Equivalence of Definitions of T3 Space

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

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:

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:


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

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:

Therefore by construction of $C_H$:
 * $\relcomp S V \in C_H$

Then:

That is:

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:


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

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:

As $F$ and $x$ are arbitrary:


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