Topologically Distinguishable Points are Distinct

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

Let $x, y \in X$ be topologically distinguishable.

Then the singleton sets $\set x$ and $\set y$ are disjoint and so:
 * $x \ne y$

Proof
Let $x$ and $y$ be topologically distinguishable.

Then either:
 * $\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$

or:
 * $\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$

$x = y$.

Then:
 * $\map \neg {\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x}$

and:
 * $\map \neg {\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y}$

Hence the result by Proof by Contradiction:
 * $x \ne y$

and so by Singleton Equality:
 * $\set x \ne \set y$