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I hope you would read my approach.

First, choose an increasing sequence, $\{a_n\}$ of irrational numbers, the limit of which is $1$. Then, the open covering, $\{(0,a_1),(a_1,a_2),(a_2,a_3),\ldots\}$ has no finite subcovering.

Therefore, $(0,1)\cap \mathbb{Q}$ is not compact.

user406753
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  • That certainly works. – Aweygan Jan 14 '17 at 02:06
  • This is compact because it consist of all rational points from $(0,1)$ and all this points are boundary, so this set is closed and bounded – openspace Jan 14 '17 at 02:06
  • @openspace The Heine-Borel theorem doesn't apply. – Aweygan Jan 14 '17 at 02:08
  • @Aweygan why Heine-Borell theorem doesn't apply? – openspace Jan 14 '17 at 02:09
  • @openspace the set is not closed in $\Bbb R$. Remember that $\Bbb Q\subseteq\Bbb R$ is dense. – Stahl Jan 14 '17 at 02:11
  • @Stahl why not closed? This set consist of rational points, and all of this points are boundary, so because of that set is closed. – openspace Jan 14 '17 at 02:13
  • @openspace The closure of $\Bbb Q$ is $\Bbb R$. The generalized Heine-Borel (for metric spaces) does not apply either, as the space in question is not complete, as Open Ball's answer notes. – Stahl Jan 14 '17 at 02:15
  • @Stahl okay, why could we numerate all irrational points in solution? – openspace Jan 14 '17 at 02:17
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    @openspace I don't know what you mean. If you're not familiar with the rationals being dense in the reals, I suggest you brush up on your analysis. Explanations of this can be found many places online, and I'm sure there are some questions about this on this very site. – Stahl Jan 14 '17 at 02:18
  • Let us continue this discussion in chat. – openspace Jan 14 '17 at 02:19
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    @openspace Not to be rude, but I don't really have the time to explain this topic in chat right now. A google search for "rationals dense in reals" should give you plenty to start with, and if you're confused by a specific part of something you find, you can ask another question here. – Stahl Jan 14 '17 at 02:21
  • Openspace. The definition of closed is not "contains nothing but limit points". It is "contains all the limit points". In R the set is certainly not closed because 0,1 and all the irratioals are also boundary points not in the set. If we assume our metric space is Q (not stated in OP) the set is still not closed as 0 and 1 are not in it. If we include points 0 and 1, the set is closed in Q (but not R) but not compact as Heine Borel doesn't apply to Q as Q isn't complete. The OP argument is spot on correct, in either Q or R. – fleablood Jan 14 '17 at 04:33

1 Answers1

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I suppose you meant an increasing sequence of irrational numbers. Yes, your approach works. Here's another one: it's not compact because it's not closed (take $x_n = 1/n$).