Compact Complement Topology is Compact
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Theorem
Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a compact space.
Proof
Let $\CC$ be an open cover of $\R$.
Let $U \in \CC$.
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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $4$