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