Question about finite sets/compactness

Question

I understand that every finite subset of a metric space is compact. But are there any topological spaces where finite sets are not compact? Is that even possible? I don't think it is but I just want to be sure.

Answer 1

It's not possible. If a finite set $F = \{x_1,\ldots,x_n\}$ is covered by a family of sets $\mathcal{S}$ (open or otherwise) then it is covered by a finite subfamily. Just pick $S_1 \in \mathcal{S}$ containing $x_1$, $S_2 \in \mathcal{S}$ containing $x_2$ and so forth.

Answer 2

If you know the general definition of compactness, this isn't too hard to prove. Let's do so thoroughly:

Proposition Let $X$ be a finite topological space, $S=(s_1,...,s_n)$ a finite subset. Then $S$ is compact.

Proof $\ $ Let $\mathcal{U}=\{U_\alpha\}$ be an open cover of $S$. Then each $U_\alpha$ is open and each $s_i\in U_\alpha$ for some $\alpha$, say $U_{\alpha_i}$. Then $(U_{\alpha_1},...,U_{\alpha_n})$ is a finite subcover of $S$.