Characterization of Paracompactness in T3 Space/Lemma 17

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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:

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


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

Lemma 16

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


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 $\AA_n$

$\Box$

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:

\(\ds V_{n + 1} \sqbrk {\map {V_{n+1} } x \cap \map {A_n} y}\) \(\subseteq\) \(\ds V_{n + 1} \sqbrk {\map {V_{n+1} } x } \cap V_{n + 1} \sqbrk {\map {A_n} y}\) Image of Intersection under Relation
\(\ds \) \(=\) \(\ds V_{n + 1} \sqbrk {\map {V_{n+1}^{-1} } x } \cap V_{n + 1} \sqbrk {\map {A_n} y}\) Definition of Symmetric Relation
\(\ds \) \(=\) \(\ds \map {\paren{V_{n + 1} \circ V_{n+1}^{-1} } } x \cap V_{n + 1} \sqbrk {\map {A_n} y}\) Image of Element under Composite Relation with Common Codomain and Domain
\(\ds \) \(=\) \(\ds \paren{V_{n + 1} \circ V_{n+1}^{-1} } \sqbrk {\set x} \cap V_{n + 1} \sqbrk {\map {A_n} y}\) Image of Singleton under Relation
\(\ds \) \(\subseteq\) \(\ds \set x \cap V_{n + 1} \sqbrk {\map {A_n} y}\) Image of Preimage under Relation is Subset


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:

\(\ds V_{n + 1} \sqbrk {\set x}\) \(\subseteq\) \(\ds V_{n + 1} \sqbrk {V_{n + 1} \sqbrk {\map {A_n} y} }\) Image of Subset under Relation is Subset of Image
\(\ds \) \(=\) \(\ds V_{n + 1} \sqbrk {V_{n + 1}^{-1} \sqbrk {\map {A_n} y} }\) Definition of Symmetric Relation
\(\ds \) \(=\) \(\ds \paren{V_{n + 1} \circ V_{n + 1}^{-1} } \sqbrk {\map {A_n} y}\) Image of Element under Composite Relation with Common Codomain and Domain
\(\ds \) \(\subseteq\) \(\ds \map {A_n} y\) Image of Preimage under Relation is Subset
\(\ds \leadsto \ \ \) \(\ds \map {V_{n + 1} } x\) \(\subseteq\) \(\ds \map {A_n} y\) Image of Singleton under Relation
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \map {V_{n + 1} } x \subseteq \map {A_n} y\) Definition of Reflexive Relation


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 $\AA_n$.

$\Box$


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

$\blacksquare$

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