# Subset of Indiscrete Space is Dense-in-itself

## Theorem

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ 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.

$\blacksquare$