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I don't think so, this is my argument: $[0,1] \cap \mathbb{Q} $ isn't closed, so it can't be compact. But I don't found an example.

bob
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1 Answers1

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Let $i\in [0,1]$ be an irrational number. The covering $$\{([0,1]\cap\Bbb Q)\setminus [i-\frac{1}{n}, i+\frac{1}{n}]:\,\,i\in\Bbb n\}$$ has no finite subcovering.

Peter Franek
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  • Yes of course :). But are they open? – bob Jun 10 '15 at 12:20
  • Yes, they are open. Each element of such set has distance from $i$ strictly larger then $\frac{1}{n}$ so the same is true for some neighborhood. – Peter Franek Jun 10 '15 at 12:31
  • But in an interval there are only elements of $\mathbb{Q}$, I don't understand how that can be open? If you take an element in such an interval, there are always irrational numbers near that point. – bob Jun 10 '15 at 12:40
  • Open in the metric that you described, of course. Everything lives in $\Bbb Q$. (The sets of the covering in my answer above are subsets of $\Bbb Q$ and open subsets of $\Bbb Q$ are such that they contain a neighborhood (in $\Bbb Q$) with each element.) – Peter Franek Jun 10 '15 at 12:40