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Suppose $T$ is a metric space. A subset $R$ of $T$ is called regular open if the interior of the closure of $R$ is equal to $R$ itself:

$$ R = \text{int}(\text{cl}(R)).$$

Suppose $R$ and $S$ are two (non empty) regular open sets with $S \subset R$ (strict inclusion). Is it then necessarily the case that

$$ \text{cl}( R \setminus S) = \text{cl}( \text{cl}(R ) \setminus \text{cl} (S) )?$$

(I worked out this was false in general, my question is whether being regularly open is the right property to ensure it's true)

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Yes, even for a topological space $T$.

Proof of $\supset$. Let $R$ and $S$ be any subsets of $T$. Let $x\in \text{cl}( (\text{cl } R) \setminus \text{cl } S )$ be an arbitrarily point and $U$ be an arbitrary open neighborhood of the point $x$. Then there exists a point $x’\in U\cap ((\text{cl }R) \setminus \text{cl } S)$. A set $V=U\setminus \text{cl } S$ is a neighborhood of the point $x’$. Since $x’\in \text{cl }R$, there exists a point $x’’\in R\cap V\subseteq R\setminus \text{cl } S $. Thus $x\in \text{cl}(R\setminus \text{cl } S )\subseteq \text{cl}(R\setminus S)$.

Proof of $\subset$. Let $R$ be an open and $S$ be a canonically open subsets of $T$. Let $x\in \text{cl}( R \setminus S)$ be an arbitrarily point and $U$ be an arbitrary open neighborhood of the point $x$. Then there exists a point $x’\in U\cap (R \setminus S)$. Since the set $R$ is open, $V=U\cap R$ is an open neighborhood of the point $x'$. If the set $S$ is dense in $V$ then $x'\in V\subseteq\text{int cl } S=S$, a contradiction. Therefore there exists a point $x''\in V\setminus\text{cl } S\subseteq R\setminus\text{cl } S$. Thus $x\in \text{cl}(R\setminus \text{cl } S)\subseteq \text{cl}((\text{cl }R)\setminus (\text{cl } S))$.

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