Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 2

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

Theorem

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

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

Let $\AA$ be a closed locally finite refinement of $\UU$.

For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.

Let $T \times T = \struct {X \times X, \tau_{X \times X} }$ denote the product space of $T$ with itself.

For each $A \in \AA$, let:

$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A} }$

Let:

$V = \ds \bigcap_{A \mathop \in \AA} V_A$

Let $\Delta_X$ denote the diagonal on $X$.

Then:

$V$ is a neighborhood of the diagonal $\Delta_X$ in $T \times T$.

Proof

Let $x \in X$.

By definition of locally finite:

$\exists W \in \tau : x \in W : \set {A \in \AA : W \cap A \ne \O}$ is finite.

Let:

$A \in \AA : W \cap A = \O$

From Subset of Set Difference iff Disjoint Set:

$W \subseteq X \setminus A$

From Cartesian Product of Subsets:

$W \times W \subseteq \paren {X \setminus A} \times \paren {X \setminus A} \subseteq V_A$

Since $A$ was arbitrary, we have established:

$\forall A \in \AA : W \cap A = \O \leadsto W \times W \subseteq V_A$

From Set is Subset of Intersection of Supersets:

$W \times W \subseteq \bigcap \set {V_A : A \in \AA : W \cap A = \O}$

From Intersection with Subset is Subset:

$(1): \quad W \times W = \paren {W \times W} \cap \bigcap \set {V_A : A \in \AA : W \cap A = \O}$

We have:

\(\ds \paren {W \times W} \cap V\)

\(=\)

\(\ds \paren {W \times W} \cap \bigcap \set {V_A : A \in \AA}\)

Definition of $V$

\(\ds \)

\(=\)

\(\ds \paren {W \times W}\)

\(\ds \)

\(\)

\(\ds \cap \bigcap \set {V_A : A \in \AA : W \cap A = \O}\)

\(\ds \)

\(\)

\(\ds \cap \bigcap \set {V_A : A \in \AA : W \cap A \ne \O}\)

\(\ds \)

\(=\)

\(\ds \paren {W \times W} \cap \bigcap \set {V_A : A \in \AA : W \cap A \ne \O}\)

from $(1)$

By definition of product topology:

$W \times W$ is open in $T \times T$

Lemma 4

$\forall A \in \AA : V_A$ is an open neighborhood of the diagonal $\Delta_X$ in $T \times T$

$\Box$

Recall:

$\set{A \in \AA : W \cap A \ne \O}$ is finite

By Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets:

$\paren{W \times W} \cap V$ is open in $T \times T$

By definition of Cartesian product:

$\tuple{x, x} \in W \times W$

By definition of diagonal $\Delta_X$:

$\tuple{x, x} \in \Delta_X$

By definition of open neighborhood:

$\forall A \in \AA : \tuple{x, x} \in \Delta_X \subseteq V_A$

By definition of set intersection:

$\tuple{x, x} \in \paren{W \times W} \cap V$

Hence:

$V$ is a neighborhood of $\tuple{x, x}$ in $T \times T$ by definition.

Since $x \in X$ was arbitrary, then:

$\forall x \in X : V$ is a neighborhood of $\tuple {x, x}$ in $T \times T$

From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:

$V$ is a neighborhood of the diagonal $\Delta_X$ in $T \times T$.

$\blacksquare$