Local Basis Test for Limit 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 a limit point of $H$ if and only if:

$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$

Proof

Necessary Condition

Let $x \in S$ be a limit point of $H$.

By definition of a limit point of $H$:

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

By definition of a local basis of $T$:

$\BB_x \subseteq \tau$

The result follows.

$\Box$

Sufficient Condition

Let $x$ satisfy:

$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$

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:

$H \cap U \setminus \set x \ne \O$

From the contrapositive statement of Subsets of Disjoint Sets are Disjoint:

$H \cap V \setminus \set x \ne \O$

Thus $x$ is a limit point of $H$ by definition.

$\blacksquare$