Section 2.8: Problem 1 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

( $\mathbb{Q}$ is dense in $\mathbb{R}$ .) Let $\mathbb{Q}$ be the set of rational numbers. Show that every member of $^{∗}\mathbb{R}$ is infinitely close to some member of $^{∗}\mathbb{Q}$ .

Let $Q$ be the relation that defines $\mathbb{Q}$ in $\mathbb{R}$ ( $x\in P_{Q}^{\mathfrak{R}}$ iff $x\in\mathbb{Q}$ ). The following sentence is true in $\mathfrak{R}$ , $\forall v_{1}(v_{1}>0\rightarrow\forall v_{2}\exists v_{3}(P_{Q}v_{3}\wedge v_{2}<v_{3}<v_{2}+v_{1}))$ . Therefore, it is true in $^{*}\mathfrak{R}$ . In particular, if $s(v_{1})=\epsilon$ such that $\epsilon\simeq0$ and $\epsilon>0$ , and $s(v_{2})=x\in{}^{*}\mathbb{R}$ , then $\vDash_{^{*}\mathfrak{R}}\exists v_{3}(P_{Q}v_{3}\wedge v_{2}<v_{3}<v_{2}+v_{1})[[\epsilon,x]]$ , implying that there is $q\in{}^{*}\mathbb{Q}$ such that $x<q<x+\epsilon$ . Therefore, $0<q-x<\epsilon<r$ for every $r\in\mathbb{R}$ , $r>0$ , and $q\simeq x$ .

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Section 2.8: Problem 1 Solution

Section 2.8: Problem 1 Solution