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

**Answer**

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.

**Attribution***Source : Link , Question Author : John Buchta , Answer Author : Community*