Separable Discrete Space is Countable/Proof 1

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Theorem

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

Let $T$ be separable.


Then $S$ is countable.


Proof

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.

$\blacksquare$


Sources