Equivalence of Definitions of T2 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== 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:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
* : Problems: $\S 2: \ 8$
Category:Hausdorff Spaces
Category:Separation Axioms