# Compact set

1. Aug 5, 2009

### 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 definiton 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.

2. Aug 5, 2009

### 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".

3. Aug 5, 2009

### 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?

4. Aug 5, 2009

### 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?).

5. Aug 5, 2009

### 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 :-)