Intersection of Regular Closed Sets is not necessarily Regular Closed

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Theorem

Let $T = \left({S, \tau}\right)$ 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:

$\left[{0 \,.\,.\, \dfrac 1 2}\right], \left[{\dfrac 1 2 \,.\,.\, 1}\right]$

are both regular closed sets of $\R$.


Consider $A$, the intersection of the two half-unit closed intervals:

$A := \left[{0 \,.\,.\, \dfrac 1 2}\right] \cap \left[{\dfrac 1 2 \,.\,.\, 1}\right] = \left\{{\dfrac 1 2}\right\} = \left[{\dfrac 1 2 \,.\,.\, \dfrac 1 2}\right]$

From Interior of Closed Real Interval is Open Real Interval:

$A^\circ = \left({\dfrac 1 2 \,.\,.\, \dfrac 1 2}\right) = \varnothing$

From Closure of Empty Set is Empty Set:

$A^{\circ -} = \varnothing \ne A$

Thus $A$ is not a regular closed set of $\R$.

$\blacksquare$


Sources