Uncountable Discrete Space is not Separable
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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
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: $8$