Characterization of Closure by Basis
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB \subseteq \tau$ be a basis.
Let $A$ be a subset of $S$.
Let $x$ be a point of $S$.
Then $x \in A^-$ if and only if:
- $\forall U \in \BB: x \in U \implies A \cap U \ne \O$
where:
- $A^-$ denotes the closure of $A$
Proof
Sufficient Condition
Let $x \in A^-$.
Let $U \in \BB$.
By definition of basis, $U$ is an open set of $T$.
Thus from Condition for Point being in Closure:
- if $x \in U$ then $A \cap U \ne \O$.
$\Box$
Necessary Condition
Let $x$ be such that for every $U \in \BB$:
- if $x \in U$
- then $A \cap U \ne \O$.
By Condition for Point being in Closure, to prove that $x \in \operatorname{Fr} A$ it is enough to prove that:
- for every open set $U$ of $T$:
- if $x \in U$ then $A \cap U \ne \O$.
Let $U$ be an open set of $T$.
Let $x \in U$.
By definition of (analytic) basis, there exists $V \in \BB$ such that:
- $x \in V \subseteq U$
By assumption:
- $A \cap V \ne \O$
From the corollary to Set Intersection Preserves Subsets:
- $A \cap V \subseteq A \cap U$
So:
- $A \cap U \ne \O$
and hence the result.
$\blacksquare$
Sources
- Mizar article YELLOW_9:37