Talk:Point Finite Set of Open Sets in Separable Space is Countable

Explanation
The explanation is a basic fact about sets, and has nothing to do with topology:

Suppose that:

$\mathcal F$ is a set of sets,

$S$ is a set, and

Every element of $\mathcal F$ contains an element of $S$ (that is, $\forall V \in \mathcal F: \exists x: x \in V \cap S$).

Then $\mathcal F = \bigcup_{x \in S} \{ V \in \mathcal F: x \in V \}$.

This can be proved just from the definition of union and such, but it hardly seems to belong in the middle of a topological proof, and it's not so easy to come up with a name for this particular triviality.

As for denseness, another characterization is that every non-empty open set in the space contains an element of the dense set. --Dfeuer (talk) 21:08, 10 May 2013 (UTC)


 * Oh okay, you've done it. --prime mover (talk) 22:18, 10 May 2013 (UTC)