Open Cover with Closed Locally Finite Refinement is Even Cover

Theorem
Let $T = \struct{X, \tau}$ be a topological Space.

Let $\UU$ be an open cover of $T$ with a closed locally finite refinement.

Then:
 * $\UU$ is an even cover

Proof
Let $\AA$ be a closed locally finite refinement of $\UU$.

By definition of refinement:
 * $\forall A \in \AA : \exists U \in \UU : A \subseteq U$

For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.

For each $A \in \AA$, let:
 * $V_A = \paren{U_A \times U_A} \cup \paren{\paren{X \setminus A} \times \paren{X \setminus A}}$

Let $T \times T = \struct{X \times X, \tau_{X \times X}}$ denote the product space of $T$ with itself.

Lemma 1
For each $x \in X, A \in \AA$, let:
 * $V_A \sqbrk x =\set{ y∈X : \tuple{x,y} ∈ V_A}$

Lemma 2
Let:
 * $V = \ds \bigcap_{A \in \AA} V_A$

For each $x \in X$, let:
 * $V \sqbrk x =\set{ y∈X : \tuple{x,y} ∈ V}$

Lemma 4
It follows that $\UU$ is an even cover by definition.