Union of Closures of Elements of Locally Finite Set is Closed

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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:

$\ds \paren {\bigcup \AA}^- = \bigcup \set{A^- : A \in \AA}$

where $A^-$ denotes the closure of $A$ in $T$.

Proof

From Closures of Elements of Locally Finite Set is Locally Finite:

$\set{A^- : A \in \AA}$ is also locally finite

From Union of Closed Locally Finite Set of Subsets is Closed:

$\bigcup \set{A^- : A \in \AA}$ is closed in $T$

We have:

\(\ds \forall A \in \AA: \, \)

\(\ds A\)

\(\subseteq\)

\(\ds A^-\)

Set is Subset of its Topological Closure

\(\ds \)

\(=\)

\(\ds \bigcup \set{A^- : A \in \AA}\)

Set is Subset of Union

\(\ds \leadsto \ \ \)

\(\ds \bigcup \AA\)

\(\subseteq\)

\(\ds \bigcup \set{A^- : A \in \AA}\)

Union of Subsets is Subset

\(\ds \leadsto \ \ \)

\(\ds \paren{\bigcup \AA}^-\)

\(\subseteq\)

\(\ds \bigcup \set{A^- : A \in \AA}\)

Closure of Subset of Closed Set of Topological Space is Subset

From Closure of Union contains Union of Closures

$\ds \bigcup \set{A^- : A \in \AA} \subseteq \paren {\bigcup \AA}^-$

By definition of set equality:

$\ds \paren {\bigcup \AA}^- = \bigcup \set{A^- : A \in \AA}$

$\blacksquare$