar{x}) > \beta - \varepsilon/2$. Since $\bar{x}$ is in $[a_1, x_0)$, this proves (b).

Now we show that there cannot be more than one hyperreal number with properties (a) and (b). Suppose that $\beta_1 < \beta_2$ and $\beta_2$ has property (b); thus, if $\varepsilon \approx 0, \varepsilon > 0$ and $a_1$ is in $[a, x_0)$ there is an $\bar{x}$ in $[a_1, x_0)$ such that $f(\bar{x}) > \beta_2 - \varepsilon$. Letting $\varepsilon = \beta_2 - \beta_1$, we see that there is an $\bar{x}$ in $[a_1, b)$ such that $f(\bar{x}) > \beta_2 - (\beta_2 - \beta_1) = \beta_1$ so $\beta_1$ cannot have property (a). Therefore, there cannot be more than one hyperreal number that satisfies both (a) and (b).

**Theorem 15.6.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then $\alpha = \underline{\lim}_{x \to x_0} f(x)$ exists and there is the unique hyperreal number with the following properties:

(a) If $\varepsilon \approx 0, \varepsilon > 0$ there is an $a_1$ in $[a, x_0)$ such that

$f(x) > \alpha - \varepsilon$ if $a_1 \le x < x_0$.

(b) If $\varepsilon \approx 0, \varepsilon > 0$ and $a_1$ is in $[a, x_0)$, then

$f(\bar{x}) < \alpha + \varepsilon$ for some $\bar{x} \in [a, x_0)$.

**Theorem 15.7.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then

(i) $\#\overline{\lim}_{x \to \# x_0} f(x) \le \#\overline{\lim}_{x \to \# x_0} f(x)$;

(ii) $\#\overline{\lim}_{x \to \# x_0} f(-x) = \#\overline{\lim}_{x \to \# x_0} f(x)$;

(iii) $\#\overline{\lim}_{x \to \# x_0} f(-x) = \#\overline{\lim}_{x \to \# x_0} f(x)$;

(iv) $\#\overline{\lim}_{x \to \# x_0} f(x) = \#\overline{\lim}_{x \to \# x_0} f(x)$ if and only if $\#\overline{\lim}_{x \to \# x_0} f(x)$ exists, in which case

$\#\overline{\lim}_{x \to \# x_0} f(x) = \#\overline{\lim}_{x \to \# x_0} f(x) = \#\overline{\lim}_{x \to \# x_0} f(x)$

**Theorem 15.8.** Suppose that $f(x)$ and $g(x)$ are bounded or hyperbounded on $[a, x_0)$.

Then: (i) $\#\overline{\lim}_{x \to \# x_0} (f + g)(x) \le \#\overline{\lim}_{x \to \# x_0} f(x) + \#\overline{\lim}_{x \to x_0} g(x)$;

(ii) $\#\overline{\lim}_{x \to \# x_0} (f + g)(x) \ge \#\overline{\lim}_{x \to \# x_0} f(x) + \#\overline{\lim}_{x \to \# x_0} g(x)$.

**Theorem 15.9.** The $\alpha = \lim_{x \to x_0} f(x)$ exists i.e., $\alpha$ is finite or hyperfinite
$$f(b-) = f(b) \qquad (15.11)$$

or

$$f(a+) = f(a) \qquad (15.12)$$

then $f$ is #-continuous on $(a, b]$ or $[a, b)$, respectively. If $f$ is #-continuous on $(a, b)$ and (15.11) and (15.12) both hold, then $f$ is #-continuous on $[a, b]$. More generally, if $S$ is a subset of $\text{dom}(f)$ consisting of finitely or countably or hyper finitely or hyper infinitely many disjoint intervals, then $f$ is #-continuous on $S$ if $f$ is #-continuous on every interval in $S$.

**Definition 15.11.** A function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is piecewise #-continuous on $[a, b]$ if

(i) $f(x_0 +)$ exists for all $x_0$ in $[a, b)$;

(ii) $f(x_0 -)$ exists for all $x_0$ in $(a, b]$;

(iii) $f(x_0 +) = f(x_0 -) = f(x_0)$ for all but except finitely or hyper finitely many points $x_0$ in $(a, b)$.

If (iii) fails to hold at some $x_0$ in $(a, b)$, $f$ has a jump #-discontinuity at $x_0$. Also, $f$ has a jump #-discontinuity at $a$ if $f(a+) \neq f(a)$ or at $b$ if $f(b-) \neq f(b)$.

**Theorem 15.12.** If $f$ and $g$ are #-continuous on a set $S$, then so are $f \pm g$, and $fg$. In addition, $f/g$ is #-continuous at each $x_0$ in $S$ such that $g(x_0) \neq 0$.

By hyper infinite induction, it can be shown that if $\forall n \in \mathbb{N}^\#$ $f_n(x)$ are #-continuous on a set $S$, then so are $\sum_{i \le n} f_n(x)$. Therefore, $\forall n, m \in \mathbb{N}^\#$ any rational function

$$r(x) = \sum_{i \le n} a_i x^i / \sum_{i \le m} b_i x^i, b_i \ne 0 \text{ is #-continuous for all values of } x \text{ except those for which its denominator vanishes.}$$

## 15.5. Removable #-discontinuities.

**Definition 15.12.** Let $f(x)$ be defined on a deleted #-neighborhood of $x_0$ and #-discontinuous (perhaps even undefined) at $x_0$. Then we say that $f(x)$ has a removable #-discontinuity at $x_0$ if $\lim_{x \to x_0} f(x)$ exists. In this case, the function

$$g(x) = \begin{cases} f(x) & \text{if } x \in \text{dom}(f) \text{ and } x \neq x_0 \\ \lim_{x \to x_0} f(x_0) & \text{if } x = x_0 \end{cases} \qquad (15.13)$$

is #-continuous at $x_0$.

## 15.6. Composite Functions $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Definition 15.13.** Suppose that $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ and $g: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ are functions with domains $\text{dom}(f)$ and $\text{dom}(g)$ correspondingly. If $\text{dom}(g)$ has a nonempty subset $T$ such that $g(x) \in \text{dom}(g)$ whenever $x \in T$, then the composite function $f \circ g: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is defined

on $T$ by $(f \circ g)(x) = f(g(x))$

**Theorem 15.10.** Suppose that $g$ is #-continuous at $x_0$, $g(x_0)$ is an #-interior point of $\text{dom}(f)$ and $f$ is #-continuous at $g(x_0)$. Then $f \circ g$ is #-continuous at $x_0$.

**Proof.** Suppose that $\varepsilon \approx 0, \varepsilon > 0$. Since $g(x_0)$ is an #-interior point of $\text{dom}(f)$ and $f(x)$ is #-continuous at $g(x_0)$, there is a $\delta_1 \approx 0, \delta_1 > 0$ such that $f(t)$ is defined and
if and only if for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$
such that $|f(x_1) - f(x_2)| < \varepsilon$ if $x_0 - \delta < x_1, x_2 < x_0$.

**Theorem 15.10.** (i) Suppose that $f(x)$ is bounded or hyperbounded on an interval $(x_0, b]$, then $\overline{\lim}_{x \to \# x_0+} f(x) = \overline{\lim}_{x \to \# x_0+} f(x)$ if and only if $\underline{\lim}_{x \to x_0+} f(x)$ exists, in which case $\overline{\lim}_{x \to \# x_0+} f(x) = \underline{\lim}_{x \to \# x_0+} f(x) = \overline{\lim}_{x \to \# x_0+} f(x)$.

(ii) Suppose that $f(x)$ is bounded or hyperbounded on an open interval containing $x_0$, then $\underline{\lim}_{x \to x_0} f(x)$ exists if and only if $\overline{\lim}_{x \to x_0} f(x) = \overline{\lim}_{x \to \# x_0+} f(x) = \overline{\lim}_{x \to \# x_0+} f(x) = \underline{\lim}_{x \to \# x_0} f(x) = \underline{\lim}_{x \to \# x_0+} f(x)$.

## 15.4. The #-continuity of a function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Definition 15.9.** (i) We say that a function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$. is #-continuous at $x_0$ if $f$ is defined on an open interval $(a, b)$ containing $x_0$ and $\lim_{x \to \# x_0} f(x_0) = x_0$.

(ii) We say that $f$ is #-continuous from the left at $x_0$ if $f$ is defined on an open interval $(a, x_0)$ and $f(x_0 -) = f(x_0)$.

(iii) We say that $f$ is #-continuous from the right at $x_0$ if $f$ is defined on an open interval $(x_0, b)$ and $f(x_0 +) = f(x_0)$.

**Theorem 15.11.** (i) A function $f$ is #-continuous at $x_0$ if and only if $f$ is defined on an open interval $(a, b)$ containing $x_0$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that

$$|f(x) - f(x_0)| < \varepsilon \qquad (15.10)$$

whenever $|x - x_0| < \delta$.

(ii) A function $f$ is #-continuous from the right at $x_0$ if and only if $f$ is defined on an interval $[x_0, b)$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that (17.10) holds whenever $x_0 \le x < x_0 + \delta$.

(iii) A function $f$ is #-continuous from the left at $x_0$ if and only if $f$ is defined on an interval $(a, x_0]$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that (15.10) holds whenever $x_0 - \delta < x \le x_0$.

Note that from Definition 15.9 and Theorem 15.8, $f$ is #-continuous at $x_0$ if and only if $f(x_0 +) = f(x_0 -) = f(x_0)$ or, equivalently, if and only if it is #-continuous from the right and left at $x_0$.

**Definition 15.10.** A function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is #-continuous on an open interval $(a, b)$ if it is

#-continuous at every point in $(a, b)$. If, in addition,

$$f(b -) = f(b) \qquad (15.11)$$

or

$$f(a +) = f(a) \qquad (15.12)$$

then $f$ is #-continuous on $(a, b]$ or $[a, b)$, respectively. If $f$ is #-continuous on $(a, b)$ and (15.11) and (15.12) both hold, then $f$ is #-continuous on $[a, b]$. More generally, if $S$ is a subset of $\text{dom}(f)$ consisting of finitely or countably or hyper finitely or hyper infinitely many disjoint intervals, then $f$ is #-continuous on $S$ if $f$ is #-continuous on every interval in $S$.

**Definition 15.11.** A function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is piecewise #-continuous on $[a, b]$ if

(i) $f(x_0 +)$ exists for all $x_0$ in $[a, b)$;

(ii) $f(x_0 -)$ exists for all $x_0$ in $(a, b]$;

(iii) $f(x_0 +) = f(x_0 -) = f(x_0)$ for all but except finitely or hyper finitely many points $x_0$
(i) $|f(t) - f(gx_0)| < \varepsilon$ if $|t - g(x_0)| < \delta_1$.

Since $g(x)$ is #-continuous at $x_0$, there is a $\delta \approx 0, \delta > 0$ such that $g(x)$ is defined and

(ii) $|g(x) - g(x_0)| < \delta_1$ if $|x - x_0| < \delta$.

Now (i) and (ii) imply that $|f(g(x)) - f(g(x_0))| < \delta$ if $|x - x_0| < \varepsilon$. Therefore, $f \circ g$ is #-continuous at $x_0$.

