Equivalence of Definitions of Local Basis/Neighborhood Basis of Open Sets Implies Local Basis for Open Sets

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

Let $x$ be an element of $S$. Let $\mathcal B$ be a set of open neighborhoods of $x$ such that:
 * every neighborhood of $x$ contains a set in $\mathcal B$.

Then $\mathcal B$ satisfies:
 * $\forall U \in \tau: x \in U \implies \exists H \in \mathcal B: H \subseteq U$

Proof
Let $U \in \tau$ such that $x \in U$.

From Set is Open iff Neighborhood of all its Points then $U$ is a neighborhood of $x$.

By assumption, there exists $H \in \mathcal B$ Such that $H \subseteq U$.

The result follows.