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 $X$ every non-empty open set of $\mathcal B$ contains an element of $S$.

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

By Open Set Characterization of Denseness then every non-empty open set contains an element of $S$.

Every non-empty set of an analytic basis is an open set by definition.

Hence every non-empty set of non-empty open set of $\mathcal B$ contains an element of $S$.

Sufficient Condition
Suppose that every non-empty open set of $\mathcal B$ contains an element of $S$.

Let $U$ be any non-empty open set.

Let $x \in U$.

By the definition of an analytic basis then:
 * $\exists B \in \mathcal B: x \in B \subseteq U$

By assumption $B$ contains an element $s$ of $S$.

Hence $s \in U$.

Since $U$ was an arbitrary non-empty open set then every non-empty open set contains an element of $S$

By Open Set Characterization of Denseness, $S$ is everywhere dense in $X$.