## 15.7. Bounded and Hyperbounded Functions $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Definition 15.14.** (i) A function $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is bounded below on a set $S \subset \mathbb{R}_c^\#$ if there is a finite or hyperfinite hyperreal number $m \in \mathbb{R}_c^\#_{\text{fin}}$ such that $f(x) \ge m$ for all $x \in S$. If in this case the set $V = \{f(x)|x \in S\}$ has infimum $\alpha$, we write $\alpha = \inf_{x \in S} f(x)$. If there is a point $x_1 \in S$ such that $f(x_1) = \alpha_>$, we say that $\alpha$ is the minimum of $f(x)$ on $S$, and write $\alpha = \min_{x \in S} f(x)$

(ii) A function $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is bounded above on $S \subset \mathbb{R}_c^\#$ if there is a finite or hyperfinite hyperreal number $M \in \mathbb{R}_c^\#_{\text{fin}}$ such that $f(x) \le M$ for all $x \in S$. If in this case, $V$ has a supremum $\beta$, we write $\beta = \sup_{x \in S} f(x)$. If there is a point $x_1 \in S$ such that $f(x_1) = \beta_>$, we say that $\beta$ is the maximum of $f(x)$ on $S$, and write $\alpha = \max_{x \in S} f(x)$.

(iii) If $f$ is bounded above and below on a set $S$, we say that $f$ is bounded on $S$.

**Theorem 15.11.** If $f$ is #-continuous on a finite or hyperfinite #-closed interval $[a, b]$, then $f$ is bounded or hyperbounded on $[a, b]$.

**Proof.** Suppose that $t \in [a, b]$. Since $f$ is #-continuous at $t$, there is an open interval $I_t$ containing $t$ such that

$$|f(x) - f(t)| < 1 \text{ if } x \in I_t \cap [a, b] \qquad (15.14)$$

To see this, set $\varepsilon = 1$ in (15.10), Theorem 15.11. The collection $H = \{I_t|a \le t \le b\}$ is an open covering of $[a, b]$. Since $[a, b]$ is #-compact, the generalized Heine-Borel theorem implies that there are hyper finitely many points $t_1, t_2, \dots, t_n, n \in \mathbb{N}^\#$ such that the intervals $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. According to (11.14) with $t = t_i$, $|f(x) - f(t_i)| < 1$ if $x \in I_{t_i} \cap [a, b]$. Therefore,

$$f(x) = |(f(x) - f(t_i)) + f(t_i)| \le |f(x) - f(t_i)| + |f(t_i)| \le 1 + |f(t_i)| \qquad (15.15)$$

if $x \in I_{t_i} \cap [a, b]$. Let $M = 1 + \max_{1 \le i \le n}|f(t_i)|$. Since $[a, b] \subset \bigcup_{i=1}^n (I_{t_i} \cap [a, b])$, (15.15) implies that $|f(x)| \le M$ if $x \in [a, b]$.

**Theorem 15.12.** Suppose that $f$ is #-continuous on a finite or hyperfinite closed interval $[a, b]$. Let $V_{a,b} = \{f(x)|x \in [a, b]\}$. Assume that the set $V_{a,b}$ is admissible above and below. Let

$$\alpha = \inf V_{a,b} = \inf_{a \le x \le b} f(x) \text{ and } \beta = \sup V_{a,b} = \sup_{a \le x \le b} f(x). \qquad (15.16)$$

Then $\alpha$ and $\beta$ are respectively the minimum and maximum of $f$ on $[a, b]$; that is there are points $x_1$ and $x_2$ in $[a, b]$ such that $\alpha = f(x_1)$ and $\beta = f(x_2)$.

**Proof.** We show that $x_1$ exists. Note that $\alpha = \inf V_{a,b}$ and $\beta = \sup V_{a,b}$ exist since the set $V_{a,b}$ is admissible below and above. Suppose that there is no $x_1$ in $[a, b]$ such that $f(x_1) = \alpha_>$. Then $f(x) > \alpha_>$ for all $x \in [a, b]$. We will show that this leads to a contradiction. Suppose that $t \in [a, b]$. Then $f(t) > \alpha$, so $f(t) > [f(t) + \alpha]/2 > \alpha$. Since $f$ is #-continuous at $t$, there is an open interval $I_t$ about $t$ such that
in $(a, b)$.

If (iii) fails to hold at some $x_0$ in $(a, b)$, $f$ has a jump #-discontinuity at $x_0$. Also, $f$ has a jump #-discontinuity at $a$ if $f(a +) \neq f(a)$ or at $b$ if $f(b -) \neq f(b)$.

**Theorem 15.12.** If $f$ and $g$ are #-continuous on a set $S$, then so are $f \pm g$, and $fg$. In addition, $fg$ is #-continuous at each $x_0$ in $S$ such that $g(x_0) \neq 0$.

By hyper infinite induction, it can be shown that if $\forall n \in \mathbb{N}^\# f_n(x)$ are #-continuous on a set $S$, then so are $\sum_{i \le n} f_n(x)$. Therefore, $\forall n, m \in \mathbb{N}^\#$ any rational function

$r(x) = \sum_{i \le n} a_i x^i / \sum_{i \le m} b_i x^i, b_i \ne 0$ is #-continuous for all values of $x$ except those for which its denominator vanishes.

## 15.5. Removable #-discontinuities.

**Definition 15.12.** Let $f(x)$ be defined on a deleted #-neighborhood of $x_0$ and #-discontinuous (perhaps even undefined) at $x_0$. Then we say that $f(x)$ has a removable #-discontinuity at $x_0$ if $\lim_{x \to x_0} f(x_0)$ exists. In this case, the function

$$g(x) = \begin{cases} f(x) & \text{if } x \in \text{dom}(f) \text{ and } x \neq x_0 \\ \lim_{x \to x_0} f(x_0) & \text{if } x = x_0 \end{cases} \qquad (15.13)$$

is #-continuous at $x_0$.

## 15.6. Composite Functions $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Definition 15.13.** Suppose that $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ and $g : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ are functions with domains $\text{dom}(f)$ and $\text{dom}(g)$ correspondingly. If $\text{dom}(g)$ has a nonempty subset $T$ such that $g(x) \in \text{dom}(g)$ whenever $x \in T$, then the composite function $f \circ g : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is defined

on $T$ by $(f \circ g)(x) = f(g(x))$

**Theorem 15.10.** Suppose that $g$ is #-continuous at $x_0$, $g(x_0)$ is an #-interior point of $\text{dom}(f)$ and $f$ is #-continuous at $g(x_0)$. Then $f \circ g$ is #-continuous at $x_0$.

**Proof.** Suppose that $\varepsilon \approx 0, \varepsilon > 0$. Since $g(x_0)$ is an #-interior point of $\text{dom}(f)$ and $f(x)$ is #-continuous at $g(x_0)$, there is a $\delta_1 \approx 0, \delta_1 > 0$ such that $f(t)$ is defined and

(i) $|f(t) - f(gx_0)| < \varepsilon$ if $|t - g(x_0)| < \delta_1$.

Since $g(x)$ is #-continuous at $x_0$, there is a $\delta \approx 0, \delta > 0$ such that $g(x)$ is defined and

(ii) $|g(x) - g(x_0)| < \delta_1$ if $|x - x_0| < \delta$.

Now (i) and (ii) imply that $|f(g(x)) - f(g(x_0))| < \delta$ if $|x - x_0| < \varepsilon$. Therefore, $f \circ g$ is #-continuous at $x_0$.

## 15.7. Bounded and Hyperbounded Functions $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Definition 15.14.** (i) A function $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is bounded below on a set $S \subset \mathbb{R}_c^\#$ if there is a finite or hyperfinite hyperreal number $m \in \mathbb{R}_c^\#_{\text{fin}}$ such that $f(x) \ge m$ for all $x \in S$. If in this case the set $V = \{f(x) | x \in S\}$ has infimum $\alpha$, we write $\alpha = \inf_{x \in S} f(x)$.

If there is a point $x_1 \in S$ such that $f(x_1) = \alpha$, we say that $\alpha$ is the minimum of $f(x)$ on $S$, and write $\alpha = \min_{x \in S} f(x)$

(ii) A function $f : \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is bounded above on $S \subset \mathbb{R}_c^\#$ if there is a finite or hyperfinite
$$f(x) > \frac{f(t) + \alpha}{2} \qquad (15.17)$$

if $x \in I_t \cap [a, b]$. The collection $H = \{I_t|a \le t \le b\}$ is an open covering of $[a, b]$.

Since $[a, b]$ is #-compact, the generalized Heine-Borel theorem implies that there are hyper finitely many points $t_1, t_2, \dots, t_n$ such that the intervals $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$.

Define $\alpha_1 = \min_{1 \le i \le n} [f(t_i) + \alpha]/2$. Then, since $[a, b] \subset \bigcup_{i=1}^n (I_{t_i} \cap [a, b])$, (15.17) implies that

$f(t) > \alpha_1, a \le t \le b$. But $\alpha_1 > \alpha$, so this contradicts the definition of $\alpha$. Therefore, $f(x_1) = \alpha$, for some $x_1 \in [a, b]$.

## 15.8. Generalized Intermediate Value Theorem.

**Theorem 15.13. (Generalized Intermediate Value Theorem)** Suppose that:

(i) $f$ is #-continuous on $[a, b]$, (ii) $f(a) \neq f(b)$ and $f(a) < \mu < f(b)$,

(iii) the set $S = \{x|(a \le x \le b) \land (f(x) \le \mu)\}$ is admissible above. Then $f(c) = \mu$ for some $c \in (a, b)$.

**Proof.** Suppose that $f(a) < \mu < f(b)$. Note that $\sup S$ exists, since the set $S$ is admissible above. Let $c = \sup S$. We will show that $f(c) = \mu$. If $f(c) > \mu$, then $c > a$ and, since $f$ is #-continuous at $c$, there is an $\varepsilon > 0, \varepsilon \approx 0$ such that $f(x) > \mu$ if $c - \varepsilon < x \le c$. Therefore, $c$ is an upper bound for $S$, which contradicts the definition of $c$ as the supremum of $S$. If $f(c) < \mu$, then $c < b$ and there is an $\varepsilon > 0, \varepsilon \approx 0$ such that $f(x) < \mu$ for $c \le x < c - \varepsilon$, so $c$ is not an upper bound for $S$. This is also a contradiction. Therefore, $f(c) = \mu$. The proof for the case where $f(b) < \mu < f(a)$ can be obtained by applying this result to $-f(x)$.

**Lemma. 15.1.** If $f$ is #-continuous at $x_0$ and $f(x_0) > \mu$, then $f(x) > \mu$ for all $x$ in some #-neighborhood of $x_0$.

## 15.9. Uniform #-Continuity.

**Definition 15.15.** A function $f$ is uniformly #-continuous on a subset $S$ of its domain if, for every $\varepsilon > 0, \delta \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - f(x')| < \varepsilon$ whenever $|x - x'| < \delta$ and $x, x' \in S$.

We emphasize that in this definition $\delta$ depends only on $S$ and not on the particular choice of $x$ and $x'$, provided that they are both in $S$.

**Theorem 15.14.** If $f$ is #-continuous on a #-closed and bounded or hyperbounded interval $[a, b]$, then $f$ is uniformly #-continuous on $[a, b]$.

