Equivalence of Definitions of T3 Space

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Theorem

The following definitions of the concept of $T_3$ space are equivalent:


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

Definition by Open Sets

$T = \struct {S, \tau}$ is a $T_3$ space if and only if:

$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$

That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.

Definition by Closed Neighborhoods

$T = \struct {S, \tau}$ is a $T_3$ space if and only if each open set contains a closed neighborhood around each of its points:

$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$

where $N_x$ denotes a neighborhood of $x$.

Definition by Intersection of Closed Neighborhoods

$T = \struct {S, \tau}$ is a $T_3$ space if and only if each of its closed sets is the intersection of its closed neighborhoods:

$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$


Proof

Definition by Open Sets implies Definition by Closed Neighborhoods

Let $T = \struct {S, \tau}$ be a topological space for which:

$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$


Let $U \in \tau$, and let $x \in U$.

Then by Relative Complement of Relative Complement:

$\relcomp S {\relcomp S U} \in \tau$

So by definition of closed:

$\relcomp S U$ is closed

and:

$x \notin \relcomp S U$


We have by hypothesis:

$\exists A, B \in \tau: \relcomp S U \subseteq A, x \in B: A \cap B = \O$

It follows that:

\(\ds x\) \(\in\) \(\ds B\) by hypothesis
\(\ds \) \(\subseteq\) \(\ds \relcomp S A\) Empty Intersection iff Subset of Complement: from $A \cap B$
\(\ds \) \(\subseteq\) \(\ds \relcomp S {\relcomp S U}\) Set Complement inverts Subsets: from $\relcomp S U \subseteq A$
\(\ds \) \(=\) \(\ds U\) Relative Complement of Relative Complement

That is:

$x \in \relcomp S A \subseteq U$

So we have demonstrated that there exists a closed neighborhood $\relcomp S A$ of $x$ contained in $U$.


As $U, x$ are arbitrary:

$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$

$\Box$


Definition by Closed Neighborhoods implies Definition by Intersection of Closed Neighborhoods

Let $T = \struct {S, \tau}$ be a topological space for which:

$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$

where $N_x$ denotes a neighborhood of $x$.


Let $H \subseteq S$ such that:

$\relcomp S H \in \tau$

Let $C_H$ be the set of all closed neighborhoods of $H$:

$C_H = \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$

By construction:

$H \subseteq \bigcap C_H$


Let:

$x \notin H$

Then:

$x \in \relcomp S H \in \tau$


We have by hypothesis:

$\exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq \relcomp S H$

It follows that:

\(\ds H\) \(=\) \(\ds \relcomp S {\relcomp S H}\) Relative Complement of Relative Complement
\(\ds \) \(\subseteq\) \(\ds \relcomp S {N_x}\) Set Complement inverts Subsets: from $N_x \subseteq \relcomp S H$
\(\ds \) \(\subseteq\) \(\ds \relcomp S V\) Set Complement inverts Subsets: from $V \subseteq N_x$


Therefore by construction of $C_H$:

$\relcomp S V \in C_H$


Then:

\(\ds x\) \(\in\) \(\ds V\) Definition of By Hypothesis
\(\ds \leadsto \ \ \) \(\ds x\) \(\notin\) \(\ds \relcomp S V\) Definition of Relative Complement
\(\ds \leadsto \ \ \) \(\ds x\) \(\notin\) \(\ds \bigcap C_H\) Definition of Set Intersection
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \relcomp S {\bigcap C_H}\) Definition of Relative Complement


That is:

\(\ds x \in \relcomp S H\) \(\implies\) \(\ds x \in \relcomp S {\bigcap C_H}\)
\(\ds \leadsto \ \ \) \(\ds \relcomp S H\) \(\subseteq\) \(\ds \relcomp S {\bigcap C_H}\) Definition of Subset
\(\ds \leadsto \ \ \) \(\ds \bigcap C_H\) \(\subseteq\) \(\ds H\) Set Complement inverts Subsets


We have shown that:

$\bigcap C_H \subseteq H$

and:

$H \subseteq \bigcap C_H$

So by definition of set equality:

$\bigcap C_H = H$


As $H$ is arbitrary:

$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$

$\Box$


Definition by Intersection of Closed Neighborhoods implies Definition by Open Sets

Let $T = \struct {S, \tau}$ be a topological space for which:

$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$


Let $F \subseteq S$ and $\relcomp S F \in \tau$.

We have by hypothesis:

$F = \bigcap \set {N_F: \relcomp S {N_F} \in \tau, \exists V \in \tau: F \subseteq V \subseteq N_F}$


Pick arbitrary $x \notin F$.

Then:

$\exists N \subseteq S: \relcomp S N \in \tau, \exists V \in \tau: F \subseteq V \subseteq N$

Because $x \notin F \subseteq N$:

$x \in \relcomp S N$

Because $V \subseteq N$, it follows from Empty Intersection iff Subset of Complement that:

$V \cap \relcomp S N = \O$


Therefore we have:

\(\ds \relcomp S N, V\) \(\in\) \(\ds \tau\)
\(\ds x\) \(\in\) \(\ds \relcomp S N\)
\(\ds F\) \(\subseteq\) \(\ds V\)
\(\ds V \cap \relcomp S N\) \(=\) \(\ds \O\)


As $F$ and $x$ are arbitrary:

$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$

$\blacksquare$


Sources

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