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As far as I can tell it is bounded, as it's within $[0, \sqrt 2]$, and is closed as there cannot be an open neighbourhood about 0, and as it's closed and bounded it is therefore compact. However I'm not sure if closed and bounded imply compact in this situation, as I've only ever used this property in $\mathbb R$. Am I wrong?

Edit: the question specifies this is $\mathbb Q$ with the Euclidean metric, so no need to allow for different ones.

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It is not compact. Note that $\left[0,\sqrt2\right]\cap\Bbb Q=\left[0,\sqrt2\right)\cap\Bbb Q$, and it can be covered by the open sets $U_n=\left[0,\sqrt2-\frac1n\right)\cap\Bbb Q$, but this cover has no finite subcover.

It closed, however, as a subset of the rationals. This shows that a closed and bounded subset of the rationals need not be compact.

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  • $\begingroup$ How can I prove it's closed as a subset of the rationals? Is there another way besides using decimal truncations of $\sqrt 2$ ? $\endgroup$ – user99777 Nov 24 '13 at 16:21
  • $\begingroup$ What is your definition of "closed in $\Bbb Q$"? In topology we simply define a set $A⊆\Bbb Q$ as closed in $\Bbb Q$ if it's the intersection of $\Bbb Q$ with a closed set in $X$. But maybe you are using the induced metric on $\Bbb Q$. In that case consider an adherence point $q\in\Bbb Q$ of $I\cap\Bbb Q$. That means for each $ε>0$ there is a $p\in I\cap\Bbb Q$ with $d(q,p)<ε$. But then $q\in I$ as $I$ is closed, hence $q\in I\cap\Bbb Q$. Here $I=[0,\sqrt2)$. So $I\cap\Bbb Q$ contains all adherence points in $\Bbb Q$ and is thus closed in $\Bbb Q$. $\endgroup$ – Stefan Hamcke Nov 24 '13 at 16:27
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Hint: Is there a sequence in $\mathbf{Q}$ convergent to an irrational number, $0<\alpha < \sqrt{2}$? Further, if indeed the set is not closed, can it be compact?

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The set is not compact because the sequence $(u_n)$ of decimal truncations of $\sqrt{2}$ at rank $n$ does not converge to a point in the set. Thus closed and bounded could not imply compactness in this situation.

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  • $\begingroup$ That doesn't converge to a rational number at all. Aren't we working in the subspace topology on $\mathbb{Q}$? $\endgroup$ – AGM Nov 24 '13 at 15:56
  • $\begingroup$ Sure, it's not closed as a subset of the reals. But I think OP meant as a subset of the rationals. $\endgroup$ – kahen Nov 24 '13 at 15:56
  • $\begingroup$ Yes, my question specifies it as a subset of the rationals. $\endgroup$ – user99777 Nov 24 '13 at 16:01
  • $\begingroup$ How can I use decimal truncations of $\sqrt 2$ in the rationals? $\endgroup$ – user99777 Nov 24 '13 at 16:21
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    $\begingroup$ @Isabelle, The sequence I had in mind is the sequence 1.4, 1.41, 1.414, 1.4142, etc. These are terminating decimals and therefore rational. $\endgroup$ – Mikhail Katz Nov 24 '13 at 16:23

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