Equivalence of Definitions of Analytic Basis

Theorem
The following definitions of an analytic basis are equivalent:

Definition 1 implies Definition 2
Let $\mathcal B$ be an analytic basis for $\tau$ by definition 1.

Let $U \in \tau$.

By definition 1 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:
 * $\forall B \in \mathcal A: B \subseteq U$

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

Thus $\mathcal B$ is an analytic basis for $\tau$ by definition 2.

Definition 2 implies Definition 1
Let $\mathcal B$ be an analytic basis for $\tau$ by definition 2.

Let $U \in \tau$.

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

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

Let $x \in U$ be arbitrary.

Since $\mathcal B$ is an analytic basis for $\tau$ by definition 2, there is some $B_x \in \mathcal B$ such that:


 * $x \in B_x \subseteq U$

Hence, by construction of $\mathcal A$, $B_x \in \mathcal A$.

Thus:


 * $\displaystyle x \in \bigcup \mathcal A$

and it follows that:


 * $\displaystyle U \subseteq \bigcup \mathcal A$

By Union is Smallest Superset applied to $\mathcal A$ and $U$:
 * $\displaystyle \bigcup \mathcal A \subseteq U$

By definition of set equality and definition 1 of an analytic basis, the result follows.

Thus $\mathcal B$ is an analytic basis for $\tau$ by definition 1.