**Proof.** Suppose that $\varepsilon > 0, \delta \approx 0$. Since $f$ is #-continuous on $[a, b]$, for each $t \in [a, b]$ there is a positive number $\delta_t$ such that

$$|f(x) - f(t)| < \varepsilon/2 \qquad (15.18)$$

if $|x - t| < \delta_t$ and $x \in [a, b]$. If $I_t = (t - \delta_t, t + \delta_t)$, the collection $H = \{I_t|t \in [a, b]\}$ is an open covering of $[a, b]$. Since $[a, b]$ is #-compact, the generalized Heine-Borel theorem implies that there are hyper finitely many points $t_1, t_2, \dots, t_n$ in $[a, b]$ such that $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. Now define

$$\delta = \min\{\delta_{t_1}, \delta_{t_2}, \dots, \delta_{t_n}\}. \qquad (15.19)$$

We will show that if

$$|x - x'| < \delta \text{ and } x, x' \in [a, b] \qquad (15.20)$$
hyperreal number $M \in \mathbb{R}_{c.\text{fin}}^{\#}$ such that $f(x) \le M$ for all $x \in S$. If in this case, $V$ has a supremum $\beta$, we write $\beta = \sup_{x \in S} f(x)$. If there is a point $x_1 \in S$ such that $f(x_2) = \beta_\epsilon$, we say that $\beta$ is the maximum of $f(x)$ on $S$, and write $\alpha = \max_{x \in S} f(x)$.

(iii) If $f$ is bounded above and below on a set $S$, we say that $f$ is bounded on $S$.

**Theorem 15.11.** If $f$ is #-continuous on a finite or hyperfinite #-closed interval $[a, b]$, then $f$ is bounded or hyperbounded on $[a, b]$.

**Proof.** Suppose that $t \in [a, b]$. Since $f$ is #-continuous at $t$, there is an open interval $I_t$ containing $t$ such that

$$|f(x) - f(t)| < 1 \text{ if } x \in I_t \cap [a, b] \qquad (15.14)$$

To see this, set $\varepsilon = 1$ in (15.10), Theorem 15.11. The collection $H = \{I_t | a \le t \le b\}$ is an open covering of $[a, b]$. Since $[a, b]$ is #-compact, the generalized Heine-Borel theorem implies that there are hyper finitely many points $t_1, t_2, \dots, t_n, n \in \mathbb{N}^\#$ such that the intervals $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. According to (11.14) with $t = t_i$, $|f(x) - f(t_i)| < 1$ if $x \in I_{t_i} \cap [a, b]$. Therefore,

$$f(x) = |(f(x) - f(t_i)) + f(t_i)| \le |f(x) - f(t_i)| + |f(t_i)| \le 1 + |f(t_i)| \qquad (15.15)$$

if $x \in I_{t_i} \cap [a, b]$. Let $M = 1 + \max_{1 \le i \le n} |f(t_i)|$. Since $[a, b] \subset \bigcup_{i=1}^n (I_{t_i} \cap [a, b])$, (15.15) implies that $|f(x)| \le M$ if $x \in [a, b]$.

**Theorem 15.12.** Suppose that $f$ is #-continuous on a finite or hyperfinite closed interval $[a, b]$. Let $V_{a,b} = \{f(x) | x \in [a, b]\}$. Assume that the set $V_{a,b}$ is admissible above and below. Let

$$\alpha = \inf V_{a,b} = \inf_{a \le x \le b} f(x) \text{ and } \beta = \sup V_{a,b} = \sup_{a \le x \le b} f(x). \qquad (15.16)$$

Then $\alpha$ and $\beta$ are respectively the minimum and maximum of $f$ on $[a, b]$; that is there are points $x_1$ and $x_2$ in $[a, b]$ such that $\alpha = f(x_1)$ and $\beta = f(x_2)$.

**Proof.** We show that $x_1$ exists. Note that $\alpha = \inf V_{a,b}$ and $\beta = \sup V_{a,b}$ exist since the set $V_{a,b}$ is admissible below and above. Suppose that there is no $x_1$ in $[a, b]$ such that $f(x_1) = \alpha_\epsilon$. Then $f(x) > \alpha_\epsilon$ for all $x \in [a, b]$. We will show that this leads to a contradiction. Suppose that $t \in [a, b]$. Then $f(t) > \alpha$, so $f(t) > [f(t) + \alpha]/2 > \alpha$. Since $f$ is #-continuous at $t$, there is an open interval $I_t$ about $t$ such that

$$f(x) > \frac{f(t) + \alpha}{2} \qquad (15.17)$$

if $x \in I_t \cap [a, b]$. The collection $H = \{I_t | a \le t \le b\}$ is an open covering of $[a, b]$.

Since $[a, b]$ is #-compact, the generalized Heine-Borel theorem implies that there are hyper finitely many points $t_1, t_2, \dots, t_n$ such that the intervals $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$.

Define $\alpha_1 = \min_{1 \le i \le n} [f(t_i) + \alpha]/2$. Then, since $[a, b] \subset \bigcup_{i=1}^n (I_{t_i} \cap [a, b])$, (15.17) implies that $f(t) > \alpha_1, a \le t \le b$. But $\alpha_1 > \alpha$, so this contradicts the definition of $\alpha$. Therefore, $f(x_1) = \alpha_\epsilon$ for some $x_1 \in [a, b]$.

# 15.8. Generalized Intermediate Value Theorem.

**Theorem 15.13. (Generalized Intermediate Value Theorem)** Suppose that:

(i) $f$ is #-continuous on $[a, b]$, (ii) $f(a) \ne f(b)$ and $f(a) < \mu < f(b)$,

(iii) the set $S = \{x | (a \le x \le b) \land (f(x) \le \mu)\}$ is admissible above. Then $f(c) = \mu$ for
then $|f(x) - f(x')| < \varepsilon$. From the triangle inequality one obtains:

$$|f(x) - f(x')| = |(f(x) - f(t_r)) + (f(t_r) - f(x'))| \le |f(x) - f(t_r)| + |f(t_r) - f(x')| \quad (15.21)$$

Since $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$, $x$ must be in one of these intervals. Suppose that $x \in I_{t_r}$ that is,

$$|x - t_r| \le \delta_{t_r}. \quad (15.22)$$

From (11.18) with $t = t_r$,

$$|f(x) - f(t_r)| \le \frac{\varepsilon}{2}. \quad (15.23)$$

From (11.20), (11.22), and the triangle inequality,

$$|x' - t_r| = |(x' - x) + (x - t_r)| \le |x' - x| + |x - t_r| < \delta + \delta_{t_r} \le 2\delta_{t_r}. \quad (15.24)$$

Therefore, (11.18) with $t = t_r$ and $x$ replaced by $x'$ implies that

$$|f(x') - f(t_r)| \le \frac{\varepsilon}{2}. \quad (15.25)$$

Thus (11.25), (11.21) and (11.23) imply that $|f(x') - f(t_r)| \le \varepsilon/2$.

## 15.10. Monotonic External Functions $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Theorem 15.15.** If $f$ is monotonic and nonconstant on $[a, b]$, then $f$ is #-continuous on $[a, b]$ if and only if its range **range** ($f) = \{f(x)|x \in [a, b]\}$ is the #-closed interval with endpoints $f(a)$ and $f(b)$.

**Theorem 15.16.** Suppose that $f$ is increasing and #-continuous on $[a, b]$ and let $f(a) = c$ and $f(b) = d$. Then there is a unique function $g$ defined on $[c, d]$ such that

$$g(f(x)) = x, a \le x \le b, \quad (15.26)$$

and

$$f(g(y)) = y, c \le y \le d. \quad (15.27)$$

Moreover, $g$ is #-continuous and increasing on $[c, d]$:

The function $g$ of Theorem 15.16 is the inverse of $f$, denoted by $f^{-1}$. Since (15.26) and (15.27) are symmetric in $f$ and $g$, we can also regard $f$ as the inverse of $g$, and denote it by $g^{-1}$.

## 15.11. The #-derivative of a $\mathbb{R}_c^\#$-valued function $f: D \to \mathbb{R}_c^\#$.

A function $f: D \to \mathbb{R}_c^\#$, $D \subset \mathbb{R}_c^\#$ is differentiable at an #-interior point $x_0 \in D$ of its domain

$D \subset \mathbb{R}_c^\#$ if the difference quotient

$$\frac{f(x) - f(x_0)}{x - x_0}, x \ne x_0 \quad (15.28)$$

approaches a #-limit as $x$ approaches $x_0$, in which case the #-limit is called the #-derivative

of $f$ at $x_0$, and is denoted by $f^{#'}(x_0)$ or by $f'^#(x_0)$ or by $d^#f(x_0)/d^#x$ i.e.,

$$d^#f(x_0)/d^#x \triangleq f'^#(x_0) = \#\text{-lim}_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} \quad (15.29)$$

If $f$ is defined on an #-open set $S \subset \mathbb{R}_c^\#$, we say that $f$ is #-differentiable on $S$ if $f$ is #-differentiable at every point of $S$. If $f$ is #-differentiable on $S$, then $f'^#$ is a function on $S$.

We say that $f$ is #-continuously #-differentiable on $S$ if $f'^#(x)$ is #-continuous on $S$. If $f$ is
some $c \in (a, b)$.

**Proof.** Suppose that $f(a) < \mu < f(b)$. Note that $\sup S$ exists, since the set $S$ is admissible above. Let $c = \sup S$. We will show that $f(c) = \mu$. If $f(c) > \mu$, then $c > a$ and, since $f$ is #-continuous at $c$, there is an $\varepsilon > 0, \varepsilon \approx 0$ such that $f(x) > \mu$ if $c - \varepsilon < x \le c$. Therefore, $c$ is an upper bound for $S$, which contradicts the definition of $c$ as the supremum of $S$. If $f(c) < \mu$, then $c < b$ and there is an $\varepsilon > 0, \varepsilon \approx 0$ such that $f(x) < \mu$ for $c \le x < c - \varepsilon$, so $c$ is not an upper bound for $S$. This is also a contradiction. Therefore, $f(c) = \mu$. The proof for the case where $f(b) < \mu < f(a)$ can be obtained by applying this result to $-f(x)$.

**Lemma. 15.1.** If $f$ is #-continuous at $x_0$ and $f(x_0) > \mu$, then $f(x) > \mu$ for all $x$ in some #-neighborhood of $x_0$.

## 15.9. Uniform #-Continuity.

**Definition 15.15.** A function $f$ is uniformly #-continuous on a subset $S$ of its domain if, for every $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - f(x')| < \varepsilon$ whenever $|x - x'| < \delta$ and $x, x' \in S$.

We emphasize that in this definition $\delta$ depends only on and $S$ and not on the particular choice of $x$ and $x'$, provided that they are both in $S$.

**Theorem 15.14.** If $f$ is #-continuous on a #-closed and bounded or hyperbounded interval $[a, b]$, then $f$ is uniformly #-continuous on $[a, b]$.

