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 non-empty finite subset of $\R$.

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

Since $H$ is finite, it has no $\omega$-accumulation points, that is, $\Omega = \O$.

Therefore $H \cup \Omega = H$ 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 $H \cup \Omega$ is finite, it is not a closed set of $T$.

Also see

 * Definition:Closure (Topology)


 * Set together with Condensation Points is not necessarily Closed