Closures of Elements of Locally Finite Set is Locally Finite
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $\AA$ be a locally finite set of subsets of $T$.
Then:
- $\set{A^- : A \in \AA}$ is locally finite in $T$
where $A^-$ denotes the closure of $A$ in $T$.
Proof
Let $\BB = \set{A^- : A \in \AA}$.
Let $x \in S$.
By definition of locally finite set of subsets:
- $\exists U_x \in \tau : x \in U_x :$ the set $\AA_x = \set{A \in \AA : A \cap U_x \ne \O}$ is finite
Let $\BB_x = \set{A^- \in \BB : A^- \cap U_x \ne \O}$
Let $A^- \in \BB_x$.
From Open Set Disjoint from Set is Disjoint from Closure:
- $U_x \cap A \ne \O$
Hence:
- $A \in \AA_x$
It follows that $\BB_x$ is finite.
Since $x$ was arbitrary, we have shown that:
- $\forall x \in S : \exists U_x \in \tau : x \in U_x :$ the set $\BB_x = \set{A^- \in \BB : A^- \cap U_x \ne \O}$ is finite
Hence $\BB$ is locally finite set of subsets by definition.
$\blacksquare$