Intersection of Regular Closed Sets is not necessarily Regular Closed

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

Proof by Counterexample:

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: $6$