Set together with Omega-Accumulation Points is not necessarily Closed

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

Let $H \subseteq S$.

Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.

Then it is not necessarily the case that $H \cup \Omega$ 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 finite subset of $\R$.

Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.

From Finite Set of Right Order Topology with Omega-Accumulation Points is not Closed, $H \cup \Omega$ is not a closed set of $T$.

Also see

 * Definition:Closure (Topology)


 * Set together with Condensation Points is not necessarily Closed