Equivalence of Definitions of T3 Space

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

The following three conditions defining a $T_3$ space are logically equivalent:

Definition by Open Sets
$T$ is a $T_3$ space iff:


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

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
$(a): \quad T$ is a $T_3$ space iff:
 * Each open set contains a closed neighborhood around each of its points:


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

$(b): \quad T$ is a $T_3$ space iff:
 * Each of its closed sets is the intersection of its closed neighborhoods:


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

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


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

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


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