# Equivalence of Definitions of T3 Space

## 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 $T_3$ 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 $T_3$ if and only if each of its closed sets is the intersection of its closed neighborhoods:

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

## 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:

$\relcomp S U$ 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:

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

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

Therefore by construction of $C_H$:

$\relcomp S V \in C_H$

Then:

 $\displaystyle x$ $\in$ $\displaystyle V$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle x$ $\notin$ $\displaystyle \relcomp S V$ Definition of Relative Complement $\displaystyle \leadsto \ \$ $\displaystyle x$ $\notin$ $\displaystyle \bigcap C_H$ Definition of Set Intersection $\displaystyle \leadsto \ \$ $\displaystyle x$ $\in$ $\displaystyle \relcomp S {\bigcap C_H}$ Definition of Relative Complement

That is:

 $\displaystyle x \in \relcomp S H$ $\implies$ $\displaystyle x \in \relcomp S {\bigcap C_H}$ $\displaystyle \leadsto \ \$ $\displaystyle \relcomp S H$ $\subseteq$ $\displaystyle \relcomp S {\bigcap C_H}$ Definition of Subset $\displaystyle \leadsto \ \$ $\displaystyle \bigcap C_H$ $\subseteq$ $\displaystyle H$ 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:

 $\displaystyle \relcomp S N, V$ $\in$ $\displaystyle \tau$ $\displaystyle x$ $\in$ $\displaystyle \relcomp S N$ $\displaystyle F$ $\subseteq$ $\displaystyle V$ $\displaystyle V \cap \relcomp S N$ $=$ $\displaystyle \O$

As $F$ and $x$ are arbitrary:

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

$\blacksquare$