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I am wanting to show that $\mathbb{Q}$ is not compact on $[0,2]$ by describing an open cover for which there is no finite subcover. The cover I thought of is:

$\left(-1, \sqrt{2}-\frac{1}{n}\right) \cup\left(\sqrt{2}+\frac{1}{n} ,3\right)$ for $ n \in \mathbb{N}$

Does this cover work? Because there would be always be a rational number between$\sqrt{2}-\frac{1}{n}$ and $\sqrt{2}$ for any $ n \in \mathbb{N}$

Note I am defining $\mathbb{N}$ as all integers $\ge1$

Carsten S
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user345
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  • Note that without subcovers, you can just consider an increasing sequence of rationals that converge to $\sqrt 2$. – Gabriel Romon Jul 20 '17 at 17:55
  • It is a consensus on this site that $\mathbb N$ is the positive integers. I don't like it but I conform to it. – DanielWainfleet Jul 20 '17 at 18:09
  • https://math.stackexchange.com/questions/1127512/show-that-s-mathbbq-cap-0-2-is-not-compact – Carsten S Jul 20 '17 at 18:20
  • Your cover is perfect. The set is not closed. So there is a limit point p that in not in the set. So there is a sequence $q_n$ that converge to the the point p. We can make a cover of open sets $S_n$ so that $S_n$ contains $q_n$ but no $q_i; i > n$. The cover can have no finite subcover. You did exactly that. – fleablood Jul 20 '17 at 19:11

2 Answers2

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Yes, your open cover works just fine.

I must say that I don't like the title of your question or the description of the problem. What you want to show is that $\mathbb{Q}\cap[0,2]$ is not compact.

José Carlos Santos
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Consider a sequence that "wants to" converge to, say $\sqrt 2$ ( Any irrational in [0,2] will do), like, e.g. {$ 1, 1.4, 1.41.., ....$}. It does not have a convergent subsequence, so it cannot be compact, since in a compact metric space every sequence has a convergent subsequence. Basically this results from the fact that $\mathbb Q$ is not complete; for metric spaces we have : Compact = Complete + Totally - Bounded.

gary
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