Basis induces Local Basis

Theorem
Let $\struct {S, \tau}$ be a topological space.

Let $\BB$ be an analytic basis for $\tau$.

Let $x \in S$, and denote with $\BB_x$ the set $\set {B \in \BB: x \in B}$.

Then $\BB_x$ is a local basis at $x$.

Proof
Let $x \in U$, with $U$ an open set of $\tau$.

Since $\BB$ is an analytic basis for $\tau$, we have:


 * $U = \ds\bigcup \AA$

for some $\AA \subseteq \BB$.

Since $x \in \ds \bigcup \AA$, there is a $B \in \AA$ such that $x \in B$, by definition of set union.

Hence, by definition of $\BB_x$, $B \in \BB_x$.

From Set is Subset of Union: General Result, we also have:


 * $B \subseteq U$

Hence $\BB_x$ is a local basis at $x$.

Also see

 * Union of Local Bases is Basis