Intersection of Regular Closed Sets is not necessarily Regular Closed
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular closed sets of $T$.
Then $U \cap V$ is not also necessarily a regular closed set of $T$.
Proof
By Closed Real Interval is Regular Closed, the closed real intervals:
- $\closedint 0 {\dfrac 1 2}, \closedint {\dfrac 1 2} 1$
are both regular closed sets of $\R$.
Consider $A$, the intersection of the two half-unit closed intervals:
- $A := \closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2} = \closedint {\dfrac 1 2} {\dfrac 1 2}$
From Interior of Closed Real Interval is Open Real Interval:
- $A^\circ = \openint {\dfrac 1 2} {\dfrac 1 2} = \O$
From Closure of Empty Set is Empty Set:
- $A^{\circ -} = \O \ne A$
Thus $A$ is not a regular closed 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: $6$