$\mathbb{Q}$ with the metric $d(p,q) = |p-q|$

Question

Consider $\mathbb{Q}$ ,the set of rational number , with metric $d(p,q) = |p-q|$.then which of the following statement is true ?

$a)$$\{q \in \mathbb{Q} : 2< q^2 < 3 \}$ is closed

$b)$ $\{ q \in \mathbb{Q} : 2 \le q^2 \le 4\}$ is compact

$c)$ $\{ q\in \mathbb{Q} : 2 \le q^2 \le 4 \}$ is closed

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My attempt : I think option $b)$ and option $c)$ is correct by Heine Boral theorem : closed + bounded = compact

Is this True ?

Any hints/solution will be appreciated

Thank you!

Answer 1

Hints.

(a) True. First try to check whether the set $\{q \in \Bbb{Q}|\sqrt 2< q< \sqrt 3\}$ is closed or not. Well, it is closed in $\Bbb{Q}$, w.r.to the subspace topology from $\Bbb{R}$. Since see that, $\{q \in \Bbb{Q}|\sqrt 2< q< \sqrt 3\}=[\sqrt 2, \sqrt 3]\cap \Bbb{Q}$ where $[\sqrt 2, \sqrt 3]$ is closed in $\Bbb{R}$.

(b) False. See that there is a sequence in this set that converges to $\sqrt 2$, which is not in the set (because of density property of $\Bbb{Q}$ in that closed interval). So, the set is not even Sequentially Compact.

(c) True. Same logic as in (a). Think about the set $[\sqrt 2, 2]\cap \Bbb{Q}$

Answer 2

Hints:

The Heine-Borel Theorem only applies to $\mathbb R$. Since $\mathbb Q$ is a different topological space, you cannot characterize the compact subspaces of $\mathbb Q$ using the same criteria. Instead, use the fact that a compact metric space must be complete. Or you could use the fact that every sequence in a compact set has a convergent subsequence. Can you find a sequence in the set in (b) that does not have a convergent subsequence?

For a set to be closed, its complement must be open. What do the complements of the sets in (a) and (c) look like? Are they open in $\mathbb Q$?