Indiscrete Space is Separable

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Theorem

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space such that $S$ has more than one element.

Then $T$ is 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 $x \in T$.

Then $\left\{{x}\right\} \subseteq T$ and $\left\{{x}\right\}$ is (trivially) countable.

From Subset of Indiscrete Space is Everywhere Dense we have that $\left\{{x}\right\}$ is everywhere dense.

Hence the result by definition of separable space.

$\blacksquare$


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