# Basis Test for Adherent Point

## Theorem

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

Let $\BB$ be a synthetic basis of $T$.

Let $H \subseteq S$.

Then $x \in S$ is an adherent point of $H$ if and only if:

$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$

## Proof

### Necessary Condition

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

By definition of an adherent point of $H$:

$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$

By definition of a synthetic basis of $T$:

$\BB \subseteq \tau$

The result follows.

$\Box$

### Sufficient Condition

Let $x$ satisfy:

$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$

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

By definition of a synthetic basis of $T$:

$\exists U \in \BB : x \in U \subseteq V$

Then:

$H \cap U \ne \O$
$H \cap V \ne \O$

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

$\blacksquare$