Countable Excluded Point Space is Second-Countable

Theorem
Let $T = \left({S, \tau_{\bar p}}\right)$ be a countable excluded point space.

Then $T$ is a second-countable space.

Proof 1
Consider the set $\mathcal B$ defined as:
 * $\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$

From Basis for Excluded Point Space, $\mathcal B$ is a basis for $T$, and trivially has the same cardinality as $T$.

So by definition, if $T$ is countable, then $T$ is second-countable.

Proof 2
We have:


 * Countable Discrete Space is Second-Countable
 * Excluded Point Topology is Open Extension Topology of Discrete Topology

The result follows from Condition for Open Extension Space to be Second-Countable