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

Proof by Counterexample:

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