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Basicly I need to show that $\mathbb{R}\cap[0,1]\cap\mathbb{Q}$ is not compact. I was looking at some posts on this topic and all, that I found, used the finite subcover definition of compact set. I wonder if it could be done this way:

A compact set is closed and bounded. So showing that the set is not closed would be enough to see that it's not compact. To show that this set is not closed I could choose any irational number in the interval $[0,1]$ and construct a sequence of rationals that converge to it. So it would be a limit point of the set $\mathbb{R}\cap[0,1]\cap\mathbb{Q}$ that is not contained in it.

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    @Thomas Andrews thankyou. Ya that was kind of strange, :)2012-11-03

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To show that $\mathbb{Q} \cap [0,1]$ is not closed, it is sufficient to construct a sequence of rational numbers converging to an irrational one. See here for example.

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$F=\mathbb{Q} \cap [0,1]$ is dense in $[0,1]$; so if $F$ is compact, it is closed whence $\mathbb{Q} \cap [0,1]= [0,1]$. However, $[0,1]$ contains irrational numbers.

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    thankyou @Seirios, nice proof. But I was looking for a more basic one involving only sequences, limit point and so on...2012-11-03