# How to show that closed subset of $\mathbb{R}$ is not compact if restricted to $\mathbb{Q}$

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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your approach of showing that the set is not closed is right – Un Chien Andalou Nov 3 '12 at 12:31
thankyou @Flute – Mykolas Nov 3 '12 at 12:33
Your title asks to show a closed set is not compact, but then the set you are trying to prove is not compact is not closed. – Thomas Andrews Nov 3 '12 at 12:49
@Thomas Andrews thankyou. Ya that was kind of strange, :) – Mykolas Nov 3 '12 at 13:52

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.
$F=\mathbb{Q} \cap [0,1]$ 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.