Indiscrete Space is Second-Countable
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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: $7$