Point Finite Set of Open Sets in Separable Space is Countable

Theorem

Let $\struct {X, \tau}$ be a separable space.

Let $\FF$ be a point finite set of open sets of $X$.

Then $\FF$ is countable.

Proof

Since $\struct {X, \tau}$ is separable, $X$ has a countable everywhere dense subset $S$.

Without loss of generality, assume that $\O \notin \FF$.

By the definition of point finite, $\set {V \in \FF: x \in V}$ is finite for each $x \in S$.

From Open Set Characterization of Denseness, each element of $\FF$ contains an element of $S$.

From Union of Set of Sets when a Set Intersects All:

$\FF = \ds \bigcup_{x \mathop \in S} \set {V \in \FF: x \in V}$

Thus by Countable Union of Finite Sets is Countable, $\FF$ is countable.

$\blacksquare$