Equivalence of Definitions of Analytic Basis

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

Let $U \in \tau$.

By definition 1 of an analytic basis, we can choose $\AA \subseteq \BB$ such that:
 * $\ds U = \bigcup \AA$

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

By Union is Smallest Superset:
 * $\forall B \in \AA: B \subseteq U$

Since $\AA \subseteq \BB$, the result follows.

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

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

Let $U \in \tau$.

Let $\AA = \set {B \in \BB: B \subseteq U}$.

Then $\AA \subseteq \BB$.

Let $x \in U$ be arbitrary.

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


 * $x \in B_x \subseteq U$

Hence, by construction of $\AA$, $B_x \in \AA$.

Thus:


 * $\ds x \in \bigcup \AA$

and it follows that:


 * $\ds U \subseteq \bigcup \AA$

By Union is Smallest Superset applied to $\AA$ and $U$:
 * $\ds \bigcup \AA \subseteq U$

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

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