Section 2.8: Problem 2 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James $^{*}\mathfrak{R}$ . Munkres

(a) Let $A\subseteq\mathbb{R}$ and $F:A\rightarrow\mathbb{R}$ . Then $F$ is also a binary relation on $\mathbb{R}$ ; show that $^{∗}F:^{∗}A\rightarrow^{∗}\mathbb{R}$ .

(b) Let $S:\mathbb{N}\rightarrow\mathbb{R}$ . Recall that $S$ is said to converge to $b$ iff for every $\epsilon>0$ there is some $k$ such that for all $n>k$ , $|S(n)−b|<\epsilon$ . Show that this is equivalent to the condition: $^{∗}S(x)\simeq b$ for every infinite $x\in^{∗}\mathbb{N}$ .

(c) Assume that $S_{i}:\mathbb{N}\rightarrow\mathbb{R}$ and $S_{i}$ converges to $b_{i}$ for $i=1,2$ . Show that $S_{1}+S_{2}$ converges to $b_{1}+b_{2}$ and $S_{1}·S_{2}$ converges to $b_{1}·b_{2}$ .

(a) Consider the following sentence: $\forall x[\exists!yP_{F}xy[P_{A}x]\forall y\neg P_{F}xy]$ where the quantifier symbol $\exists!$ is defined in Exercise 21 of Section 2.2, and the ternary connective symbol $[[]]$ is defined in Exercise 13-A of Section 1.5. The sentence is true in $\mathfrak{R}$ and $^{*}\mathfrak{R}$ , therefore, $^{*}F$ defines a function on $^{*}A$ .

(b) According to (a), $^{*}S$ defines a function on $^{*}\mathbb{N}$ . Let $N$ be the relation defining $\mathbb{N}$ in $\mathbb{R}$ ( $<x>\in P_{N}^{\mathfrak{R}}$ iff $x\in\mathbb{N}$ ). If $S$ converges to $b$ , then for every $\epsilon\in\mathbb{R}$ , $\epsilon>0$ , and some $k_{\epsilon}\in\mathbb{N}$ , the sentence

is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ . In particular, for an infinite $n\in^{*}\mathbb{N}$ , this means $^{*}|{}^{*}S(n){}^{*}-b|<\epsilon$ for every $\epsilon\in\mathbb{R}$ , $\epsilon>0$ , i.e. $^{∗}S(n)\simeq b$ . Now, if $^{∗}S(n)\simeq b$ for every infinite $n\in^{∗}\mathbb{N}$ , then for every $\epsilon\in\mathbb{R}$ , $\epsilon>0$ , the sentence $\exists k(P_{N}k\wedge\forall n\forall s(P_{N}n\wedge n>k\wedge P_{S}ns\rightarrow|s-b|<c_{\epsilon}))$ is true in $^{*}\mathfrak{R}$ (since we can take any infinite $k\in{}^{*}\mathbb{N}$ ), and, hence, in $\mathfrak{R}$ , implying that $S$ converges to $b$ .

(c) According to (b), for every infinite $n\in{}^{*}\mathbb{N}$ , $^{*}S_{i}(n)\simeq b_{i}$ , and $^{*}(S_{1}\dotplus S_{2})(n)={}^{*}S_{1}(n)\dotplus{}^{*}S_{2}(n)\simeq b_{1}\dotplus b_{2}$ , implying $S_{1}\dotplus S_{2}$ converges to $b_{1}\dotplus b_{2}$ .

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Section 2.8: Problem 2 Solution

Section 2.8: Problem 2 Solution