Subset of Indiscrete Space is Dense-in-itself
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Theorem
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ be a subset of $S$ containing more than one point.
Then $H$ is dense-in-itself.
Proof
Let $x \in H$.
Then as $H$ is not singleton, $\exists y \in H: y \ne x$.
Then every neighborhood of $x$ contains $y$, as the only open set of $T$ is $S$, which also contains both $x$ and $y$.
Hence $x$ is not isolated by definition.
$x$ is general, so all points in $H$ are similarly not isolated.
Hence the subset $H$ is dense-in-itself 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: $5$