Characterization of Paracompactness in T3 Space/Lemma 16

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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 $\sequence{V_n}_{n \in \N}$ 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:

\(\ds V_{n + 1} \sqbrk {\map {A_n} y}\) \(=\) \(\ds V_{n + 1} \sqbrk {\map {U_n} y \setminus \bigcup_{w \preccurlyeq y, w \ne y} \map {U_{n + 1} } w}\) Definition of $\map {A_n} y$
\(\ds \) \(\subseteq\) \(\ds V_{n + 1} \sqbrk {\map {U_n} y}\) Set Difference is Subset and Image of Subset under Relation is Subset of Image
\(\ds \) \(\subseteq\) \(\ds \map {\paren{V_{n + 1} \circ U_n } } y\) Image of Element under Composite Relation with Common Codomain and Domain
\(\ds \) \(\subseteq\) \(\ds \map {U_{n + 1} } y\) Definition of $U_{n + 1}$
\(\ds \) \(\subseteq\) \(\ds \bigcup_{w \preccurlyeq z, w \ne z} \map {U_{n + 1} } w\) Set is Subset of Union


From Set Intersection Preserves Subsets:

\(\ds \map {A_n} z \cap V_{n + 1} \sqbrk {\map {A_n} y}\) \(\subseteq\) \(\ds \map {A_n} z \cap \bigcup_{w \preccurlyeq z, w \ne z} \map {U_{n + 1} } w\)


From Set Difference Intersection with Second Set is Empty Set:

\(\ds \map {A_n} z \cap \bigcup_{w \preccurlyeq z, w \ne z} \map {U_{n + 1} } w\) \(=\) \(\ds \O\)


From Intersection is Empty Implies Intersection of Subsets is Empty:

\(\ds \map {A_n} z \cap V_{n + 1} \sqbrk {\map {A_n} y}\) \(=\) \(\ds \O\)

$\Box$

Case 2: $z \preccurlyeq y$

Applying the same argument used in Case 1 with $y$ and $z$ swapped around:

$(1):\quad\map {A_n} y \cap V_{n + 1} \sqbrk {\map {A_n} z} = \O$


We have:

\(\ds V_{n + 1} \sqbrk{\map {A_n} z \cap V_{n + 1} \sqbrk {\map {A_n} y} }\) \(\subseteq\) \(\ds V_{n + 1} \sqbrk {\map {A_n} z} \cap V_{n + 1} \sqbrk {V_{n + 1} \sqbrk {\map {A_n} y} }\) Image of Intersection under Mapping
\(\ds \) \(=\) \(\ds V_{n + 1} \sqbrk {\map {A_n} z} \cap V_{n + 1} \sqbrk {V_{n + 1}^{-1} \sqbrk {\map {A_n} y} }\) Definition of Symmetric Relation
\(\ds \) \(=\) \(\ds V_{n + 1} \sqbrk {\map {A_n} z} \cap \paren {V_{n + 1} \circ V_{n + 1}^{-1} } \sqbrk {\map {A_n} y}\) Image of Element under Composite Relation with Common Codomain and Domain
\(\ds \) \(\subseteq\) \(\ds V_{n + 1} \sqbrk {\map {A_n} z} \cap \map {A_n} y\) Image of Preimage under Relation is Subset
\(\ds \) \(=\) \(\ds \O\) From Case $(1)$ above


Hence:

$V_{n + 1} \sqbrk{\map {A_n} z \cap V_{n + 1} \sqbrk {\map {A_n} y} } = \O$


From Image under Left-Total Relation is Empty iff Subset is Empty:

$\map {A_n} z \cap V_{n + 1} \sqbrk {\map {A_n} y} = \O$

$\Box$


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:

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

$\blacksquare$

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