Uncountable Excluded Point Space is not Separable

Theorem

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

Then $T$ is not separable.

Proof

Let $H \subseteq S$ such that $H$ is countable.

Then $H \ne S$ as $S$ is uncountable by hypothesis.

From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.

So, by definition, the closure of $H$ is $H \cup \set p$.

From Countable Union of Countable Sets is Countable we have that $H \cup \set p$ is countable.

So $H \cup \set p \ne S$.

So $H$ is not everywhere dense in $T$.

Thus $T$ can have no countable subset of $S$ which 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: $15$. Uncountable Excluded Point Topology: $6$