Is the set $\{ x\in \mathbb{Q}: 2< x^2 <3\}$ closed, bounded, compact in $\mathbb{Q}$? Is the set $\{ x\in \mathbb{Q}: 2< x^2 <3\}$ closed, bounded, compact in $\mathbb{Q}$ ?
I think $\{ x\in \mathbb{Q}: 2< x^2 <3\}=\{ x\in \mathbb{Q}: 2\leq x^2 \leq 3\}$, so it is bounded and closed in $Q$, is that right?
 A: The set is not compact. The open cover $U_n=\left((-\sqrt{3},-\sqrt{2})\cup \left(\sqrt{2}+
\frac{1}{n},\sqrt{3}-\frac{1}{n}\right)\right)\cap\mathbb{Q},$ $n\ge 7,$ has no any finite subcover.
A: 
Theorem: Suppose $K\subset Y\subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$.

Using the theorem, if $K$ is compact relative to $\mathbb{Q}$ then it is compact relative to $\mathbb{R}$. Clearly the set is not closed in $\mathbb{R}$.
A: Let $S = \{ x \in \mathbb{Q}_+ : 2 < x^2 < 3\}$. 
Let $\{a_n\}_{n=1}^\infty \subset \mathbb{Q}$ be monotonic decreasing and $\lim\limits_{n \to \infty}a_n = \sqrt{2}$ (e.g., $\{1.5; 1.42; 1.415; \ldots\}$)
Let $\{b_n\}_{n=1}^\infty \subset \mathbb{Q}$ be monotonic increasing and $\lim\limits_{n \to \infty}b_n = \sqrt{3}$ (e.g., $\{1.7; 1.73; 1.732; 1.7320; \ldots \}$)
Then, the open cover $\mathcal{O}=(a_n + \dfrac{1}{n}, b_n - \dfrac{1}{n})$ for $n \geq 4$ is an open cover of $S$ from which one cannot extract a finite subcover. Hence it is not compact.
But note that the set is open, closed and bounded in $\mathbb{Q}$.
