Finite Complement Topology is Separable
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Theorem
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a separable space.
Proof
Let $H$ be a countably infinite subset of $S$.
From Closure of Infinite Subset of Finite Complement Space, the closure of $H$ is $S$.
So by definition $H$ is everywhere dense in $T$.
Hence the result by definition of separable space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $1$