Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 16

Problem

Regard $\mathbb{Q}$, the set of all rational numbers, as a metric space, with $d(p,q)=|p-q|$. Let $E$ be the set of all $p \in \mathbb{Q}$ such that $2 < p^2 < 3$. Show that $E$ is closed and bounded in $\mathbb{Q}$, but that $E$ is not compact. Is $E$ open in $\mathbb{Q}$?

Answer

(1) Closed: $E=\mathbb{Q} \cap ((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3})) = \mathbb{Q} \cap ([-\sqrt{3},-\sqrt{2}] \cup [\sqrt{2},\sqrt{3}])$. Define $F:=[-\sqrt{3},-\sqrt{2}] \cup [\sqrt{2},\sqrt{3}]$  $\Rightarrow$  $F^c$ is open in $\mathbb{R}$  $\Rightarrow$  $\mathbb{Q} \cap F^c$ is open in $\mathbb{Q}$ by Theorem 2.30  $\Rightarrow$  $\mathbb{Q} \cap F$ is closed in $\mathbb{Q}$.

(2) Bounded: $d(p,0) < \sqrt{3}$ for all $p \in E$, which means $E$ is bounded in $\mathbb{Q}$.

(3) Not compact: $E$ is not compact in $\mathbb{R}$, and hence is not compact in $\mathbb{Q}$ by Theorem 2.33.

(4) Open: Since $((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3}))$ is open in $\mathbb{R}$, $E=\mathbb{Q} \cap ((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3}))$ is open in $\mathbb{Q}$.

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