Separable Discrete Space is Countable

Theorem

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

Let $T$ be separable.

Then $S$ is countable.

Proof 1

Let $T$ be separable.

By Space is Separable iff Density not greater than Aleph Zero:

$\map d T \le \aleph_0$

where:

$\map d T$ denotes the density of $T$

$\aleph$ denotes the aleph mapping.

By definition of density:

$\exists A \subseteq S: A$ is everywhere dense and $\map d T = \card A$

where $\card A$ denotes the cardinality of $A$.

By definition of everywhere dense set:

$A^- = S$

where $A^-$ denotes the closure of $A$.

By Set in Discrete Topology is Clopen:

$A$ is closed

Then by Set is Closed iff Equals Topological Closure:

$A^- = A$

Thus by Countable iff Cardinality not greater than Aleph Zero:

$S$ is countable.

$\blacksquare$

Proof 2

Let $T$ be separable.

Aiming for a contradiction, suppose $S$ is uncountable.

Then by Uncountable Discrete Space is not Separable, $T$ is not separable.

Hence the result by Proof by Contradiction.

$\blacksquare$