Characterization of Paracompactness in T3 Space/Lemma 21

Theorem
Let $X$ be a set.

Let $X \times X$ denote the Cartesian product of $X$ with itself.

Let $\BB$ be a set of subsets of $X$.

Let $W \subseteq X \times X$ be symmetric as a relation on $X \times X$.

Then:
 * $\forall B \in \BB, x \in X : \map W x \cap W \sqbrk B \ne \O \leadsto \map {W \circ W} x \cap B \ne \O$

where:
 * $\map W x$ denotes the image of $x$ under the relation $W$
 * $W \sqbrk B$ denotes the image of $B$ under the relation $W$
 * $W \circ W$ denotes the composite relation of $W$ with itself

Proof
Let $B \in \BB$.

Let $x \in X$.

Let $y \in \map W x \cap W \sqbrk B$.

By definition of intersection:
 * $y \in \map W x$

and
 * $y \in W \sqbrk B$

By definition of image of element:
 * $\tuple{x, y} \in W$

By definition of symmetric:
 * $\tuple{y, x} \in W$

By definition of image of subset:
 * $\exists z \in B : \tuple{z, y} \in W$

By definition of composite relation:
 * $\exists z \in B : \tuple{z, x} \in W \circ W$

By definition of image of element:
 * $z \in \map {W \circ W} x \cap B$

Hence:
 * $\map {W \circ W} x \cap B \ne \O$

Since $B$ and $x$ were arbitrary, then:
 * $\forall B \in \BB, x \in X : \map W x \cap W \sqbrk B \ne \O \leadsto \map {W \circ W} x \cap B \ne \O$