Characterization of Paracompactness in T3 Space/Lemma 17

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:
 * $\map {A_n} x = \map {U_n} x \setminus \ds \bigcup_{y \preccurlyeq x, y \ne x} \map {U_{n + 1}} y$

For each $n \in \N_{> 0}$, let:
 * $\AA_n = \set{\map {A_n} x : x \in X}$

Then:
 * $\forall n \in \N_{> 0} : \AA_n$ is a discrete set of subsets.

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

Let $x \in X$.

Case: $\forall y : \map {V_{n+1} } x \cap \map {A_n} y = \O$
For all $y \in X$, let:
 * $\map {V_{n+1} } x \cap \map {A_n} y = \O$

From Image of Point under Neighborhood of Diagonal is Neighborhood of Point:
 * $\map {V_{n+1} } x$ is a neighborhood of $x$ in $T$ that intersects no element of $\AA_n$

Case: $\exists y : \map {V_{n+1} } x \cap \map {A_n} y \ne \O$
Let $y \in X$:
 * $\map {V_{n+1} } x \cap \map {A_n} y \ne \O$

From Image under Left-Total Relation is Empty iff Subset is Empty:
 * $V_{n + 1} \sqbrk {\map {V_{n+1} } x \cap \map {A_n} y} \ne \O$

We have:

From Subset of Empty Set:
 * $\set x \cap V_{n + 1} \sqbrk {\map {A_n} y} \ne \O$

Hence:
 * $\set x \subseteq V_{n + 1} \sqbrk {\map {A_n} y}$

We have:

From Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset:
 * $V_{n + 1} \sqbrk {\map {A_n} y}$ is a neighborhood of $\map {A_n} y$

From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
 * $V_{n + 1} \sqbrk {\map {A_n} y}$ is a neighborhood of $x$

From Lemma 16:
 * $\forall z \in X : y \ne z \leadsto \map {A_n} z \cap V_{n+1} \sqbrk {\map {A_n} y} = \O$

Hence $V_{n+1} \sqbrk {\map {A_n} y}$ is a neighborhood of $x$ in $T$ that intersects at most one element of $\AA_n$.

In either case:
 * there exists a neighborhood of $x$ in $T$ that intersects at most one element of $\AA_n$.

Since $x$ was arbitrary:
 * For all $x \in X$ there exists a neighborhood of $x$ in $T$ that intersects at most one element of $\AA_n$.

Hence $\AA_n$ is a discrete set of subsets by definition.

Since $n$ was arbitrary:
 * $\forall n \in \N_{>0} : \AA_n$ is a discrete set of subsets