Union of Closed Locally Finite Set of Subsets is Closed

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Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let $\FF$ be a closed locally finite set of subsets of $T$.

Let $E = \ds \bigcup \FF$.

Then:

$E$ is closed in $T$.

Proof

Let:

$\UU = \ds \leftset{U \in \tau : \set{F \in \FF : U \cap F \ne \O}}$ is finite $\ds \rightset{}$

By definition of closed locally finite set of subsets:

$\forall x \in S : \exists U \in \tau : x \in U : \set{F \in \FF : U \cap F \ne \O}$ is finite

That is:

$\forall x \in S : \exists U \in \UU : x \in U$

Hence

$\UU$ is a open cover of $T$ by definition.

Let $U \in \UU$.

Let $\FF_U= \set{F \in \FF : F \cap U \ne \O}$

We have:

\(\ds U \cap E\)

\(=\)

\(\ds U \cap \paren{\cup \set{F : F \in \FF} }\)

\(\ds \)

\(=\)

\(\ds \cup \set{ U \cap F : F \in \FF}\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \cup \set{ U \cap F : F \in \FF_U}\)

Union with Empty Set

From Closed Set in Topological Subspace:

$\forall F \in \FF_U : U \cap F$ is closed in the subspace topology on $U$

From Closed Set Axiom $\paren {\text C 2 }$: Finite Union of Closed Sets:

$U \cap E$ is closed in the subspace topology on $U$

Since $U$ was an arbitrary element of $\UU$:

$\forall U \in \UU : U \cap E$ is closed in the subspace topology on $U$

From Characterization of Closed Set by Open Cover:

$E$ is closed in $T$

$\blacksquare$