# 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:

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

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms - 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.): Problems: $\S 2: \ 9$