Uncountable Discrete Space is not Separable

Theorem

Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.

Then $T$ is not separable.

Proof

By definition, $T$ is separable if and only if there exists a countable subset of $S$ which is everywhere dense in $T$.

Let $H \subseteq S$ be everywhere dense in $T$.

Then by definition of everywhere dense, $H^- = S$ where $H^-$ denotes the closure of $H$.

However, as $T$ is a discrete space, $H^- = H$ from Interior Equals Closure of Subset of Discrete Space.

So $H^- = S \implies H = S$.

But $S$ is uncountable.

So there exists no $H \subseteq S$ such that $H$ is both countable and everywhere dense.

Hence by definition of separable space, if $T$ is an uncountable discrete space it can not be separable.

$\blacksquare$

Also see

• Countable Discrete Space is Separable

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $3$. Uncountable Discrete Topology: $8$