Local Basis Test for Adherent Point
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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 $H$ if and only if:
- $\forall U \in \BB_x : H \cap U \ne \O$
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 local basis of $T$:
- $\BB_x \subseteq \tau$
The result follows.
$\Box$
Sufficient Condition
Let $x$ satisfy:
- $\forall U \in \BB_x : H \cap U \ne \O$
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.
$\blacksquare$