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Letting $\mathbb{Q}$ be equipped with the Euclidean metric.

What I can work out is that it is bounded as its contained in the closed ball of radius ${\sqrt2}/{2}$ centred at ${\sqrt2}/{2} $.

Its not compact as it can be expressed as union of the two disjoint open sets $[0,{\sqrt2}/{2}) $and$ ({\sqrt2}/{2}, \sqrt2)$ (though I'm not sure if this makes it not compact in $\mathbb{Q}$ or just $\mathbb{R}$).

And its not closed as the sequence of truncations of $\sqrt2$ in $\mathbb{Q}$ converges to $\sqrt2$

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  • $\begingroup$ I'm unsure about the open cover/finite subcover definition and how to apply it, (how to prove there is no finite subcover). A good explanation of this would be greatly appreciated $\endgroup$
    – 123454321
    Nov 25, 2014 at 12:26
  • $\begingroup$ It is the intersection of a closed set in $\;\Bbb R\supset\Bbb Q\;$, and obvioulsy bounded as it is fully contained in $\;[0,\sqrt2]\;$ . $\endgroup$
    – Timbuc
    Nov 25, 2014 at 12:26
  • $\begingroup$ What is your definition of compactness? Is it sequential compactness? $\endgroup$
    – user99914
    Nov 25, 2014 at 12:31
  • $\begingroup$ @John It is that for every open cover, there exists a finite subcover. $\endgroup$
    – 123454321
    Nov 25, 2014 at 12:33
  • $\begingroup$ @123454321: Thanks. Then Timbuc's answer give you good example of open cover. $\endgroup$
    – user99914
    Nov 25, 2014 at 12:34

2 Answers 2

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It is not closed (in $\Bbb R$) because of what you said. It is not compact because not closed in this space (and compactness is not related to the ambient subspace).

It is obviously bounded.

But what you wrote on compactness refers not to compactness, but connectedness.

edit: let us prove it is closed in $\Bbb Q$:

consider a sequence of rational numbers converging to some $x\in\Bbb Q$. As $x_n\in [0,\sqrt 2]:=C$ and $C$ is closed, $x\in C$. Hence $x\in C\cap\Bbb Q$ (this is also the general proof that the intersection of any subspace with a closed/open set is closed/open in the subset).

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  • $\begingroup$ Oh of course, forgot about connectedness. $\endgroup$
    – 123454321
    Nov 25, 2014 at 12:31
  • $\begingroup$ I think it is closed? $\endgroup$
    – user99914
    Nov 25, 2014 at 12:31
  • $\begingroup$ It is closed in $\;\Bbb Q\;$ as an intersection of closed (in $\;\Bbb R\;$ and the wholse set $\;\Bbb Q\;$ . $\endgroup$
    – Timbuc
    Nov 25, 2014 at 12:32
  • $\begingroup$ well ,it would be nice if the downvoters could talk. $\endgroup$
    – mookid
    Nov 25, 2014 at 12:36
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    $\begingroup$ @123454321: Can you tell us what is your definition of closedness? $\endgroup$
    – user99914
    Nov 25, 2014 at 12:45
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About compact: take the open cover (and prove it is such)

$$\left\{\;\left(\frac1n\;,\;\;\sqrt2-\frac1n\right)\cap\Bbb Q\;\right\}_{n\in\Bbb N\setminus\{1\}}$$

Resuming (see the comments): it is closed, bounded and not compact.

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  • $\begingroup$ It is easy to prove that this is an open cover for the subset, but how do I prove there is no finite subcover? $\endgroup$
    – 123454321
    Nov 25, 2014 at 12:38
  • $\begingroup$ @123454321, assume there there is... but observe the sets in this cover are embedded in each other: $$\left(\frac12\,,\,\sqrt2-\frac12\right)\subset\left(\frac13\,,\,\sqrt2-\frac13\right)\subset\ldots$$ so if there's a finite cover then... $\endgroup$
    – Timbuc
    Nov 25, 2014 at 12:42
  • $\begingroup$ But why can't we say that the single open set at n = ∞ covers the set, this is a finite number of sets and hence a finite open cover. I know you're right, I'm just a bit foggy on the definition. $\endgroup$
    – 123454321
    Nov 25, 2014 at 12:46
  • $\begingroup$ @123454321 "At $\;n=\infty\;$"? There's no such a thing: infinity is not a number (at least not here and in this context). $\endgroup$
    – Timbuc
    Nov 25, 2014 at 13:33
  • $\begingroup$ This fails to be a cover, as $0$ does not lie in any of the sets. $\endgroup$
    – Cameron Buie
    Aug 23, 2017 at 11:36

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