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

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## Contents

## 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

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

- 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