**Proof.** Suppose that $\varepsilon > 0, \varepsilon \approx 0$. Since $f$ is #-continuous on $[a, b]$, for each $t \in [a, b]$ there is a positive number $\delta_t$ such that

$$|f(x) - f(t)| < \varepsilon/2 \qquad (15.18)$$

if $|x - t| < \delta_t$ and $x \in [a, b]$. If $I_t = (t - \delta_t, t + \delta_t)$, the collection $H = \{I_t | t \in [a, b]\}$ is an open covering of $[a, b]$. Since $[a, b]$ is #-compact, the generalized Heine-Borel theorem implies that there are hyperfinitely many points $t_1, t_2, \dots, t_n$ in $[a, b]$ such that $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. Now define

$$\delta = \min\{\delta_{t_1}, \delta_{t_2}, \dots, \delta_{t_n}\}. \qquad (15.19)$$

We will show that if

$$|x - x'| < \delta \text{ and } x, x' \in [a, b] \qquad (15.20)$$

then $|f(x) - f(x')| < \varepsilon$. From the triangle inequality one obtains:

$$|f(x) - f(x')| = |(f(x) - f(t_r)) + (f(t_r) - f(x'))| \le |f(x) - f(t_r)| + |f(t_r) - f(x')| \quad (15.21)$$

Since $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$, $x$ must be in one of these intervals. Suppose that $x \in I_{t_r}$ that is,

$$|x - t_r| \le \delta_{t_r}. \qquad (15.22)$$

From (11.18) with $t = t_r$,

$$|f(x) - f(t_r)| \le \frac{\varepsilon}{2}. \qquad (15.23)$$

From (11.20), (11.22), and the triangle inequality,

$$|x' - t_r| = |(x' - x) + (x - t_r)| \le |x' - x| + |x - t_r| < \delta + \delta_{t_r} \le 2\delta_{t_r}. \qquad (15.24)$$

Therefore, (11.18) with $t = t_r$ and $x$ replaced by $x'$ implies that

$$|f(x') - f(t_r)| \le \frac{\varepsilon}{2}. \qquad (15.25)$$
#-differentiable on a #-neighbourhood of $x_0$, it is reasonable to ask if $f'^#(x)$ is
#-differentiable at $x_0$. If so, we denote the #-derivative of $f'^#$ at $x_0$ by $f''^#(x_0)$. This is
the

second #-derivative of $f$ at $x_0$, and it is also denoted by $f^{(2)}(x_0)$. Continuing inductively, if $f^{(n-1)}$ is defined on a #-neighborhood of $x_0$, then the $n$-th #-derivative of $f$ at $x_0$, denoted by $f^{(n)}(x_0)$, where $n \in \mathbb{N}^\#$ or by $d^{n\#}f(x_0)/d^\#x^n$ is the #-derivative of $f^{(n-1)}(x)$ at $x_0$. For convenience we define the zeroth #-derivative of $f$ to be $f$ itself; thus $f^{(0)} = f$.

**Example15.1** If $n \in \mathbb{N}^\# \setminus \mathbb{N}$ is a positive hyperinteger and $f(x), x^n$ then

$$
\frac{f(x) - f(x_0)}{x - x_0} = \frac{x^n - x_0^n}{x - x_0} = \frac{x - x_0}{x - x_0} \left( \operatorname{Ext} - \sum_{k=0}^{n-1} x^{n-k-1} \right). \quad (15.30)
$$

Thus $f'^{\#}(x_0) = \#\text{-lim}_{x \to x_0} \text{Ext}-\sum_{k=0}^{n-1} x^{n-k-1} = nx^{n-1}$.

**Lemma 15.2.** If $f$ is $\#$-differentiable at $x_0$; then

$$
f(x) = f(x_0) + [f'^{\#}(x_0) + E(x)](x - x_0), \quad (15.31)
$$

where $E(x)$ is defined on a $\#$-neighborhood of $x_0$ and $\#\text{-lim}_{x \to x_0} E(x) = E(x_0) = 0$.

**Proof.** Define

$$
E(x) = \begin{cases} \frac{f(x) - f(x_0)}{x - x_0} - f'^{\#}(x_0) & x \in \text{Dom}(f) \text{ and } x \neq x_0 \\ 0 & x = x_0 \end{cases} \tag{15.32}
$$

Solving (15.32) for $f(x)$ yields (15.31) if $x \neq x_0$, and (15.31) is obvious if $x = x_0$.

Definition 15.29 implies that $\#\text{-lim}_{x \to x_0} E(x) = 0$. We defined $E(x_0) = 0$ to make $E(x)$
\#-continuous at $x_0$. Since the right side of (15.32) is \#-continuous at $x_0$, so is the left.
This yields the following theorem.

**Theorem 15.17.** If $f$ is $\#$-differentiable at $x_0$; then $f$ is $\#$-continuous at $x_0$.

**Theorem 15.18.** If $f$ and $g$ are $\#$-differentiable at $x_0$, then so are $f \pm g$ and $fg$ with

$$
(a) (f + g)^{\prime\#}(x_0) = f^{\prime\#}(x_0) + g^{\prime\#}(x_0);
$$

$$
(b) (f - g)^{\prime\#}(x_0) = f^{\prime\#}(x_0) - g^{\prime\#}(x_0);
$$

$$
(c) (fg)^{\prime\#}(x_0) = f^{\prime\#}(x_0)g(x_0) + f(x_0)g^{\prime\#}(x_0);
$$

(d) The quotient $f/g$ is $\#$-differentiable at $x_0$ if $g(x_0) \neq 0$ with

$$
\left(\frac{f}{g}\right)^{\prime\#}(x_0) = \frac{f^{\prime\#}(x_0)g(x_0) - g^{\prime\#}(x_0)f(x_0)}{[g(x_0)]^2}.
$$

(e) If $n \in \mathbb{N}^\#$ and $f_i, 1 \le i \le n$ are $\#$-differentiable at $x_0$, then so are $\text{Ext}-\sum_{i=1}^n f_i$ and

$$
\left( \text{Ext} - \sum_{i=1}^{n} f_i(x_0) \right)^{\prime\#} = \text{Ext} - \sum_{i=1}^{n} f_i^{\prime\#}(x_0).
$$

(f) If $n \in \mathbb{N}^\#$ and $f^{(n)}(x_0), g^{(n)}(x_0)$ exist, then so does $(f \times g)^{(n)}(x_0)$ and

$$
(fg)^{(n)}(x_0) = \text{Ext} - \sum_{i=0}^{n} \binom{n}{i} f^{(i)}(x_0) g^{(n-i)}(x_0).
$$

**Proof.** For the statements (a)-(d) the proof is straightforward. For the statements (e) and (f) immediately by hyper infinite induction.

**Theorem 15.19.** (The Chain Rule) Suppose that $g$ is $\#$-differentiable at $x_0$ and $f$
is $\#$-differentiable at $g(x_0)$. Then the composite function $h = f \circ g$ defined by
$h(x) = f(g(x))$ is $\#$-differentiable at $x_0$ with $h'^{\#}(x_0) = f'^{\#}(g(x_0))g'^{\#}(x_0)$.
Thus (11.25),(11.21) and (11.23) imply that $|f(x') - f(t_r)| \le \varepsilon/2$.

### 15.10. Monotonic External Functions $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$.

**Theorem 15.15.** If $f$ is monotonic and nonconstant on $[a, b]$, then $f$ is #-continuous on $[a, b]$ if and only if its range **range** ($f) = \{f(x)|x \in [a, b]\}$ is the #-closed interval with endpoints $f(a)$ and $f(b)$.

**Theorem 15.16.** Suppose that $f$ is increasing and #-continuous on $[a, b]$ and let $f(a) = c$ and $f(b) = d$. Then there is a unique function $g$ defined on $[c, d]$ such that

$$g(f(x)) = x, a \le x \le b, \qquad (15.26)$$

and

$$f(g(y)) = y, c \le y \le d. \qquad (15.27)$$

Moreover, $g$ is #-continuous and increasing on $[c, d]$:

The function $g$ of Theorem 15.16 is the inverse of $f$, denoted by $f^{-1}$. Since (15.26) and (15.27) are symmetric in $f$ and $g$, we can also regard $f$ as the inverse of $g$, and denote it by $g^{-1}$.

### 15.11. The #-derivative of a $\mathbb{R}_c^\#$-valued function $f: D \to \mathbb{R}_c^\#$.

A function $f: D \to \mathbb{R}_c^\#$, $D \subset \mathbb{R}_c^\#$ is differentiable at an #-interior point $x_0 \in D$ of its domain

$D \subset \mathbb{R}_c^\#$ if the difference quotient

$$\frac{f(x) - f(x_0)}{x - x_0}, x \neq x_0 \qquad (15.28)$$

approaches a #-limit as $x$ approaches $x_0$, in which case the #-limit is called the #-derivative

of $f$ at $x_0$, and is denoted by $f'''(x_0)$ or by $f'''(x_0)$ or by $d^\#f(x_0)/d^\#x$ i.e.,

$$d^\#f(x_0)/d^\#x \triangleq f'''(x_0) = \#\text{-lim}_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} \qquad (15.29)$$

If $f$ is defined on an #-open set $S \subset \mathbb{R}_c^\#$, we say that $f$ is #-differentiable on $S$ if $f$ is #-differentiable at every point of $S$. If $f$ is #-differentiable on $S$, then $f'''$ is a function on $S$.

We say that $f$ is #-continuously #-differentiable on $S$ if $f'''(x)$ is #-continuous on $S$. If $f$ is #-differentiable on a #-neighbourhood of $x_0$, it is reasonable to ask if $f'''(x)$ is #-differentiable at $x_0$. If so, we denote the #-derivative of $f'''$ at $x_0$ by $f'''(x_0)$. This is the

second #-derivative of $f$ at $x_0$, and it is also denoted by $f^{(2)\#}(x_0)$. Continuing inductively, if $f^{(n-1)\#}$ is defined on a #-neighborhood of $x_0$, then the $n$-th #-derivative of $f$ at $x_0$, denoted by $f^{(n)\#}(x_0)$, where $n \in \mathbb{N}^\#$ or by $d^{n\#}f(x_0)/d^\#x^n$ is the #-derivative of $f^{(n-1)\#}(x)$ at $x_0$. For convenience we define the zeroth #-derivative of $f$ to be $f$ itself; thus $f^{(0)\#} = f$.

**Example 15.1** If $n \in \mathbb{N}^\# \mathbb{N}$ is a positive hyperinteger and $f(x) = x^n$ then

$$\frac{f(x) - f(x_0)}{x - x_0} = \frac{x^n - x_0^n}{x - x_0} = \frac{x - x_0}{x - x_0} \left( \text{Ext} - \sum_{k=0}^{n-1} x^{n-k-1} \right). \qquad (15.30)$$

$$\text{Thus } f'''(x_0) = \#\text{-lim}_{x \to x_0} \text{Ext} - \sum_{k=0}^{n-1} x^{n-k-1} = nx^{n-1}.$$

**Lemma 15.2.** If $f$ is #-differentiable at $x_0$; then
**Definition 15.16.** If $f(x)$ is defined on $[x_0, b)$, the right-hand derivative of $f(x)$ at $x_0$ is defined to be

$$f_{+}^{\prime\#}(x_0) = \#\lim_{x \to x_0+} \frac{f(x) - f(x_0)}{x - x_0}, \qquad (15.33)$$

if the #-limit exists, while if $f$ is defined on $(a, x_0]$, the left-hand derivative of $f(x)$ at $x_0$ is defined to be

$$f_{-}^{\prime\#}(x_0) = \#\lim_{x \to x_0-} \frac{f(x) - f(x_0)}{x - x_0}, \qquad (15.34)$$

if the #-limit exists.

