Characterization of Paracompactness in T3 Space/Lemma 10

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

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

Let $\AA = \ds \bigcup_{n \in \N} \AA_n$ be a $\sigma$-discrete refinement of $\UU$:
 * $\forall n \in \N : \AA_n$ is a discrete set of subsets

For each $n \in \N$, let $V_n$ be an open neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$:
 * $\forall x \in X : \card {\set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \ne \O}} \le 1$

For each $A \in \AA$ let:
 * $U_A \in U : A \subseteq U_A$

For each $n \in \N$, let:
 * $\WW_n = \set{U_A \cap V_n \sqbrk A : A \in \AA_n}$

Let:
 * $\WW = \ds \bigcup_{n \in \N} \WW_n$

Then:
 * $\WW$ is an open $\sigma$-discrete refinement of $\UU$

$\WW$ is Set of Open Sets
Let:
 * $W \in \WW$

By definition of $\WW$:
 * $\exists n \in \N, A \in \AA : W = U_A \cap V_n \sqbrk A$

We have :
 * $U_A \in \tau$

From Image of Subset under Open Neighborhood of Diagonal is Open Neighborhood of Subset:
 * $V_n \sqbrk A \in \tau$

By :
 * $W \in \tau$

Since $W$ was arbitrary, it follows that:
 * $\WW$ is a set of open sets

$\WW$ is a Cover of $X$
Let:
 * $x \in X$

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

Let:
 * $n = \min \set{m \in \N : A \in \AA_m}$

By definition of neighborhood:
 * $\Delta_X \subseteq V_n$

Hence:
 * $\forall a \in A : \tuple{a, a} \in V_n$

By definition of image:
 * $\forall a \in A : a \in V_n \sqbrk A$

By definition of subset:
 * $A \subseteq V_n \sqbrk A$

We have :
 * $A \subseteq U_A$

From Set is Subset of Intersection of Supersets:
 * $A \subseteq U_A \cap V_n \sqbrk A$

By definition of subset:
 * $x \in U_A \cap V_n \sqbrk A$

By definition of $\WW_n$:
 * $U_A \cap V_n \sqbrk A \in \WW_n \subseteq \WW$

Hence:
 * $\exists W \in \WW : x \in W$

Since $x$ was arbitrary, it follows that $\WW$ is a cover of $X$.

$\WW$ is a Refinement of $\UU$
Let:
 * $W \in \WW$

By definition of $\WW$:
 * $\exists n \in \N, A \in \AA : W = U_A \cap V_n \sqbrk A$

From Intersection is Subset:
 * $W \subseteq U_A$

Hence:
 * $\exists U \in \UU : W \subseteq U$

Since $W$ was arbitrary, it follows that $\WW$ is a refinement of $\UU$ by definition.

$\WW_k$ is Discrete
Let $n \in \N$.

Let $x \in X$.

We have :
 * $\card {\set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \ne \O}} \le 1$

From Subsets of Disjoint Sets are Disjoint:
 * $\forall A \in A_n : \map {V_n} x \cap V_n \sqbrk A \cap U_A \ne \O \leadsto \map {V_n} x \cap V_n \sqbrk A \ne \O$

Hence:
 * $\set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \cap U_A \ne \O} \subseteq \set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \ne \O}$

From Cardinality of Subset of Finite Set:
 * $\card{\set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \cap U_A \ne \O}} \le 1$

The mapping $f: \set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \cap U_A \ne \O} \to \set{W \in \WW_n : \map {V_n} x \cap W \ne \O}$ defined by:
 * $\map f A = V_n \sqbrk A \cap U_A$

is surjective.

From Cardinality of Surjection:
 * $\card {\set{W \in \WW_n : \map {V_n} x \cap W \ne \O}} \le \card {\set{A \in \AA_n : \map {V_n} x \cap V_n \sqbrk A \cap U_A \ne \O}}$

Hence:
 * $\card {\set{W \in \WW_n : \map {V_n} x \cap W \ne \O}} \le 1$

From Image of Point under Open Neighborhood of Diagonal is Open Neighborhood of Point:
 * $\map {V_n} x \in \tau$ is an open neighborhood of $x$

Since $x$ was arbitrary, it follows that:
 * $\forall x \in X : \map {V_n} x \in \tau : x \in \map {V_n} x : \card {\set{W \in \WW_n : \map {V_n} x \cap W \ne \O}} \le 1$

It follows that $\WW_n$ is a discrete set of subsets by definition.

Since $n$ was arbitrary, it follows that:
 * $\forall n \in \N : \WW_n$ is a discrete set of subsets

It follows that $\WW$ is an open $\sigma$-discrete refinement of $\UU$ by definition.