Equivalence of Definitions of Completely Hausdorff Space

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

Definition 1 implies Definition 2
Let $\struct {S, \tau}$ satisfy:
 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$

Let $x, y \subseteq S, x \ne y $ be arbitrary.

Then:
 * $\exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$

Let $N_x = U$ and $N_y = V$.

From Set is Subset of Itself then:
 * $\exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \in U \subseteq N_x, y \in V \subseteq N_y: N_x^- \cap N_y^- = \O$

Because $x, y \in S$ were 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 2 implies Definition 1
Let $\struct {S, \tau}$ satisfy:
 * $\forall x, y \in S, x \ne y : \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x^- \cap N_y^- = \O$

Let $x, y \subseteq S, x \ne y $ be arbitrary.

Then:
 * $\exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \in U \subseteq N_x, y \in V \subseteq N_y: N_x^- \cap N_y^- = \O$

From Topological Closure of Subset is Subset of Topological Closure:
 * $U^- \subseteq N_x^-, V^- \subseteq N_y^-$

From Subsets of Disjoint Sets are Disjoint then:
 * $U^- \cap V^- = \O$

Thus:
 * $\exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$

Because $x, y \in S$ were arbitrary:
 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$