Countable Excluded Point Space is Separable

Theorem

Let $T = \struct {S, \tau_{\bar p} }$ be a countable excluded point space.


Then $T$ is a separable space.


Proof

The closure of the set $S$ (trivially) equals $S$.

That is, $S$ is everywhere dense in $T$.

But as $S$ is countable it follows by definition that $T$ is separable.

$\blacksquare$


Sources

  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $14$. Countable Excluded Point Topology: $6$
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