Characterization of Paracompactness in T3 Space/Lemma 11

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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:

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

From Set is Subset of Union:

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

We have:

\(\ds V^*\)

\(=\)

\(\ds X \setminus \bigcup \set {A \in \AA : A \cap V = \O }\)

definition of $V^*$

\(\ds \)

\(\subseteq\)

\(\ds X \setminus B\)

Set Difference with Subset is Superset of Set Difference

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.

$\blacksquare$