Indiscrete Space is Second-Countable

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Theorem

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Then $T$ is a second-countable space.


Proof

The only basis for $T$ is $\left\{{S}\right\}$ which is trivially countable.

Hence the result.

$\blacksquare$


Sources