Compact Complement Topology is Compact

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Theorem

Let $T = \left({\R, \tau}\right)$ be the compact complement topology on $\R$.


Then $T$ is a compact space.


Proof

Let $\mathcal C$ be an open cover of $\R$.

Let $U \in \mathcal C$.

Then by definition $\R \setminus U$ is compact in the usual (Euclidean) topology.

Since each $U$ is open in the Euclidean topology, a finite number must cover $\R \setminus U$.

$\blacksquare$


Sources