Limit Points of Indiscrete Space

Theorem

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space consisting of at least two points.

Let $H$ be a subset of $T$ such that $H \ne \O$.

Then every point of $T$ is a limit point of $H$.

Proof

By definition, $x \in \struct {S, \tau}$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.

Here, of course, there is only one open set that contains any points at all, and that is $S$.

So as $S$ contains more than one point, it follows that every point of $T$ is a limit point of $H$.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $4$