Subset of Indiscrete Space is Everywhere Dense
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Theorem
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ such that $H \ne \O$.
Then $H$ is everywhere dense.
Proof
From Limit Points of Indiscrete Space, every point of $T$ is a limit point of $H$.
Hence $H$ is everywhere dense by definition.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $7$