Characterization of Paracompactness in T3 Space/Lemma 11

Theorem
Let $X$ be a set.

Let $\AA$ and $\VV$ be sets of subsets of $X$.

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

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

Then:
 * $\forall A \in \AA, V^* \in \VV^* : A \cap V^* \ne \O \implies A \cap V \ne \O$

Proof
We prove the contrapositive statement:
 * $\forall A \in \AA, V^* \in \VV^* : A \cap V = \O \implies A \cap V^* = \O$

Let $B \in \AA, V^* \in \VV^* : B \cap V = \O$.

Hence:
 * $B \in \set{A \in \AA : A \cap V = \O }$

So:
 * $B \subseteq \ds \bigcup \set{A \in \AA : A \cap V = \O }$

We have:

From Subset of Set Difference iff Disjoint Set:
 * $V^* \cap B = \O$

Since $B$ and $V^*$ were arbitrary:
 * $\forall A \in \AA, V^* \in \VV^* : A \cap V = \O \implies A \cap V^* = \O$

The result follows from Rule of Transposition.