Union of Closed Locally Finite Set of Subsets is Closed

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:

From Closed Set in Topological Subspace:
 * $\forall F \in \FF_U : U \cap F$ is closed in the subspace topology on $U$

From :
 * $U \cap E$ 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 User:Leigh.Samphier/Topology/Characterization of Closed Set by Open Cover:
 * $E$ is closed in $T$