Local Basis Test for Adherent Point

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

Let $H \subseteq S$.

Let $x \in S$.

Let $\BB_x$ be a local basis of $x$.

Then $x \in S$ is an adherent point of $H$ :
 * $\forall U \in \BB_x : H \cap U \ne \O$

Necessary Condition
Let $x \in S$ be an adherent point of $H$.

By definition of an adherent point of $H$:
 * $\forall U \in \tau : x \in U \implies H \cap U \ne \O$

By definition of a local basis of $T$:
 * $\BB_x \subseteq \tau$

The result follows.

Sufficient Condition
Let $x$ satisfy:
 * $\forall U \in \BB_x : H \cap U \ne \O$

Let $V$ be any open neighborhood of $x$.

By definition of a local basis of $T$:
 * $\exists U \in \BB : x \in U \subseteq V$

Then:
 * $H \cap U \ne \O$

From the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
 * $H \cap V \ne \O$

Thus $x$ is an adherent point of $H$ by definition.