Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 1
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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:
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:
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
- $\forall A \in \AA, x \in A : \map {V_A} x = U_A$
$\Box$
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.
$\blacksquare$