Point Finite Set of Open Sets in Separable Space is Countable

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Theorem

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

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


Then $\mathcal F$ is countable.


Proof

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

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

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

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

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

$\mathcal F = \displaystyle \bigcup_{x \mathop \in S} \set {V \in \mathcal F: x \in V}$

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

$\blacksquare$