How to index compact set problem?

Question

For any α∈I, if Aα is compact set then ∩(α∈I)Aα is compact set.

Tomorrow, I will mid-exam, i'm very dizzy.

Please, prove that problem.

Answer 1

The intersection of compact sets in a space $X$ is not necessarily compact, in fact not even finite intersections.

If $X$ is Hausdorff, i.e., $T_2$, then arbitrary intersections of compact sets is compact.

Hint: Use that a compact set in a Hausdorff space is closed and a closed subset of a compact space is compact.

In general it's false, consider the discrete topology on $\mathbb{N}$. We add two points $\{-\infty,\infty\}$ to the space, where the only open sets that contain these are either $\mathbb{N}\cup\{\infty\},\ \mathbb{N}\cup\{-\infty\}$ and $\mathbb{N}\cup\{\infty,-\infty\}$. You should verify that all subsets of $\mathbb{N}$ together with these three members gives a topology on $\mathbb{N}_\infty:=\mathbb{N}\cup\{\infty,-\infty\}$.

Now, $X_1:=\mathbb{N}\cup\{\infty\}$ and $X_2:=\mathbb{N}\cup\{-\infty\}$ are both compact. $X_1$ is compact since every open cover of $X_1$ must contain $\mathbb{N}\cup\{\infty\}$ or $\mathbb{N}\cup\{\infty,-\infty\}$ (as these are the only open sets that contain $\infty$). Thus, each open cover of $X_1$ has a finite subcover containing one element only. Similarly for $X_2$.

But $X_1\cap X_2 = \mathbb{N}$ which is an infinite discrete space and is obviously not compact.

Maybe it could be helpful to note that $\mathbb{N}_\infty$ is not Hausdorff. There are no disjoint open sets containing $\infty$ and $-\infty$. The space is however $T_1$, since $X_1$ is an open set containing $\infty$, thus $\mathbb{N}_\infty \setminus X_1 = \{-\infty\}$ is a closed set. Similarly $\{\infty\}$ is closed.