Union of Regular Open Sets is not necessarily Regular Open
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular open sets of $T$.
Then $U \cup V$ is not also necessarily a regular open set of $T$.
Proof
By Open Real Interval is Regular Open, the open real intervals:
- $\openint 0 {\dfrac 1 2}, \openint {\dfrac 1 2} 1$
are both regular open sets of $\R$.
Consider $A$, the union of the adjacent open intervals:
- $A := \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1$
From Interior of Closure of Interior of Union of Adjacent Open Intervals:
- $A^{- \circ} = \openint 0 1$
Thus $A$ is not a regular open set of $\R$.
$\blacksquare$
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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(c)}$