# Section 2.8: Problem 4 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 $A\subseteq\mathbb{R}$ . Show that $A=^{∗}A$ iff $A$ is finite.
If $A$ is finite, then it can be described by a sentence $\forall v_{1}(P_{A}v_{1}\rightarrow(v_{1}=a_{1}\vee\ldots\vee v_{1}=a_{n}))$ , which is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ , implying $^{*}A=A$ . If $A$ is unbounded, then the sentence $\forall v_{1}\exists v_{2}(P_{A}v_{2}\wedge|v_{2}|>|v_{1}|)$ is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ , implying, in particular, that for any infinite $x\in{}^{*}\mathfrak{R}$ , there is $b\in{}^{*}A$ such that $|b|>|x|$ , i.e. $b\notin\mathbb{R}$ , and $b\in{}^{*}A-A$ . Finally, if $A$ is bounded but infinite, then, according to Exercise 5, there is a point $p\in\mathbb{R}$ such that $p$ is infinitely close to some $b\in{}^{*}A$ and $p\neq b$ , implying that $b\notin\mathbb{R}$ , and $b\in{}^{*}A-A$ .