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Compact set

  1. Aug 5, 2009 #1
    I can't understand why the set [tex]\mathcal{A}=\left\{\frac{1}{2^n};\,n\in\mathbb{N}\right\}[/tex] is not compact, while [tex]\mathcal{A}\cup\{0\}[/tex] is. I know that set is compact if and only if it's closed and bounded, so in order to make set [tex]\mathcal{A}[/tex] closed, we need to include zero, as it's condesation point of this set. Another definiton of compact set tells me, that the set [tex]\mathcal{B}[/tex] is compact if we are able to construct finite cover of all open covers for this set. Now I'm constructing open cover for set [tex]A[/tex]. For example open intervals
    [tex]\left(\frac{1}{5},2\right),\left(\frac{1}{4},\frac{3}{2}\right),\left(-1,\frac{1}{3}\right)[/tex]
    and I'm able to find finite cover
    [tex]\left(\frac{1}{5},2\right),\left(-1,\frac{1}{3}\right)[/tex]
    It's true that
    [tex]\mathcal{A}\subset\left(\frac{1}{5},2\right)\cup\left(-1,\frac{1}{3}\right)[/tex]
    Where's the problem? Thank you.
     
  2. jcsd
  3. Aug 5, 2009 #2
    The problem is you misunderstood the second definition. It's not "compact iff you can find an open cover with a finite subcover", it's "compact iff EVERY open cover has a finite subcover".
     
  4. Aug 5, 2009 #3
    aha, thanx. So, for example for cover
    [tex]\bigcup_{n=1}^{\infty}\left(\frac{1}{2^n},\frac{1}{2^n}+\frac{1}{10}\right)[/tex]
    there's no finite subcover. Is it correct?
     
  5. Aug 5, 2009 #4
    Indeed. In this case, for example, the open cover [tex]\left\{ \left( 1/n, 1 \right) \; : \; n \in \mathbb{N} \right\}[/tex] does not have a finite subcover (why?).
     
  6. Aug 5, 2009 #5
    If such finite cover exists, then for some N
    [tex]\mathcal{A}\subset\bigcup_{n=1}^{N}\left(\frac{1}{n},1\right)=\left(\frac{1}{N},1\right)[/tex]
    That's not true, because there exists [tex]\frac{1}{N+1}[/tex] which is not covered.

    ThanX :-)
     
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