Characterization of Paracompactness in T3 Space/Lemma 18

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 $\sequence{U_n}_{n \in \N}$ be a sequence of subsets of $X \times X$ containing the diagonal $\Delta_X$ of $X \times X$.

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:
 * $\AA_n = \set{\map {A_n} x : x \in X}$

Let:
 * $\AA = \ds \bigcup_{n \in \N, n \ne 0} \AA_n$

Then:
 * $\AA$ is a cover of $X$

Proof
Let $x \in X$.

We have :
 * $\forall n \in \N_{>0} : \Delta_X \subseteq U_n$

By definition of image:
 * $\forall n \in \N_{>0} : x \in \map {U_n} x$

Let:
 * $y = \min \set{z \in X : x \in \ds \bigcup_{n \in N} \map {U_n} z}$

with respect to the well-ordering $\preccurlyeq$.

By choice of $y$
 * $\exists n \in \N_{>0}$: $x \in \map {U_n} y$

and
 * $\forall z \preccurlyeq y, z \ne y : x \notin \map {U_{n + 1}} z$

By definition of set union:
 * $x \notin \ds \bigcup_{z \preccurlyeq y, z \ne y} \map {U_{n + 1}} z$

By definition of set difference
 * $x \in \map {U_n} y \setminus \ds \bigcup_{z \preccurlyeq y, z \ne y} \map {U_{n + 1}} z$

That is:
 * $x \in \map {A_n} y$

Since $x$ was arbitrary:
 * $\forall x \in X : \exists y \in X, n \in \N_{>0} : x \in \map {A_n} y$

Hence $\AA$ is a cover of $X$ by definition.