Characterization of Paracompactness in T3 Space/Lemma 18
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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 by hypothesis:
- $\forall n \in \N_{>0} : \Delta_X \subseteq U_n$
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.
$\blacksquare$