Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
your approach of showing that the set is not closed is right –  El Angel Exterminador 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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer

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

share|improve this answer
    
thankyou @Seirios, nice proof. But I was looking for a more basic one involving only sequences, limit point and so on... –  Mykolas Nov 3 '12 at 12:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.