Why is $\mathcal{A} \cup \{0\}$ Compact but $\mathcal{A}$ is Not?

[lukaszh]

I can't understand why the set \mathcal{A}=\left\{\frac{1}{2^n};\,n\in\mathbb{N}\right\} is not compact, while \mathcal{A}\cup\{0\} is. I know that set is compact if and only if it's closed and bounded, so in order to make set \mathcal{A} closed, we need to include zero, as it's condesation point of this set. Another definition of compact set tells me, that the set \mathcal{B} 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 A. For example open intervals
\left(\frac{1}{5},2\right),\left(\frac{1}{4},\frac{3}{2}\right),\left(-1,\frac{1}{3}\right)
and I'm able to find finite cover
\left(\frac{1}{5},2\right),\left(-1,\frac{1}{3}\right)
It's true that
\mathcal{A}\subset\left(\frac{1}{5},2\right)\cup\left(-1,\frac{1}{3}\right)
Where's the problem? Thank you.

 

[Dragonfall]

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".

 

[lukaszh]

aha, thanx. So, for example for cover
\bigcup_{n=1}^{\infty}\left(\frac{1}{2^n},\frac{1}{2^n}+\frac{1}{10}\right)
there's no finite subcover. Is it correct?

 

[Moo Of Doom]

Indeed. In this case, for example, the open cover \left\{ \left( 1/n, 1 \right) \; : \; n \in \mathbb{N} \right\} does not have a finite subcover (why?).

 

[lukaszh]

If such finite cover exists, then for some N
\mathcal{A}\subset\bigcup_{n=1}^{N}\left(\frac{1}{n},1\right)=\left(\frac{1}{N},1\right)
That's not true, because there exists \frac{1}{N+1} which is not covered.

ThanX :-)