Characterization of T3 Space

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


 * $(1) \quad T$ is a $T_3$ space
 * $(2) \quad \forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F = \O$
 * $(3) \quad \forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$

where $V^-$ denotes the closure of $V$ in $T$

Statement $(1)$ implies Statement $(2)$
Let $T$ be a $T_3$ space.

Let $F \subseteq S : S \setminus F \in \tau$.

Let $y \in S \setminus F$.

By definition of $T_3$ space:
 * $\exists V, W \in \tau : y \in V, F \subseteq W : V \cap W = \O$

From Subset of Set Difference iff Disjoint Set:
 * $V \subseteq S \setminus W$

By definition of closed set:
 * $S \setminus W$ is closed in $T$

From Closure of Subset of Closed Set of Topological Space is Subset:
 * $V^- \subseteq S \setminus W$

From Set Difference with Subset is Superset of Set Difference:
 * $S \setminus W \subseteq S \setminus F$

From Subset Relation is Transitive:
 * $V^- \subseteq S \setminus F$

From Subset of Set Difference iff Disjoint Set:
 * $V^- \cap F = \O$

The result follows.

Statement $(2)$ implies Statement $(3)$
Let $T$ satisfy:
 * $\forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F = \O$

Let $U \in \tau$.

Let $y \in U$.

Let $F = S \setminus U$.

From Set Difference with Set Difference:
 * $S \setminus F = U$

Hence:
 * $S \setminus F \in \tau$

and
 * $y \in S \setminus F$

From $(2)$:
 * $\exists V \in \tau : y \in V, V^- \cap F = \O$

Hence:
 * $V^- \cap S \setminus U = \O$

From Subset of Set Difference iff Disjoint Set:
 * $V^- \subseteq U$

Since $U$ and $y$ were arbitrary, we have shown:
 * $\forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$

Statement $(3)$ implies Statement $(1)$
Let $T$ satisfy:
 * $\forall U \in S, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$

Let $U \in \tau$.

Let $y \in U$.

From $(3)$:
 * $\exists V \in \tau : y \in V, V^- \subseteq U$

Let $N = V^-$.

By definition of closure:
 * $V \subseteq N$

From Topological Closure is Closed:
 * $S \ N \in \tau$

Hence:
 * $\exists N \subseteq S : S \setminus N \in \tau : \exists V \in \tau : y \in V \subseteq N \subseteq U$

Since $U$ and $y$ were arbitrary, we have shown:
 * $\forall U \in \tau, y \in U : \exists N \subseteq S : S \setminus N \in \tau : \exists V \in \tau : y \in V \subseteq N \subseteq U$

By definition, $T$ is a $T_3$ space