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Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.

I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.

Thank you for the help

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What is your definition of compactness? – Phira Nov 5 '12 at 16:41
up vote 40 down vote accepted

For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.

Take $\mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be $\{x_i\}\cup \mathbb{N}$ and $\{x_1 , x_2\}\cup \mathbb{N}$. (If you can't see it immediately, check this gives a topology on $\{x_1 , x_2\}\cup \mathbb{N}$).

Now $\{x_i\}\cup \mathbb{N}$ is compact for $i=1,2$, since any open cover must contain $\{x_i\}\cup \mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $\mathbb{N}$, is infinite and discrete, so definitely not compact.

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How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right? – verticese Apr 7 at 15:56
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain $\{x_i\}\cup \mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy. – MightyJohnson Apr 7 at 21:30
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks – verticese Apr 14 at 12:55
@verticese The open cover $\{x_i\}\cup\Bbb N$ must contain an open member $U$ with $x_1\in U$. But then necessarily $\Bbb N\subset U$ as well. – Hagen von Eitzen Apr 14 at 13:06
@HagenvonEitzen But then $U$ would be infinite, as $\mathbb{N}$ is. How can it be a finite subcover then? – verticese Apr 14 at 13:09

In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.

Now $G=\cap F$ is closed and $G^c=S\backslash G$ is open. If $V$ is a family of open sets with $\cup V\supset G,$ then $V'=V\cup \{G^c\}$ is an open cover of $f.$ (Indeed, $\cup V'=S.$)

Since $f$ is compact, there exists a finite $H\subset V'$ with $\cup H\supset f.$ Then $H'=H\backslash \{G^c\}$ is a finite subset of $V$, and $\cup H'\supset G.$

So $G$ is compact.

Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.

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