Open Set Characterization of Denseness/Analytic Basis

Theorem
Let $\left({X, \tau}\right)$ be a topological space.

Let $S \subseteq X$. Let $\mathcal B \subseteq \tau$ be an analytic basis for $\tau$.

Then $S$ is (everywhere) dense in $X$ iff every non-empty open set of $\mathcal B$ contains an element of $S$.

Necessary Condition
Let $S$ be everywhere dense in $X$.

Let $U$ be open and non-empty.

Then $U$ has an element $x$.

Since $S$ is everywhere dense in $X$, $x \in S^-$, the closure of $S$.

By Equivalence of Definitions of Adherent Point, every open neighborhood of $x$ contains an element of $S$.

Thus in particular, $U$ contains an element of $S$.

Sufficient Condition
Suppose that every non-empty open set in $X$ contains an element of $S$.

Let $x \in X$.

Let $U$ be an open neighborhood of $x$.

Then $U$ contains an element $s$ of $S$.

As this holds for all open neighborhoods of $x$, Equivalence of Definitions of Adherent Point shows that $x \in S^-$.

As this holds for all $x \in X$, $S$ is everywhere dense in $X$.