**Remark 15.2.** Note that $f(x)$ is #-differentiable at $x_0$ if and only if $f_{+}^{\prime\#}(x_0)$ and $f_{-}^{\prime\#}(x_0)$ exist and are equal, in which case $f^{\prime\#}(x_0) = f_{+}^{\prime\#}(x_0) = f_{-}^{\prime\#}(x_0)$.

**Definition 15.16' (1)** We say that $f$ is #-differentiable on the #-closed interval $[a, b]$ if $f$ is #-differentiable on the #-open interval $(a, b)$ and $f_{+}^{\prime\#}(a)$ and $f_{-}^{\prime\#}(b)$ both exist. (2) We say that $f$ is #-continuously #-differentiable on $[a, b]$ if $f$ is #-differentiable on $[a, b]$, $f^{\prime\#}$ is #-continuous on $(a, b)$, $f_{+}^{\prime\#}(a) = f^{\prime\#}(a +)$, and $f_{-}^{\prime\#}(b) = f^{\prime\#}(b +)$.

**Definition 15.17.** We say that $f(x_0)$ is a local extreme value of $f(x)$ if there is a $\delta > 0$, $\delta \approx 0$ such that $f(x) - f(x_0)$ does not change sign on

$$(x_0 - \delta x_0 + \delta) \cap \text{dom}(f). \qquad (15.35)$$

More specifically, $f(x_0)$ is a local maximum value of $f(x)$ if

$$f(x) \le f(x_0) \qquad (15.36)$$

or a local minimum value of $f(x)$ if

$$f(x) \ge f(x_0) \qquad (15.37)$$

for all $x \in (x_0 - \delta x_0 + \delta) \cap \text{dom}(f)$. The point $x_0$ is called a local extreme point of $f(x)$, or, more specifically, a local maximum or local minimum point of $f(x)$.

**Theorem 15.20.** If $f(x)$ is #-differentiable at a local extreme point $x_0 \in \text{dom}(f)$ then

$$f^{\prime\#}(x_0) = 0.$$

**Proof.** We will show that $x_0$ is not a local extreme point of $f$ if $f^{\prime\#}(x_0) \ne 0$. From Lemma 15.2 we get

$$\frac{f(x) - f(x_0)}{x - x_0} = f^{\prime\#}(x_0) + E(x), \qquad (15.37')$$

where $\#\lim_{x \to x_0} E(x) = 0$. Therefore, if $f^{\prime\#}(x_0) \ne 0$, there is a $\delta > 0$, $\delta \approx 0$, such that $|E(x)| < |f^{\prime\#}(x_0)|$, and the right side of (15.37') must have the same sign as $f^{\prime\#}(x_0)$ for $|x - x_0| < \delta$. Since the same is true of the left side, $f(x) - f(x_0)$ must change sign in every neighborhood of $x_0$ (since $x - x_0$ does). Therefore, neither (15.36) nor (15.37) can hold for all $x$ in any interval about $x_0$.

**Theorem 15.21. (Generalized Rolle's Theorem)** Suppose that:

(i) $f$ is #-continuous on the #-closed interval $[a, b]$,

(ii) $f$ is #-differentiable on the #-open interval $(a, b)$,

(iii) the set $V_{a,b} = \{f(x)|x \in [a,b]\}$ is admissible above and below and

(iv) $f(a) = f(b)$.

Then $f^{\prime\#}(c) = 0$ for some $c \in (a, b)$.

**Proof.** Since $f$ is #-continuous on $[a, b]$ and the set $V_{a,b} = \{f(x)|x \in [a,b]\}$ is
$$f(x) = f(x_0) + [f^{\prime\#}(x_0) + E(x)](x - x_0), \qquad (15.31)$$

where $E(x)$ is defined on a $\#$-neighborhood of $x_0$ and $\#\text{-lim}_{x\to x_0} E(x) = E(x_0) = 0$.

**Proof.** Define

$$E(x) = \begin{cases} \frac{f(x) - f(x_0)}{x - x_0} - f^{\prime\#}(x_0) & x \in \text{Dom}(f) \text{ and } x \neq x_0 \\ 0 & x = x_0 \end{cases} \qquad (15.32)$$

Solving (15.32) for $f(x)$ yields (15.31) if $x \neq x_0$, and (15.31) is obvious if $x = x_0$.

Definition 15.29 implies that $\#\text{-lim}_{x\to x_0} E(x) = 0$. We defined $E(x_0) = 0$ to make $E(x)$ $\#$-continuous at $x_0$. Since the right side of (15.32) is $\#$-continuous at $x_0$, so is the left. This yields the following theorem.

**Theorem 15.17.** If $f$ is $\#$-differentiable at $x_0$; then $f$ is $\#$-continuous at $x_0$.

**Theorem 15.18.** If $f$ and $g$ are $\#$-differentiable at $x_0$, then so are $f \pm g$ and $fg$ with

(a) $(f+g)^{\prime\#}(x_0) = f^{\prime\#}(x_0) + g^{\prime\#}(x_0)$;

(b) $(f-g)^{\prime\#}(x_0) = f^{\prime\#}(x_0) - g^{\prime\#}(x_0)$;

(c) $(fg)^{\prime\#}(x_0) = f^{\prime\#}(x_0)g(x_0) + f(x_0)g^{\prime\#}(x_0)$;

(d) The quotient $f/g$ is $\#$-differentiable at $x_0$ if $g(x_0) \neq 0$ with

$$\left(\frac{f}{g}\right)^{\prime\#}(x_0) = \frac{f^{\prime\#}(x_0)g(x_0) - g^{\prime\#}(x_0)f(x_0)}{[g(x_0)]^2}.$$

(e) If $n \in \mathbb{N}^\#$ and $f_i$, $1 \le i \le n$ are $\#$-differentiable at $x_0$, then so are $\text{Ext}-\sum_{i=1}^n f_i$ and

$$\left( \text{Ext} - \sum_{i=1}^{n} f_i(x_0) \right)^{\prime\#} = \text{Ext} - \sum_{i=1}^{n} f_i^{\prime\#}(x_0).$$

(f) If $n \in \mathbb{N}^\#$ and $f^{(n)\#}(x_0)$, $g^{(n)\#}(x_0)$ exist, then so does $(f \times g)^{(n)\#}(x_0)$ and

$$(fg)^{(n)\#}(x_0) = \text{Ext} - \sum_{i=0}^{n} \binom{n}{i} f^{(i)\#}(x_0)g^{(n-i)\#}(x_0).$$

**Proof.** For the statements (a)-(d) the proof is straightforward. For the statements (e) and (f) immediately by hyper infinite induction.

**Theorem 15.19.** (The Chain Rule) Suppose that $g$ is $\#$-differentiable at $x_0$ and $f$ is $\#$-differentiable at $g(x_0)$. Then the composite function $h = f \circ g$ defined by $h(x) = f(g(x))$ is $\#$-differentiable at $x_0$ with $h^{\prime\#}(x_0) = f^{\prime\#}(g(x_0))g^{\prime\#}(x_0)$.

**Definition 15.16.** If $f(x)$ is defined on $[x_0, b)$, the right-hand derivative of $f(x)$ at $x_0$ is defined to be

$$f_{+}^{\prime\#}(x_0) = \#\text{-lim}_{x\to x_0} \frac{f(x) - f(x_0)}{x - x_0}, \qquad (15.33)$$

if the $\#$-limit exists, while if $f$ is defined on $(a, x_0]$, the left-hand derivative of $f(x)$ at $x_0$ is defined to be

$$f_{-}^{\prime\#}(x_0) = \#\text{-lim}_{x\to x_0} \frac{f(x) - f(x_0)}{x - x_0}, \qquad (15.34)$$

if the $\#$-limit exists.

**Remark 15.2.** Note that $f(x)$ is $\#$-differentiable at $x_0$ if and only if $f_+^{\prime\#}(x_0)$ and $f_-^{\prime\#}(x_0)$ exist and are equal, in which case $f^{\prime\#}(x_0) = f_-^{\prime\#}(x_0) = f_+^{\prime\#}(x_0)$.

**Definition 15.16' (1)** We say that $f$ is $\#$-differentiable on the $\#$-closed interval $[a, b]$ if $f$ is $\#$-differentiable on the $\#$-open interval $(a, b)$ and $f_+^{\prime\#}(a)$ and $f_-^{\prime\#}(b)$ both exist.

(2) We say that $f$ is $\#$-continuously $\#$-differentiable on $[a, b]$ if $f$ is $\#$-differentiable on
admissible above and below, $f$ attains a maximum and a minimum value on $[a,b]$ (Theorem 15.12). If these two extreme values are the same, then $f$ is constant on $(a,b)$, so $f^{\#}(x) = 0$ for all $x \in [a,b]$. If the extreme values differ, then at least one must be attained at some point $c$ in the \#-open interval $(a,b)$, and $f^{\#}(c) = 0$, by Theorem 15.20.

**Theorem 15.22. (Intermediate Value Theorem for \#-Derivatives)** Suppose that:

(i) $f(x)$ is \#-differentiable on $[a,b]$,

(ii) the set $V_{a,b}[f] = \{f(x)|x \in [a,b]\}$ is admissible above and below,

(iii) $f^{\#}(a) \neq f^{\#}(b)$ and $f^{\#}(a) < \mu < f^{\#}(b)$. Then

$f^{\#}(c) = \mu$ for some $c \in (a,b)$.

**Proof.** Suppose first that: (1) $f^{\#}(a) < \mu < f^{\#}(b)$ and define $g(x) = f(x) - \mu x$. Then (2) $g^{\#}(x) = f^{\#}(x) - \mu, a \le x \le b$, and (1) implies that: (3) $g^{\#}(a) < 0$ and $g^{\#}(b) > 0$. Notice (ii) implies that $V_{a,b}[g] = \{x(x)|x \in [a,b]\}$ is admissible above and below. Since $g$ is \#-continuous on $[a,b]$, $g$ attains a minimum at some point $c \in [a,b]$. Lemma 15.2 and (3) implies that there is a $\delta > 0, \delta \approx 0$, such that $g(x) < g(a), a < x < a + \delta$ and $g(x) < g(b), b - \delta < x < b$, and therefore $c \neq a$ and $c \neq b$. Hence, $a < c < b$, and therefore $g^{\#}(c) = 0$ by Theorem 11.20. From (2) $f^{\#}(c) = \mu$. The proof for the case where $f^{\#}(b) < \mu < f^{\#}(a)$ can be obtained by applying this result to $-f(x)$.

**Theorem 15.23. (Generalized Mean Value Theorem)** Assume that:

(i) $f$ and $g$ are \#-continuous on the \#-closed interval $[a,b]$ and \#-differentiable on the open interval $(a,b)$, (ii) the set $V_{a,b}[f]$ and $V_{a,b}[g]$ are admissible above and below,

(iii) let $h(x) = [g(b) - g(a)]f(x) - [f(b) - f(a)]g(x)$, the set $V_{a,b}[h]$ admissible above and below, then

$$[g(b) - g(a)]f^{\#}(c) = [f(b) - f(a)]g^{\#}(c) \quad (15.38)$$

for some $c \in (a,b)$.

