Intersection of arbitrary union of compact subsets.

Question

My textbooks asks to prove that arbitrary intersection of compact subsets in hausdorff space is again compact.

I've kinda found the counterexample $\bigcap_{1\leq x<2} [x,3]=(2,3]$, and can't find where I have mistaken.

Answer 1

First: your intersection $\cap_{1 \le x < 2 }[x,3] = [2,3]$: The right is included in the left as $[2,3] \subset [x,3]$ for all $x < 2$. And if $p$ is in the left hand intersection, then $p \le 3$ is clear and if $p < 2$ would hold, $p$ would not be in the set $[\max(1,\frac{p+2}{2}), 3]$, which is a set the left hand set takes the intersection over, and this is a contradiction, so $p \ge 2$ and so $p \in [2,3]$. But it's good to try to come up with potential counterexamples and see what goes wrong.

Second, the statement is in fact true. As a hint: use that compact sets are closed (which uses the Hausdorffness!).