Equivalence of Definitions of T3 Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Definition by Open Sets implies Definition by Closed Neighborhoods
Let $T = \left({S, \tau}\right)$ be a topological space for which:


 * $\forall F \subseteq S: \complement_S \left({F}\right) \in \tau, y \in \complement_S \left({F}\right): \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \varnothing$

Let $U\in\tau$, and let $x\in U$.

Then $\complement_S(U)$ is closed and $x\not\in\complement_S(U)$.

We have by hypothesis:


 * $\exists A,B\in\tau: \complement_S(U)\subseteq A, x\in B:A\cap B=\varnothing$

It follows that $x\in B\subseteq\complement_S(A)\subseteq\complement_S(\complement_S(U))=U$.

So $\complement_S(A)$ is a closed neighborhood of $x$ contained in $U$.

As $U,x$ are arbitrary:


 * $\forall U \in \tau: \forall x \in U: \exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$

Definition by Closed Neighborhoods implies Definition by Intersection of Closed Neighborhoods
Let $T = \left({S, \tau}\right)$ be a topological space for which:


 * $\forall U \in \tau: \forall x \in U: \exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$

Let $H\subseteq S$ and $\complement_S(H)\in\tau$.

Let $C_H$ be the set of all closed neighborhoods of $H$:


 * $C_H=\{N_H:\complement_S(N_H)\in\tau,\exists V\in\tau:H\subseteq V\subseteq N_H\}$

Clearly $H\subseteq \bigcap C_H$.

Now let $x\not\in H$. Then $x\in\complement_S(H)\in\tau$.

We have by hypothesis:


 * $\exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq \complement_S(H)$

It follows that $H=\complement_S(\complement_S(H))\subseteq\complement_S(N_x)\subseteq\complement_S(V)$.

Therefore $\complement_S(V)\in C_H$.

Since $x\not\in\complement_S(V)$, $x\not\in\bigcap C_H$.

Therefore $\bigcap C_H\subseteq H$, and thus $\bigcap C_H=H$.

As $H$ is arbitrary:


 * $\forall H \subseteq S: \complement_S \left({H}\right) \in \tau: H = \bigcap \left\{{N_H: \complement_S \left({N_H}\right) \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}\right\}$

Definition by Intersection of Closed Neighborhoods implies Definition by Open Sets
Let $T = \left({S, \tau}\right)$ be a topological space for which:


 * $\forall H \subseteq S: \complement_S \left({H}\right) \in \tau: H = \bigcap \left\{{N_H: \complement_S \left({N_H}\right) \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}\right\}$

Let $F\subseteq S$ and $\complement_S(F)\in\tau$.

We have by hypothesis $F = \bigcap \left\{{N_F: \complement_S \left({N_F}\right) \in \tau, \exists V \in \tau: F \subseteq V \subseteq N_F}\right\}$

Pick arbitrary $x\not\in F$.

Then $\exists N\subseteq S: \complement_S(N) \in \tau, \exists V \in \tau: F \subseteq V \subseteq N$

Since $x\not\in F\subseteq N$, $x\in\complement_S(N)$.

Since $V\subseteq N$, $V\cap \complement_S(N)=\varnothing$.

Therefore we have $\complement_S(N),V\in \tau$, $x\in\complement_S(N)$, $F\subseteq V$, $V\cap \complement_S(N)=\varnothing$.

As $F,x$ is arbitrary:


 * $\forall F \subseteq S: \complement_S \left({F}\right) \in \tau, y \in \complement_S \left({F}\right): \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \varnothing$