**Proof.** The function $h(x) = [g(b) - g(a)]f(x) - [f(b) - f(a)]g(x)$ is \#-continuous on $[a,b]$ and \#-differentiable on $(a,b)$, and $h(a) = h(b) = g(b)f(a) - f(b)g(a)$. Note that the set $V_{a,b}[h]$ is admissible above and below. Therefore, Rolle's theorem (Theorem 11.21) implies that $h^{\#}(c) = 0$ for some $c \in (a,b)$. Since

$h^{\#}(c) = [g(b) - g(a)]f^{\#}(c) - [f(b) - f(a)]g^{\#}(c)$, this implies Eq. (15.38).

The following special case of Theorem 15.23 is important enough to be stated separately.

**Theorem 15.24. (Mean Value Theorem)** Assume that: (i) $f$ is \#-continuous on the \#-closed interval $[a,b]$, (ii) \#-differentiable on the \#-open interval $(a,b)$, (iii) the set $V_{a,b}[f]$ is admissible above and below, then

$$f^{\#}(c) = \frac{f(b) - f(a)}{b - a} \quad (15.39)$$

for some $c \in (a,b)$.

**Proof.** Apply Theorem 15.24 with $g(x) = x$.

$$h(x) = [b - a]f(x) - [f(b) - f(a)]x$$

$$h(a) = h(b) = bf(a) - f(b)g(a)$$

**Remark 15.3.** Assume that the set $V_{a,b}[f]$ is admissible above and below. If $f$ is \#-differentiable on $(a,b)$ and $x_1, x_2 \in (a,b)$ then $f$ is \#-continuous on the \#-closed interval with endpoints $x_1$ and $x_2$ and \#-differentiable on its interior. Hence, the mean
$[a,b]$, $f'^{\#}$ is #-continuous on $(a,b)$, $f_+^{\#}(a) = f'^{\#}(a+)$, and $f_-^{\#}(b) = f'^{\#}(b+)$.

**Definition 15.17.** We say that $f(x_0)$ is a local extreme value of $f(x)$ if there is a $\delta > 0$, $\delta \approx 0$ such that $f(x) - f(x_0)$ does not change sign on

$$ (x_0 - \delta x_0 + \delta) \cap \mathrm{dom}(f). \qquad (15.35) $$

More specifically, $f(x_0)$ is a local maximum value of $f(x)$ if

$$ f(x) \leq f(x_0) \qquad (15.36) $$

or a local minimum value of $f(x)$ if

$$ f(x) \geq f(x_0) \qquad (15.37) $$

for all $x \in (x_0 - \delta x_0 + \delta) \cap \mathrm{dom}(f)$. The point $x_0$ is called a local extreme point of $f(x)$, or, more specifically, a local maximum or local minimum point of $f(x)$.

**Theorem 15.20.** If $f(x)$ is #-differentiable at a local extreme point $x_0 \in \mathrm{dom}(f)$ then

$$ f'^{\#}(x_0) = 0. $$

**Proof.** We will show that $x_0$ is not a local extreme point of $f$ if $f'^{\#}(x_0) \neq 0$. From Lemma 15.2 we get

$$ \frac{f(x) - f(x_0)}{x - x_0} = f'^{\#}(x_0) + E(x), \qquad (15.37') $$

where $\#\lim_{x \to x_0} E(x) = 0$. Therefore, if $f'^{\#}(x_0) \neq 0$, there is a $\delta > 0$, $\delta \approx 0$, such that $|E(x)| < |f'^{\#}(x_0)|$, and the right side of (15.37') must have the same sign as $f'^{\#}(x_0)$ for $|x - x_0| < \delta$. Since the same is true of the left side, $f(x) - f(x_0)$ must change sign in every neighborhood of $x_0$ (since $x - x_0$ does). Therefore, neither (15.36) nor (15.37) can hold for all $x$ in any interval about $x_0$.

**Theorem 15.21. (Generalized Rolle's Theorem)** Suppose that:

(i) $f$ is #-continuous on the #-closed interval $[a,b]$,

(ii) $f$ is #-differentiable on the #-open interval $(a,b)$,

(iii) the set $V_{a,b} = \{f(x)|x \in [a,b]\}$ is admissible above and below and

(iv) $f(a) = f(b)$.

Then $f'^{\#}(c) = 0$ for some $c \in (a,b)$.

**Proof.** Since $f$ is #-continuous on $[a,b]$ and the set $V_{a,b} = \{f(x)|x \in [a,b]\}$ is admissible above and below, $f$ attains a maximum and a minimum value on $[a,b]$ (Theorem 15.12). If these two extreme values are the same, then $f$ is constant on $(a,b)$, so $f'^{\#}(x) = 0$ for all $x \in [a,b]$. If the extreme values differ, then at least one must be attained at some point $c$ in the #-open interval $(a,b)$, and $f'^{\#}(c) = 0$, by Theorem 15.20.

**Theorem 15.22. (Intermediate Value Theorem for #-Derivatives)** Suppose that:

(i) $f(x)$ is #-differentiable on $[a,b]$,

(ii) the set $V_{a,b}[f] = \{f(x)|x \in [a,b]\}$ is admissible above and below,

(iii) $f'^{\#}(a) \neq f'^{\#}(b)$ and $f'^{\#}(a) < \mu < f'^{\#}(b)$. Then

$f'^{\#}(c) = \mu$ for some $c \in (a,b)$.

**Proof.** Suppose first that: (1) $f'^{\#}(a) < \mu < f'^{\#}(b)$ and define $g(x) = f(x) - \mu x$.

Then (2) $g'^{\#}(x) = f'^{\#}(x) - \mu, a \le x \le b$, and (1) implies that: (3) $g'^{\#}(a) < 0$ and $g'^{\#}(b) > 0$. Notice (ii) implies that $V_{a,b}[g] = \{x(x)|x \in [a,b]\}$ is admissible above and below. Since $g$ is #-continuous on $[a,b]$, $g$ attains a minimum at some point
value theorem (Theorem 15.24) implies that

$$f(x_2) - f(x_1) = f^{\#}(c)(x_2 - x_1). \qquad (15.39')$$

for some $c$ between $x_1$ and $x_2$. (This is true whether $x_1 < x_2$ or $x_2 < x_1$.) The next three theorems follow from (11.39').

**Theorem 15.25.** Assume that the set $V_{a,b}[f]$ is admissible above and below. If $f^{\#}(x) = 0$ for all $x \in (a, b)$, then $f$ is constant on $(a, b)$.

**Remark 15.4.**

**Theorem 15.26.** If $f^{\#}(x)$ exists for all $x \in (a, b)$ and does not change sign on $(a, b)$, then $f(x)$ is monotonic on $(a, b)$ increasing, nondecreasing, decreasing, or nonincreasing as: (i) $f^{\#}(x) > 0$, (ii) $f^{\#}(x) \ge 0$, (iii) $f^{\#}(x) < 0$, (iv) $f^{\#}(x) \le 0$, respectively, for all $x \in (a, b)$.

**Theorem 15.27.** If $|f^{\#}(x)| < M, a < x < b$ then
$$|f(x) - f(x')| \le M|x - x'|, \qquad (15.40)$$

where $x, x' \in (a, b)$.

**Definition 15.18.** A function that satisfies an inequality like (15.40) for all $x$ and $x'$ in an interval is said to satisfy a Lipschitz condition on the interval.

**Theorem 15.28.** (Generalized L'Hospital's Rule) Suppose that $f$ and $g$ are $\#$-differentiable and $g^{\#}$ has no zeros on $(a, b)$. Let $\#\lim_{x \to \#} b-f(x) = \#\lim_{x \to \#} b-g(x)$ or $\#\lim_{x \to \#} b-f(x) = \pm\infty$ and $\#\lim_{x \to \#} b-g(x) = \pm\infty$ and suppose that

$$\# - \lim_{x \to \#} b - \frac{f^{\#}(x)}{g^{\#}(x)} = L, \qquad (15.41)$$

where $L \in \mathbb{R}_c^\#$ or $L = \pm\infty$. Then

$$\# - \lim_{x \to \#} b - \frac{f(x)}{g(x)} = L, \qquad (15.42)$$

As we saw above in Lemma 15.2 if $f$ is $\#$-differentiable at $x_0$; then

$$f(x) = f(x_0) + f^{\#}(x_0)(x - x_0) + E(x)(x - x_0), \qquad (15.43)$$

where $\#\lim_{x \to \#} x_0 E(x) = 0$. To generalize this result, we first restate it: the polynomial $P_1(x) = f(x_0) + f^{\#}(x_0)(x - x_0)$ which is of degree $\le 1$ and satisfies $P_1(x_0) = f(x_0)$, $P_1^{\#}(x) = f^{\#}(x_0)$, approximates $f(x)$ so well near $x_0$ such that

$$\#\lim_{x \to \#} x_0 \frac{f(x) - P_1(x)}{x - x_0} = 0. \qquad (15.44)$$

Now suppose that $f$ has $n$ $\#$-derivatives at $x_0$ and $P_n(x)$ is the polynomial of degree $n \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$P_n^{(r)\#}(x_0) = f^{(r)\#}(x_0), 0 \le r \le n. \qquad (15.45)$$

Since $P_n(x)$ is a polynomial of hyperfinite degree $n$, it can be written as

$$P_n(x) = \mathrm{Ext} - \sum_{i=0}^{n} a_i (x - x_0)^i \qquad (15.46)$$

where $a_0, \dots, a_n \in \mathbb{R}_c^\#$ are constants. Differentiating (11.46) gives $P_n^{(r)\#}(x_0) = r!a_r$, $0 \le r \le n$, so (15.45) determines $a_r$ uniquely as $a_r = f^{(r)\#}(x_0)/r!$, $0 \le r \le n$. Therefore,
c ∈ [a,b]. Lemma 15.2 and (3) implies that there is a δ > 0, δ ≈ 0, such that $g(x) < g(a), a < x < a + δ$ and $g(x) < g(b), b - δ < x < b$, and therefore $c \neq a$ and $c \neq b$. Hence, $a < c < b$, and therefore $g'^#(c) = 0$ by Theorem 11.20. From (2) $f'^#(c) = μ$. The proof for the case where $f'^#(b) < μ < f'^#(a)$ can be obtained by applying this result to $-f(x)$.

**Theorem 15.23. (Generalized Mean Value Theorem)** Assume that:
(i) $f$ and $g$ are #-continuous on the #-closed interval $[a,b]$ and #-differentiable on the open interval $(a,b)$, (ii) the set $V_{a,b}[f]$ and $V_{a,b}[g]$ are admissible above and below, (iii) let $h(x) = [g(b) - g(a)]f(x) - [f(b) - f(a)]g(x)$, the set $V_{a,b}[h]$ admissible above and below, then

$$[g(b) - g(a)]f'^{\#}(c) = [f(b) - f(a)]g'^{\#}(c) \qquad (15.38)$$

for some $c \in (a,b)$.

**Proof.** The function $h(x) = [g(b) - g(a)]f(x) - [f(b) - f(a)]g(x)$ is #-continuous on $[a,b]$ and #-differentiable on $(a,b)$, and $h(a) = h(b) = g(b)f(a) - f(b)g(a)$. Note that the set $V_{a,b}[h]$ is admissible above and below. Therefore, Rolle's theorem (Theorem 11.21) implies that $h'^{(c)} = 0$ for some $c \in (a,b)$. Since

$$h'^{(c)} = [g(b) - g(a)]f'^{(c)} - [f(b) - f(a)]g'^{(c)}, \text{ this implies Eq. (15.38).}$$

The following special case of Theorem 15.23 is important enough to be stated separately.

