# Local Basis Test for Isolated Point

## Theorem

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

Let $H \subseteq S$.

Let $x \in H$.

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

Then $x$ is an isolated point of $H$ if and only if:

$\exists U \in \BB_x : U \cap H = \set x$

## Proof

### Necessary Condition

Let $x \in H$ be an isolated point of $H$.

By definition of an isolated point:

$\exists U \in \tau: U \cap H = \set x$

By definition of a local basis of $T$:

$\exists V \in \BB_x : x \in V \subseteq U$
$V \cap H \subseteq U \cap H = \set x$
$\set x \subseteq V \cap H$

From set equality:

$V \cap H = \set x$

$\Box$

### Sufficient Condition

Let $U \in \BB_x : U \cap H = \set x$.

By definition of local basis of $T$:

$U \in \tau$

Then $x$ is an isolated point of $H$ by definition.

$\blacksquare$