# Compactness in finite complement topology

This has been asked a couple times, but I want to check if the following proof is correct (in the case that it is wrong, I do not want to have already read the right answer!).

Lemma. Every subset of $\mathbb R$ is compact in the cofinite topology.

Proof. Let $E \subset \mathbb R$ and let $\{U_\alpha\}$ be an open covering of $E$. Let $U_1$ be some set in the open covering. Then $U_1$ contains all but finitely many points of $\mathbb R$. In particular, it contains all but finitely many points of $E$, say $x_2, \dots, x_n$. Since the $\{U_\alpha\}$ cover $E$, there exist sets $U_2, \dots, U_n \in \{U_\alpha\}$ such that $x_i \in U_i$ for each $i$. Then $\{U_1, U_2, \dots, U_n\}$ is a finite subcover for $E$.

• This is correct. – Matt Samuel Feb 16 '16 at 3:08