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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
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1 Answer 1

up vote 35 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 there is only one possible open cover to consider. However, their intersection, $\mathbb{N}$, is infinite and discrete, so definitely not compact.

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