Closures of Elements of Locally Finite Set is Locally Finite

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.