Set together with Omega-Accumulation Points is not necessarily Closed

From ProofWiki
Jump to navigation Jump to search

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 $T$.

$\blacksquare$


Also see


Sources