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.

$V_A$ is an Open Neighborhood of the Diagonal $\Delta_X$ in $T_{X \times X}$
Let $A \in \AA$.

By definition of closed set:
 * $X \setminus A$ is open in $T$

By definition of product topology:
 * $U_A \times U_A, \paren{X \setminus A} \times \paren{X \setminus A}$ are open in $T \times T$

By :
 * $V_A = \paren{U_A \times U_A} \cup \paren{\paren{X \setminus A} \times \paren{X \setminus A}}$ is open in $T \times T$

Let $x \in X$.

By definition of set difference:
 * either $x \in A$ or $x \in X \setminus A$.

By definition of subset:
 * either $x \in U_A$ or $x \in X \setminus A$.

By definition of cartesian product:
 * either $\tuple{x, x} \in U_A \times U_A$ or $\tuple{x, x} \in \paren{X \setminus A} \times \paren{X \setminus A}$

By definition of set union:
 * $\tuple{x, x} \in V_A$

Since $x$ was arbitrary:
 * $\forall x \in X : \tuple{x, x} \in V_A$

By definition of diagonal:
 * $\Delta_X \subseteq V_A$

Since $A$ was arbitrary:
 * $\forall A \in \AA : V_A$ is an open neighborhood of the diagonal $\Delta_X$ in $T \times T$

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

For all $A \in \AA, x \in A$ we have $V_A \sqbrk x = U_A$

 * $\forall A \in \AA, x \in A$, $V_A \sqbrk x = U_A$

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

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

The Set $\set{V \sqbrk x : x \in X}$ is a Refinement of $\UU$
Let $x \in X$.

By definition of refinement:
 * $\AA$ is a cover of $X$

By definition of cover:
 * $\exists A \in \AA : x \in A$

Hence:
 * $V \sqbrk x \subseteq V_A \sqbrk x = U_A \in \UU$

Since $x$ was arbitrary, then:
 * $\forall x \in X : \exists U \in \UU : V \sqbrk x \subseteq U$

Also, we have:
 * $\forall x \in X : x \in V \sqbrk x$

It follows that $\set{V \sqbrk x : x \in X}$ is a refinement of $\UU$.

$V$ is an Open Neighborhood of the Diagonal $\Delta_X$ in $X \times X$
Let $x \in X$.

By definition of locally finite:
 * $\exists W \in \tau : x \in W : \set{A \in \AA : W \cap A \ne \O}$ is finite

Let:
 * $A \in \AA : W \cap A = \O$

From Subset of Set Difference iff Disjoint Set:
 * $W \subseteq X \setminus A$

From Cartesian Product of Subsets:
 * $W \times W \subseteq \paren{X \setminus A} \times \paren{X \setminus A} \subseteq V_A$

Since $A$ was arbitrart, we have established:
 * $\forall A \in \AA : W \cap A = \O \leadsto W \times W \subseteq V_A$

From Set is Subset of Intersection of Supersets:
 * $W \times W \subseteq ds \bigcap \set{V_A : A \in \AA : W \cap A = \O}$

From Intersection with Subset is Subset:
 * $(1):\quad W \times W = \paren{W \times W} \cap \bigcap \set{V_A : A \in \AA : W \cap A = \O}$

We have:

By definition of product topology:
 * $W \times W$ is open in $T \times T$

Recall:
 * $\set{A \in \AA : W \cap A \ne \O}$ is finite

By :
 * $\paren{W \times W} \cap V$ is open in $T \times T$

By definition of Cartesian product:
 * $\tuple{x, x} \in W \times W$

By definition of diagonal $\Delta_X$:
 * $\tuple{x, x} \in \Delta_X$

By definition of open neighborhood:
 * $\forall A \in \AA : \tuple{x, x} \in \Delta_X \subseteq V_A$

By definition of set intersection:
 * $\tuple{x, x} \in \paren{W \times W} \cap V$

Hence:
 * $V$ is a neighborhood of $\tuple{x, x}$ in $T \times T$ by definition.

Since $x \in X$ was arbitrary, then:
 * $\forall x \in X : V$ is a neighborhood of $\tuple{x,x}$ in $T \times T$

From User:Leigh.Samphier/Topology/Set Neighborhood of Subset iff Neighborhood of all Points of Subset:
 * $V$ is an neighborhood of the diagonal $\Delta_X$ in $T \times T$

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