Section 2.8: Problem 3 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

Let $F:A\rightarrow\mathbb{R}$ be one-to-one, where $A\subseteq\mathbb{R}$ . Show that if $x\in^{∗}A$ but $x\notin A$ , then $^{∗}F(x)\notin\mathbb{R}$ .

According to Exercise 2(a), $^{*}F:{}^{*}A\rightarrow{}^{*}\mathbb{R}$ . Suppose $^{*}F(x)=r\in\mathbb{R}$ . $r$ cannot be in $\mathbb{R}-F(A)$ , because the sentence $\forall v_{1}\forall v_{2}(P_{F}v_{1}v_{2}\rightarrow v_{2}\neq c_{r})$ is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ . But then, $r$ is in $F(A)$ , and there is $a\in A\subseteq\mathbb{R}$ such that the sentence $\forall v_{1}(P_{F}v_{1}r\rightarrow v_{1}=c_{a})$ is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ , implying that $x=a\in A$ .

Note. This exercise also implies that for any $A\subseteq\mathbb{R}$ , $^{*}A-A\cap\mathbb{R}=\emptyset$ . Indeed, consider $F:A\rightarrow\mathbb{R}$ such that $F(x)=x$ , for which $\forall v_{1}\forall v_{2}(P_{F}v_{1}v_{2}\rightarrow v_{1}=v_{2})$ is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ . And if $x\in{}^{*}A-A$ , then $^{*}F(x)=x\notin\mathbb{R}$ .

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Section 2.8: Problem 3 Solution

Section 2.8: Problem 3 Solution