Subset of Indiscrete Space is Everywhere Dense

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: $7$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.