Characterization of Paracompactness in T3 Space/Lemma 8

Theorem
Let $T = \struct{X, \tau}$ be a topological space.

Let every open cover of $T$ be even.

Let $\UU$ be an open cover of $T$.

Then:
 * there exists a $\sigma$-discrete refinement $\AA$ of $\UU$

Proof
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.

By definition of even cover there exists a neighborhood $V$ of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$:
 * $\set{\map V x : x \in X}$ is a refinement of $\UU$

where:
 * $V$ is seen as a relation on $X \times X$
 * $\map V x$ denotes the image of $x$ under $V$.

In what follows subsets of $X \times X$ will be treated at times as a relation on $X \times X$.

Lemma 13
For all $n \in \N_{> 0}$, let:
 * $U_n = V_n \circ V_{n - 1}, \circ \cdots \circ V_1$

Then:
 * $\forall n \in \N_{> 0} : U_{n + 1} = V_{n + 1} \circ U_n$

Lemma 15
We have :
 * $\set{\map {V_0} x : x \in X}$ refines $\UU$

From Refinement of a Refinement is Refinement of Cover:
 * $(1)\quad\forall n \in \N_{>0} : \set{\map {U_n} x : x \in X}$ refines $\UU$

From Zermelo's Well-Ordering Theorem, let:
 * $\preccurlyeq$ well-order $X$

For each $n \in \N_{> 0}, x \in X$, let:
 * $\map {A_n} x = \map {U_n} x \setminus \ds \bigcup_{y \preccurlyeq x, y \ne x} \map {U_{n + 1}} y$

Lemma 16
For each $n \in \N_{> 0}$, let:
 * $\AA_n = \set{\map {A_n} x : x \in X}$

$\AA_n$ is a Discrete Set of Subsets
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 User:Leigh.Samphier/Topology/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

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

$\AA$ is a Cover of $X$
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.

$\AA$ is a Refinement of $\UU$
Let $x \in X$ and $n \in \N_{>0}$.

From Set Difference is Subset:
 * $\map {A_n} x \subseteq \map {U_n} x$

From (1) and definition of refinement:
 * $\exists U \in \UU : \map {U_n} x \subseteq U$

From Subset Relation is Transitive:
 * $\exists U \in \UU : \map {A_n} x \subseteq U$

Since $x$ and $n$ were arbitrary, then $\AA$ refines $\UU$ by definition.

It follows that $\AA$ is a $\sigma$-discrete refinement of $\UU$ by definition.