**Theorem 15.24. (Mean Value Theorem)** Assume that: (i) $f$ is #-continuous on the #-closed interval $[a,b]$, (ii) #-differentiable on the #-open interval $(a,b)$, (iii) the set $V_{a,b}[f]$ is admissible above and below, then

$$f'^{(c)}(c) = \frac{f(b) - f(a)}{b - a} \qquad (15.39)$$

for some $c \in (a,b)$.

**Proof.** Apply Theorem 15.24 with $g(x) = x$.

$$h(x) = [b - a]f(x) - [f(b) - f(a)]x$$

$$h(a) = h(b) = bf(a) - f(b)g(a)$$

**Remark 15.3.** Assume that the set $V_{a,b}[f]$ is admissible above and below. If $f$ is #-differentiable on $(a,b)$ and $x_1, x_2 \in (a,b)$ then $f$ is #-continuous on the #-closed interval with endpoints $x_1$ and $x_2$ and #-differentiable on its interior. Hence, the mean value theorem (Theorem 15.24) implies that

$$f(x_2) - f(x_1) = f'^{(c)}(c)(x_2 - x_1). \qquad (15.39')$$

for some $c$ between $x_1$ and $x_2$. (This is true whether $x_1 < x_2$ or $x_2 < x_1$.) The next three theorems follow from (11.39').

**Theorem 15.25.** Assume that the set $V_{a,b}[f]$ is admissible above and below. If $f'^{(c)}(x) = 0$ for all $x \in (a,b)$, then $f$ is constant on $(a,b)$.

**Remark 15.4.**

**Theorem 15.26.** If $f'^{(c)}(x)$ exists for all $x \in (a,b)$ and does not change sign on $(a,b)$, then $f(x)$ is monotonic on $(a,b)$ increasing, nondecreasing, decreasing, or nonincreasing as: (i) $f'^{(c)}(x) > 0$, (ii) $f'^{(c)}(x) \ge 0$, (iii) $f'^{(c)}(x) < 0$, (iv) $f'^{(c)}(x) \le 0$, respectively, for all $x \in (a,b)$.

**Theorem 15.27.** If $|f'^{(c)}(x)| < M, a < x < b$ then

$$|f(x) - f(x')| \le M|x - x'|, \qquad (15.40)$$
$$P_n(x) = \mathrm{Ext}-\sum_{r=0}^{n} \frac{(x-x_0)^r f^{(r)}(x_0)}{r!}. \qquad (15.47)$$

We call $P_n(x)$ the n-th Taylor hyper polynomial of $f(x)$ about $x_0$

**Theorem 15.29.** If $f^{(n)}(x_0)$ exists for some hyper integer $n \in \mathbb{N}\#\mathbb{N}$ and $P_n(x)$ is the n-th Taylor hyper polynomial of $f$ about $x_0$, then

$$\#\lim_{x \to \# x_0} \frac{f(x) - P_n(x)}{(x - x_0)^n} = 0. \qquad (15.48)$$

**Theorem 15.30.** (Generalized Taylor's Theorem) Suppose that $f^{(n+1)}(x)$ exists on an $\#$-open interval $I$ about $x_0$, and let $x \in I$. Then the remainder $R_n(x) = f(x) - P_n(x)$ can be written as

$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - x_0)^n, \qquad (15.49)$$

where $c$ depends upon $x$ and is between $x$ and $x_0$.

## 15.12. The Riemann integral of a $\mathbb{R}_c^\#$-valued external function $f(x)$.

The Riemann integral is defined as $\#$-limit of Riemann hyperfinite sums of functions with respect to tagged partitions of an interval $[a, b] \subset \mathbb{R}_c^\#$ A tagged hyperfinite partition $P$ of a closed interval $[a, b]$ on the real line is a hyperfinite sequence

$$a = x_0 \le t_1 \le x_1 \le t_2 \le x_2 \le \dots \le x_{n-1} \le t_n \le x_n = b, \qquad (15.50)$$

where $n \in \mathbb{N}\#\mathbb{N}$. This partitions the interval $[a, b]$ into $n$ sub-intervals $[x_{i-1}, x_i]$ indexed by $i \in \mathbb{N}^\#$, each of which is "tagged" with a distinguished point $t_i \in [x_{i-1}, x_i]$. Thus, any set of $n+1 \in \mathbb{N}\#\mathbb{N}$ points satisfying (15.50) defines a partition $P$ of $[a, b]$, which we denote by $P = \{x_0, x_1, \dots, x_n\}$. A Riemann hyperfinite sum of a function $f$ with respect to such a tagged hyperfinite partition is defined as

$$I_n = \sum_{i=1}^{n} f(t_i) \Delta_i, \qquad (15.51)$$

where $n \in \mathbb{N}\#\mathbb{N}$. Thus each term of the sum (15.51) is the area of a rectangle with height equal to the function value at the distinguished point of the given sub-interval, and width the same as the width of sub-interval, $\Delta_i = x_i - x_{i-1}$. The mesh($P$) of such a tagged partition is the width of the largest sub-interval formed by the partition, $\max_{i=1...n} \Delta_i$.

**Definition 15.19.** The Riemann integral of a function $f$ over the interval $[a, b]$ is equal to $I$ if for every $\varepsilon > 0$, $\varepsilon \approx 0$ there exists $\delta > 0$, $\delta \approx 0$ such that for any partition with distinguished points on $[a, b]$ whose mesh is less than $\delta$.

### Upper and Lower Integrals.

**Definition 15.20.** $f$ is bounded (hyperbounded) on $[a, b]$ and $P = \{x_0, x_1, \dots, x_n\}$ is a hyperfinite partition of $[a, b]$, let

$$M_j = \sup_{x_{j-1} \le x \le x_j} f(x) \qquad (15.52)$$

and

$$m_j = \inf_{x_{j-1} \le x \le x_j} f(x) \qquad (15.53)$$

The upper external hyperfinite sum of $f$ over $P$ is
where $x, x' \in (a, b)$.

**Definition 15.18.** A function that satisfies an inequality like (15.40) for all $x$ and $x'$ in an interval is said to satisfy a Lipschitz condition on the interval.

**Theorem 15.28.** (Generalized L'Hospital's Rule) Suppose that $f$ and $g$ are \#-differentiable and $g'^{\#}$ has no zeros on $(a, b)$. Let $\#\lim_{x \to \#} b-f(x) = \#\lim_{x \to \#} b-g(x)$ or $\#\lim_{x \to \#} b-f(x) = \pm\infty^{\#}$ and $\#\lim_{x \to \#} b-g(x) = \pm\infty^{\#}$ and suppose that

$$ \# - \lim_{x \to \#} b - \frac{f'^{\#}(x)}{g'^{\#}(x)} = L, \qquad (15.41) $$

where $L \in \mathbb{R}_c^\#$ or $L = \pm\infty^{\#}$. Then

$$ \# - \lim_{x \to \#} b - \frac{f(x)}{g(x)} = L, \qquad (15.42) $$

As we saw above in Lemma 15.2 if $f$ is \#-differentiable at $x_0$; then

$$ f(x) = f(x_0) + f'^{\#}(x_0)(x - x_0) + E(x)(x - x_0), \qquad (15.43) $$

where $\#\lim_{x \to \#} x_0 E(x) = 0$. To generalize this result, we first restate it: the polynomial $P_1(x) = f(x_0) + f'^{\#}(x_0)(x - x_0)$ which is of degree $\le 1$ and satisfies $P_1(x_0) = f(x_0)$, $P_1^{\#}(x) = f'^{\#}(x_0)$, approximates $f(x)$ so well near $x_0$ such that

$$ \# - \lim_{x \to \#} x_0 \frac{f(x) - P_1(x)}{x - x_0} = 0. \qquad (15.44) $$

Now suppose that $f$ has $n$ \#-derivatives at $x_0$ and $P_n(x)$ is the polynomial of degree $n \in \mathbb{N}^\# \setminus \mathbb{N}$ such that

$$ P_n^{(r)\#}(x_0) = f^{(r)\#}(x_0), \quad 0 \le r \le n. \qquad (15.45) $$

Since $P_n(x)$ is a polynomial of hyperfinite degree $n$, it can be written as

$$ P_n(x) = \mathrm{Ext} - \sum_{i=0}^{n} a_i (x - x_0)^i \qquad (15.46) $$

where $a_0, \dots, a_n \in \mathbb{R}_c^\#$ are constants. Differentiating (11.46) gives $P_n^{(r)\#}(x_0) = r!a_r$, $0 \le r \le n$, so (15.45) determines $a_r$ uniquely as $a_r = f^{(r)\#}(x_0)/r!$, $0 \le r \le n$. Therefore,

$$ P_n(x) = \mathrm{Ext} - \sum_{r=0}^{n} \frac{(x - x_0)^r f^{(r)\#}(x_0)}{r!}. \qquad (15.47) $$

We call $P_n(x)$ the $n$-th Taylor hyper polynomial of $f(x)$ about $x_0$

**Theorem 15.29.** If $f^{(n)\#}(x_0)$ exists for some hyper integer $n \in \mathbb{N}^\# \setminus \mathbb{N}$ and $P_n(x)$ is the $n$-th Taylor hyper polynomial of $f$ about $x_0$, then

$$ \# - \lim_{x \to \#} x_0 \frac{f(x) - P_n(x)}{(x - x_0)^n} = 0. \qquad (15.48) $$

**Theorem 15.30.** (Generalized Taylor's Theorem) Suppose that $f^{(n+1)\#}(x)$ exists on an \#-open interval $I$ about $x_0$, and let $x \in I$. Then the remainder $R_n(x) = f(x) - P_n(x)$ can be written as

$$ R_n(x) = \frac{f^{(n+1)\#}(c)}{(n+1)!} (x - x_0)^n, \qquad (15.49) $$

where $c$ depends upon $x$ and is between $x$ and $x_0$.

## 15.12. The Riemann integral of a $\mathbb{R}_c^\#$-valued external
$$S(P) = \mathrm{Ext} - \sum_{j=1}^{n} M_j(x_j - x_{j-1}) \qquad (15.54)$$

and the upper external integral of $f$ over $[a, b]$, denoted by

$$\mathrm{Ext} - \int_{a}^{b} f(x) d^{\#}x \qquad (15.55)$$

is the infimum of all hyperfinite upper sums. The lower external hyperfinite sum of $f$ over $P$ is

$$s(P) = \mathrm{Ext} - \sum_{j=1}^{n} m_j(x_j - x_{j-1}) \qquad (15.56)$$

and the lower external integral of $f$ over $[a, b]$, denoted by

$$\mathrm{Ext} - \int_{a}^{b} f(x) d^{\#}x \qquad (15.57)$$

is the supremum of all lower hyperfinite sums. If $m \le f(x) \le M$ for all $x \in [a, b]$, then

$$m(b-a) \le s(P) \le S(P) \le M(b-a) \qquad (15.58)$$

for every hyperfinite partition $P$; thus, the set of upper hyperfinite sums of $f$ over all partitions $P$ of $[a, b]$ is bounded, as is the set of lower hyperfinite sums. Therefore, Theorems 15.3 and 15.8 imply that: if the quantity (15.55) and (15.57) exist then both are unique, and satisfy the inequalities

$$m(b-a) \le \mathrm{Ext} - \int_{a}^{b} f(x) d^{\#}x \le M(b-a) \qquad (15.59)$$

and

$$m(b-a) \le \mathrm{Ext} - \int_{a}^{b} f(x) d^{\#}x \le M(b-a). \qquad (15.60)$$

**Theorem 15.31.** Let $f$ be bounded on $[a, b]$, and let $P$ be a hyperfinite partition of $[a, b]$. Then (i) The upper hyperfinite sum $S(P)$ of $f$ over $P$ is the supremum of the set of all hyperfinite Riemann sums of $f$ over $P$.

