Basis induces Local Basis

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

Let $\mathcal B$ be an analytic basis for $\tau$.

Let $x \in S$, and denote with $\mathcal B_x$ the set $\left\{{B \in \mathcal B: x \in B}\right\}$.

Then $\mathcal B_x$ is a local basis at $x$.

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

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


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

for some $\mathcal A \subseteq \mathcal B$.

Since $x \in \displaystyle \bigcup \mathcal A$, there is a $B \in \mathcal A$ such that $x \in B$, by definition of set union.

Hence, by definition of $\mathcal B_x$, $B \in \mathcal B_x$.

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


 * $B \subseteq U$

Hence $\mathcal B_x$ is a local basis at $x$.