Characterization of Paracompactness in T3 Space/Lemma 15

Theorem
Let $X$ be a set.

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

Let $\sequence{V_n}_{n \in \N}$ be a sequence of subsets of $X \times X$ containing the diagonal $\Delta_X$ of $X \times X$:
 * $\forall n \in \N_{> 0}$ the composite relation $V_n \circ V_n$ is a subset of $V_{n - 1}$, that is, $V_n \circ V_n \subseteq V_{n - 1}$

For all $n \in \N_{> 0}$, let:
 * $U_n = V_n \circ V_{n - 1}, \circ \cdots \circ V_1$

Then:
 * $\forall n \in \N_{>0}: \set{\map {U_n} x : x \in X}$ refines $\set{\map {V_0} x : x \in X}$

Proof
From Composite of Reflexive Relations is Reflexive:
 * $\forall n \in \N_{>0} : U_n$ is reflexive.

From User:Leigh.Samphier/Topology/Set of Images of Reflexive Relation is Cover of Set:
 * $\set{\map {V_0} x : x \in X}$ is a cover of $X$.

and
 * $\forall n \in \N_{>0} : \set{\map {U_n} x : x \in X}$ is a cover of $X$.

Lemma 14
Let $n \in \N_{>0}$.

Let $x \in X$.

We have:

By definition of subset:
 * $\map {U_n} x \subseteq \map {V_0} x$

Since $n$ and $x$ were arbitrary, we have:
 * $\forall n \in \N_{>0}, x \in X : \map {U_n} x \subseteq \map {V_0} x$

By definiton of refinement:
 * $\forall n \in \N : \set{\map {U_n} x : x \in X}$ refines $\set{\map {V_0} x : x \in X}$