Countable Excluded Point Space is Second-Countable

Theorem

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

Then $T$ is a second-countable space.

Proof 1

Consider the set $\BB$ defined as:

$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$

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

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

$\blacksquare$

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.

$\blacksquare$

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: $6$