(ii) The lower hyperfinite sum $s(P)$ of $f$ over $P$ is the infimum of the set of all hyperfinite Riemann sums of $f$ over $P$.

**Proof (a)** If $P = \{x_0, x_1, \dots, x_n\}$, then $S(P) = \mathrm{Ext} - \sum_{j=1}^{n} M_j(x_j - x_{j-1})$ where

$$M_j = \sup_{x_{j-1} \le x \le x_j} f(x).$$

An arbitrary hyperfinite Riemann sum of $f$ over $P$ is of the following form

$$\sigma = \mathrm{Ext} - \sum_{j=1}^{n} f(c_j)(x_j - x_{j-1}), \qquad (15.61)$$

where $x_{j-1} \le c_j \le x_j$. Since $f(c_j) \le M_j$, it follows that $\sigma \le S(P)$. Now let $\varepsilon > 0$, $\varepsilon \approx 0$ and choose $\bar{c}_j \in [x_{j-1}, x_j]$ so that

$$f(\bar{c}_j) > M_j - \frac{\varepsilon}{n(x_j - x_{j-1})}, \qquad (15.62)$$

where $1 \le j \le n \in \mathbb{N}^*$. The hyperfinite Riemann sum $\bar{\sigma}$ produced in this way is

$$\bar{\sigma} = \mathrm{Ext} - \sum_{j=1}^{n} f(\bar{c}_j)(x_j - x_{j-1}) > \mathrm{Ext} - \sum_{j=1}^{n} \left[ M_j - \frac{\varepsilon}{n(x_j - x_{j-1})} \right] (x_j - x_{j-1}) = S(P) - \varepsilon. \qquad (15.63)$$

Now Theorem 15.3 implies that $S(P)$ is the supremum of the set of hyperfinite
function $f(x)$.

The Riemann integral is defined as #-limit of Riemann hyperfinite sums of functions with respect to tagged partitions of an interval $[a, b] \subset \mathbb{R}_c^#$. A tagged hyperfinite partition $P$ of a closed interval $[a, b]$ on the real line is a hyperfinite sequence

$$a = x_0 \le t_1 \le x_1 \le t_2 \le x_2 \le \dots \le x_{n-1} \le t_n \le x_n = b, \qquad (15.50)$$

where $n \in \mathbb{N}^# \setminus \mathbb{N}$. This partitions the interval $[a, b]$ into $n$ sub-intervals $[x_{i-1}, x_i]$ indexed by $i \in \mathbb{N}^#$, each of which is "tagged" with a distinguished point $t_i \in [x_{i-1}, x_i]$. Thus, any set of $n+1 \in \mathbb{N}^# \setminus \mathbb{N}$ points satisfying (15.50) defines a partition $P$ of $[a, b]$, which we denote by $P = \{x_0, x_1, \dots, x_n\}$. A Riemann hyperfinite sum of a function $f$ with respect to such a tagged hyperfinite partition is defined as

$$I_n = \sum_{i=1}^{n} f(t_i) \Delta_i, \qquad (15.51)$$

where $n \in \mathbb{N}^# \setminus \mathbb{N}$. thus each term of the sum (15.51) is the area of a rectangle with height equal to the function value at the distinguished point of the given sub-interval, and width the same as the width of sub-interval, $\Delta_i = x_i - x_{i-1}$. The mesh($P$) of such a tagged partition is the width of the largest sub-interval formed by the partition, $\max_{i=1...n} \Delta_i$.

**Definition 15.19.** The Riemann integral of a function $f$ over the interval $[a, b]$ is equal to $I$ if for every $\varepsilon > 0$, $\varepsilon \approx 0$ there exists $\delta > 0$, $\delta \approx 0$ such that for any partition with distinguished points on $[a, b]$ whose mesh is less than $\delta$.

**Upper and Lower Integrals.**

**Definition 15.20.** $f$ is bounded (hyperbounded) on $[a, b]$ and $P = \{x_0, x_1, \dots, x_n\}$ is a hyperfinite partition of $[a, b]$, let

$$M_j = \sup_{x_{j-1} \le x \le x_j} f(x) \qquad (15.52)$$

and

$$m_j = \inf_{x_{j-1} \le x \le x_j} f(x) \qquad (15.53)$$

The upper external hyperfinite sum of $f$ over $P$ is

$$S(P) = \mathrm{Ext} - \sum_{j=1}^{n} M_j(x_j - x_{j-1}) \qquad (15.54)$$

and the upper external integral of $f$ over $[a, b]$, denoted by

$$\mathrm{Ext} - \int_{a}^{b} f(x) d^{\#} x \qquad (15.55)$$

is the infimum of all hyperfinite upper sums. The lower external hyperfinite sum of $f$ over $P$ is

$$s(P) = \mathrm{Ext} - \sum_{j=1}^{n} m_j(x_j - x_{j-1}) \qquad (15.56)$$

and the lower external integral of $f$ over $[a, b]$, denoted by

$$\mathrm{Ext} - \int_{a}^{b} f(x) d^{\#} x. \qquad (15.57)$$

is the supremum of all lower hyperfinite sums. If $m \le f(x) \le M$ for all $x \in [a, b]$, then

$$m(b-a) \le s(P) \le S(P) \le M(b-a) \qquad (15.58)$$

for every hyperfinite partition $P$; thus, the set of upper hyperfinite sums of $f$ over all partitions $P$ of $[a, b]$ is bounded, as is the set of lower hyperfinite sums. Therefore,
Riemann sums of $f$ over $P$.

## 15.12. The Riemann–Stieltjes Integral of a $\mathbb{R}^\#$-valued external function $f(x)$.

**Definition 15.21.** Let $f$ and $g$ be defined on $[a, b]$. We say that $f$ is Riemann–Stieltjes integrable with respect to $g$ on $[a, b]$, if there is a number $L \in \mathbb{R}^\#$ with the following property: For every $\varepsilon > 0, \varepsilon \approx 0$, there is a $\delta > 0, \delta \approx 0$ such that

$$ \left| \operatorname{Ext} - \sum_{j=1}^{n} f(c_j) [g(x_j) - g(x_{j-1})] - L \right| < \varepsilon \qquad (15.64) $$

provided only that $P = \{x_0, x_1, \dots, x_n\}, n \in \mathbb{N}^*\mathbb{N}$ is a hyperfinite partition of $[a, b]$ such that $\|P\| < \delta$ and $x_{j-1} \le c_j \le x_j, j \in n$. In this case, we say that $L$ is the external Riemann–Stieltjes integral of $f$ with respect to $g$ over $[a, b]$, and write

$$ \operatorname{Ext} - \int_a^b f(x) d^\# g(x) = L. \qquad (15.65) $$

## 15.13 Existence of the integral of a $\mathbb{R}^\#$-valued external function $f(x)$.

**Lemma 15.3** Suppose that

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

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

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

and

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

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

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

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

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

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

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

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

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

**Theorem 15.35.** A bounded (hyperbounded) function $f$ is integrable on $[a, b]$ if and only if
Theorems 15.3 and 15.8 imply that: if the quantity (15.55) and (15.57) exist then both are unique, and satisfy the inequalities

$$m(b-a) \leq \mathrm{Ext}-\int_{a}^{b} f(x) d^{\#}x \leq M(b-a) \quad (15.59)$$

and

$$m(b-a) \leq \mathrm{Ext}-\int_{a}^{b} f(x) d^{\#}x \leq M(b-a). \quad (15.60)$$

**Theorem 15.31.** Let $f$ be bounded on $[a, b]$, and let $P$ be a hyperfinite partition of $[a, b]$. Then (i) The upper hyperfinite sum $S(P)$ of $f$ over $P$ is the supremum of the set of all hyperfinite Riemann sums of $f$ over $P$.

(ii) The lower hyperfinite sum $s(P)$ of $f$ over $P$ is the infimum of the set of all hyperfinite Riemann sums of $f$ over $P$.

**Proof (a)** If $P = \{x_0, x_1, \dots, x_n\}$, then $S(P) = \mathrm{Ext}-\sum_{j=1}^{n} M_j(x_j - x_{j-1})$ where

$$M_j = \sup_{x_{j-1} \le x \le x_j} f(x).$$

An arbitrary hyperfinite Riemann sum of $f$ over $P$ is of the following form

$$\sigma = \mathrm{Ext}-\sum_{j=1}^{n} f(c_j)(x_j - x_{j-1}), \quad (15.61)$$

where $x_{j-1} \le c_j \le x_j$. Since $f(c_j) \le M_j$, it follows that $\sigma \le S(P)$. Now let $\varepsilon > 0$, $\varepsilon \approx 0$ and choose $\bar{c}_j \in [x_{j-1}, x_j]$ so that

$$f(\bar{c}_j) > M_j - \frac{\varepsilon}{n(x_j - x_{j-1})}, \quad (15.62)$$

where $1 \le j \le n \in \mathbb{N}^{\#}\mathbb{N}$. The hyperfinite Riemann sum $\bar{\sigma}$ produced in this way is

$$\bar{\sigma} = \mathrm{Ext}-\sum_{j=1}^{n} f(\bar{c}_j)(x_j - x_{j-1}) > \mathrm{Ext}-\sum_{j=1}^{n} \left[ M_j - \frac{\varepsilon}{n(x_j - x_{j-1})} \right] (x_j - x_{j-1}) = S(P) - \varepsilon. \quad (15.63)$$

Now Theorem 15.3 implies that $S(P)$ is the supremum of the set of hyperfinite Riemann sums of $f$ over $P$.

## 15.12. The Riemann–Stieltjes Integral of a $\mathbb{R}^\#$-valued external function $f(x)$.

**Definition 15.21.** Let $f$ and $g$ be defined on $[a, b]$. We say that $f$ is Riemann–Stieltjes integrable with respect to $g$ on $[a, b]$, if there is a number $L \in \mathbb{R}^\#$ with the following property: For every $\varepsilon > 0$, $\varepsilon \approx 0$, there is a $\delta > 0$, $\delta \approx 0$ such that

$$\left| \mathrm{Ext}-\sum_{j=1}^{n} f(c_j)[g(x_j) - g(x_{j-1})] - L \right| < \varepsilon \quad (15.64)$$

provided only that $P = \{x_0, x_1, \dots, x_n\}$, $n \in \mathbb{N}^\#\mathbb{N}$ is a hyperfinite partition of $[a, b]$ such that $\|P\| < \delta$ and $x_{j-1} \le c_j \le x_j, j \in n$. In this case, we say that $L$ is the external Riemann–Stieltjes integral of $f$ with respect to $g$ over $[a, b]$, and write

$$\mathrm{Ext}-\int_{a}^{b} f(x) d^{\#}g(x) = L. \quad (15.65)$$

## 15.13 Existence of the integral of a $\mathbb{R}^\#$-valued external