Equivalence of Definitions of T2 Space

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

Definition 1 implies Definition 2
Let $T = \struct {S, \tau}$ be a topological space for which:


 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$

Let us take any arbitrary $x, y \in S: x \ne y$.

Let $\CC_x$ be the set of all closed neighborhoods of $x$:
 * $\CC_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$

where $\relcomp S H$ is the complement of $H$ in $S$.

We need to demonstrate that the only element in the intersection of $\CC_x$ is $x$:
 * $\bigcap \CC_x = \set x$

and to do that we show that if $y \ne x$ then $y \notin \bigcap \CC_x$.

Let $C = \bigcap C_x$.

Clearly $x \in C$ and so $\set x \subseteq C_x$.

We have that $\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ by hypothesis.

As $x \in U$ it follows that $x \notin V$ and so $x \in \relcomp S V$.

Thus $x \in U \subseteq \relcomp S V$.

That is, $\relcomp S V$ is a closed neighborhood of $x$ and so $\relcomp S V \in \CC_x$.

As $y \in V$ it follows that $y \notin \relcomp S V$.

So $\relcomp S V$ is a closed neighborhood of $x$ which does not contain $y$.

So $y \notin \bigcap C_x$.

As $y$ is arbitrary:
 * $\forall y \in S, y \ne x: \exists H: \relcomp S H \in \tau: y \notin H$

and so $C_x \subseteq \set x$.

That is:
 * $\ds \forall x \in S: \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$

or, each point is the intersection of all its closed neighborhoods.

Definition 2 implies Definition 1
Let $T = \struct {S, \tau}$ be a topological space for which each point is the intersection of all its closed neighborhoods.

Let $x, y \in S: x \ne y$.

Let $\CC_x$ be the set of all closed neighborhoods of $x$:
 * $\CC_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$

where $\relcomp S H$ is the complement of $H$ in $S$.

This arises from the definition of a closed set as the complement in $S$ of an open set.

We have that:
 * $\ds \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$

Then as $y \notin \set x$ it is not the case that $\forall H \in C_x: y \in H$.

So for some $H \in C_x$ it must be the case that $y \in \relcomp S H = V$.

But $V = \relcomp S H \in \tau$, that is, $V$ is open in $T$.

Also, as $U \subseteq H$, it must follow that $U \cap V = \O$.

So:
 * $\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$

As $x$ and $y$ are arbitrary, it follows that:
 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$

Definition 1 implies Definition 3
Let $T = \struct {S, \tau}$ be a topological space for which:


 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$

Let $x, y \in S: x \ne y$ be arbitrary.

From Set is Open iff Neighborhood of all its Points, $U$ and $V$ are neighborhoods of $x$ and $y$.

Thus as $x$ and $y$ are arbitrary:
 * $\forall x, y \in S, x \ne y: \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$

Definition 3 implies Definition 1
Let $T = \struct {S, \tau}$ be a topological space for which:


 * $\forall x, y \in S, x \ne y: \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$

Let $x, y \in S: x \ne y$ be arbitrary.

We have that:
 * $\exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$

$\exists z \in S: z \in U \cap V$.

Then $z \in U, z \in V$

From this contradiction it follows that $U \cap V = \O$.

As $x$ and $y$ are arbitrary, it follows that:
 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$