Local Basis Test for Isolated Point
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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 $T$:
- $\exists V \in \BB_x : x \in V \subseteq U$
From Set Intersection Preserves Subsets:
- $V \cap H \subseteq U \cap H = \set x$
From Singleton of Element is Subset:
- $\set x \subseteq V \cap H$
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.
$\blacksquare$