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:

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

Aiming for a contradiction, suppose $x = y$.

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:

$\set x \ne \set y$

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): topologically distinguishable