Particular Point Space is First-Countable

Theorem

Let $T = \struct {S, \tau_p}$ be a particular point space.

Then $T$ is first-countable.

Proof

Let $x \in S: x \ne p$.

Consider the set $U_x = \set {x, p} \subseteq S$.

Now let $V \in \tau_p$ be an open set in $S$ such that $x \in V$.

So $x \in V$, by definition of $V$, and $p \in V$ as $V$ is open.

It follows directly that $U_x \subseteq V$.

So $\set {U_x}$ is a local basis at $x$ which is (trivially) countable.

If $x = p$ then $U_x = \set p$ and the same argument still applies.

So every element of $T$ has a countable local basis.

Hence $T$ is first-countable by definition.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $7$