Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 1

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

Let $\UU$ be an open cover of $T$.

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

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

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

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

where:
 * $V_A$ is seen as a relation on $X \times X$
 * $\map {V_A} x$ denotes the image of $x$ under $V_A$.

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

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

where:
 * $V$ is seen as a relation on $X \times X$
 * $\map V x$ denotes the image of $x$ under $V$.

Then:
 * $\set {\map V x : x \in X}$ is a refinement of $\UU$.

Proof
Let $x \in X$.

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

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

Lemma 3
From lemma 3
 * $\map V x \subseteq \map {V_A} x = U_A \in \UU$

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

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

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