Characterization of Paracompactness in T3 Space/Lemma 16

Theorem
Let $X$ be a set with well-ordering $\preccurlyeq$ on $X$.

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 $X \times X$:
 * $\forall n \in \N_{> 0}, V_n$ is symmetric as a relation on $X \times X$
 * $\forall n \in \N_{> 0}$, 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:
 * $U_n = V_n \circ V_{n - 1}, \circ \cdots \circ V_1$

For each $n \in \N_{> 0}, x \in X$, let:
 * $\map {A_n} x = \map {U_n} x \setminus \ds \bigcup_{y \preccurlyeq x, y \ne x} \map {U_{n + 1}} y$

Then:
 * $\forall n \in \N_{> 0}, \forall y, z \in X : y \ne z \leadsto \map {A_n} z \cap V_{n+1} \sqbrk {\map {A_n} y} = \O$

Proof
Let $n \in \N_{> 0}$.

Let $y, z \in X : y \ne z$.

Case 1: $y \preccurlyeq z$
We have:

Case 2: $z \preccurlyeq y$
In either case, we have:
 * $\map {A_n} z \cap V_{n+1} \sqbrk {\map {A_n} y} = \O$

Since $n$, $y$ and $z$ were arbitrary, we have:
 * $\forall n \in \N_{> 0}, \forall y, z \in X : y \ne z \leadsto \map {A_n} z \cap V_{n+1} \sqbrk {\map {A_n} y} = \O$