Characterization of Paracompactness in T3 Space/Statement 3 implies Statement 1

Theorem
Let $T = \struct{X, \tau}$ be a $T_3$ Space.

If every open cover of $T$ have a closed locally finite refinement then:
 * $T$ is paracompact.

Proof
Let every open cover of $T$ have a closed locally finite refinement.

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

Let $\VV$ be a closed locally finite refinement of $\UU$, which exists by assumption.

By definition of locally finite:
 * $\forall x \in X: \exists W \in \tau: x \in W$ and $\set{V \in \VV : V \cap W \ne \O}$ is finite.

For all $x \in X$, let:
 * $W_x \in \tau: x \in W_x$ and $\set{V \in \VV : V \cap W \ne \O}$ is finite

Let $\WW = \set{W_x : x \in X}$.

We have $\WW$ is an open cover of $T$, by definition.

Let $\AA$ be a closed locally finite refinement of $\WW$, which exists by assumption.

For each $V \in \VV$, let:
 * $V^* = X \setminus \ds \bigcup \set{A \in \AA : A \cap V = \O}$

Lemma 4
Let $\VV^* = \set{V^* : V \in \VV}$.

Lemma 5
From Lemma 4 and Lemma 5 it follows that $\VV$ is a refinement of $\VV^*$ by definition.

From [User:Leigh.Samphier/Topology/Common Refinement Condition for Open Locally Finite Refinement of Open Cover]:
 * there exists an open locally finite refinement $\UU^*$ of $\UU$

Since $\UU$ was arbitrary, it follows that $T$ is paracompact by definition.