Equivalence of Definitions of Analytic Basis

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

Let $\mathcal B \subseteq \tau$.

Then $\mathcal B$ is an analytic basis for $\tau$ iff:
 * $\forall U \in \tau: \forall x \in U: \exists B \in \mathcal B: x \in B \subseteq U$

Necessary Condition
Let $U \in \tau$.

Let $\mathcal A = \left\{{B \in \mathcal B: B \subseteq U}\right\}$.

Then $\mathcal A \subseteq \mathcal B$.

It follows, by hypothesis, that:
 * $\displaystyle U \subseteq \bigcup \mathcal A$

By Union is Smallest Superset: General Result, we have:
 * $\displaystyle \bigcup \mathcal A \subseteq U$

By definition of set equality and the definition of an analytic basis, the result follows.

Sufficient Condition
Let $U \in \tau$.

By the definition of an analytic basis, we can choose $\mathcal A \subseteq \mathcal B$ such that:
 * $\displaystyle U = \bigcup \mathcal A$

By the definition of union:
 * $\forall x \in U: \exists B \in \mathcal A: x \in B$

By Union is Smallest Superset: General Result, we have:
 * $\forall B \in \mathcal A: B \subseteq U$

Since $\mathcal A \subseteq \mathcal B$, the result follows.