Indiscrete Space is Second-Countable

Theorem

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.

Then $T$ is a second-countable space.


Proof

The only basis for $T$ is $\set S$ which is trivially countable.

Hence the result.

$\blacksquare$


Sources

  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $7$
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