Set together with Condensation Points is not necessarily Closed

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

Let $\CC$ denote the set of condensation points of $H$.

Then it is not necessarily the case that $H \cup \CC$ is a closed set of $T$.

Proof

Proof by Counterexample:

Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.

Let $H \subseteq \R$ be a non-empty finite subset of $\R$.

Let $\CC$ denote the set of condensation points of $H$.

Since $H$ is finite, it has no condensation points, that is, $\CC = \O$.

Therefore $H \cup \CC = H$ is finite.

From Closed Sets of Right Order Space on Real Numbers, closed sets of $T$ are either infinite or is the empty set.

Since $H \cup \CC$ is finite, it is not a closed set of $T$.

$\blacksquare$

Also see

• Definition:Closure (Topology)

• Set together with Omega-Accumulation Points is not necessarily Closed

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors