$\DeclareMathOperator\cl{cl}$Let $X$ be a topological space and let $Y$ be a dense subspace of $X$. Suppose
that $R\left( X\right) $ denotes all regular closed subsets of $X$.


Question 1: $R\left( Y\right) \longrightarrow R\left( X\right) $, $A\rightarrow \cl\_{X}A$ is bijective.


Question 2: If $A$ is regular closed in $Y$, then $\cl\_{X}A$ is the unique
regular closed subset of $X$ with $A=Y\cap \cl\_{X}A$.


Can you answer these questions?



A question about regular closed sets
The answer to both of these questions is **Yes.** And this result can generalize to point-free topology and I consider this result to be more natural in the context of point-free topology.


The following observations were produced earlier by Mehmet Onat, so let me paraphrase those arguments.


Observation: If $U$ is an open subset of $X$ and $Y$ is dense in $X$, then
$\text{Cl}\_X(U\cap Y)=\text{Cl}\_X(U)$.


Proof: Clearly $\text{Cl}\_X(U\cap Y)\subseteq\text{Cl}\_X(U)$. For the converse direction, suppose that $a\in\text{Cl}\_X(U)$. Then whenever $O$ is an open subset of $X$ that contains $a$, we have $O\cap U\neq\emptyset$. However, since $Y$ is dense in $X$, we also know that $O\cap U\cap Y\neq\emptyset$. Therefore, $a\in\text{Cl}\_X(U\cap Y)$, so we may conclude the converse $\text{Cl}\_X(U)\subseteq \text{Cl}\_X(U\cap Y).$ $\square$


Claim: Suppose that $Y$ is dense in $X$. If $C\in R(Y)$, then $\text{Cl}\_X(C)\in R(X)$.


Proof: If $C\in R(Y)$, then there is some open subset $U\subseteq X$ where
$$C=\text{Cl}\_Y(U\cap Y)=\text{Cl}\_X(U\cap Y)\cap Y=\text{Cl}\_X(U)\cap Y.$$


Therefore, $$\text{Cl}\_X(C)=\text{Cl}\_X(\text{Cl}\_X(U)\cap Y)\supseteq
\text{Cl}\_X(U\cap Y)=\text{Cl}\_X(U)=\text{Cl}\_X(\text{Cl}\_X(U))$$
$$\supseteq\text{Cl}\_X(\text{Cl}\_X(U)\cap Y)=\text{Cl}\_X(C),$$
so $\text{Cl}\_X(C)=\text{Cl}\_X(U)$ which is regular closed. $\square$


Claim: If $C\in R(X)$, then $C\cap Y\in R(Y)$


Proof: Since $C\in R(X)$, we have $C=\text{Cl}\_X(U)$ for some open $U\subseteq X$. Therefore, $$R(Y)\ni\text{Cl}\_Y(U\cap Y)=\text{Cl}\_X(U\cap Y)\cap Y=\text{Cl}\_X(U)\cap Y=C\cap Y.$$ $\square.$


If $X$ is a topological space and $Y\subseteq X$ is dense, then let
$i\_{Y,X}:R(Y)\rightarrow R(X)$ be the mapping defined by
$i\_{Y,X}(C)=\text{Cl}\_X(C)$ and define a mapping $j\_{X,Y}:R(X)\rightarrow R(Y)$ by letting $j\_{X,Y}(C)=C\cap Y$.


Claim: The mapping $i\_{Y,X}$ is injective. More generally, if $C,D$ are distinct closed subsets of $Y$, then $\text{Cl}\_X(C)\neq\text{Cl}\_X(D)$.


Proof: We can assume that
$x\_0\in C\setminus D$. Then since $Y\setminus D$ is open in $Y$, there is an open set $U\subseteq X$ where $U\cap Y=Y\setminus D$. Therefore, since
$U\cap D=\emptyset$, we have $U\cap\text{Cl}\_X(D)=\emptyset$ as well, so
$x\_0\in\text{Cl}\_X(C)$, but $x\_0\not\in\text{Cl}\_X(D)$. Therefore, the mapping $R(Y)\rightarrow R(X),C\mapsto\text{Cl}\_X(C)$ is injective. Mehmet Onat observed that injectivity also follows from the fact that $C=\text{Cl}\_Y(C)=Y\cap\text{Cl}\_X(C)$. $\square$


Claim: If $C\in R(X)$, then $i\_{Y,X}(j\_{X,Y}(C))=C$. Therefore, the mappings $i\_{Y,X},j\_{X,Y}$ are inverses.


Proof: Clearly, $\text{Cl}\_X(C\cap Y)\subseteq C$. For the converse direction, suppose that $x\_0\in C$. Suppose now that $U$ is an open subset of $X$ with $x\_0\in U$. Then since $C=\overline{C^\circ}$, we know that
$U\cap C^\circ\neq\emptyset$. Since $U\cap C^\circ\neq\emptyset$ and $U\cap C^\circ$ is open, we know that $U\cap C^\circ\cap Y\neq\emptyset$ since $Y$ is dense in $X$. Therefore, since $U$ is an arbitrary neighborhood of $x\_0$, we have $x\_0\in\overline{C^\circ\cap Y}\subseteq\overline{C\cap Y}$. Therefore,
$C=\text{Cl}\_X(C\cap Y)$. Q.E.D.


**Forcing and point-free topology**


Let $X$ be a regular space, and let $D$ be the intersection of all open dense subsets of $X$. If $X$ has no isolated points and is $T\_1$, then $D$ is empty, but $D$ has virtual points, and $R(X)$ is isomorphic to the lattice of closed subsets of the space $D$. The virtual points of the space $D$ live inside forcing extensions $V[G].$ For example, the Boolean valued model $V^{R[X]}$ always adds points to the space $D$.


If $L$ is a frame, then let $B\_L=\{x^{\*\*}\mid x\in L\}=\{x^\*\mid x\in L\}$ where $^\*$ is the pseudocomplement operation. Then $B\_L$ is a complete Boolean algebra which is the point-free analogue to the lattice of all regular open (and also the lattice of regular closed) subsets of $X$. We say that a sublocale $S$ of a frame $L$ is dense if $0\in S$. The frame $B\_L$ is also a sublocale of $L$, and $B\_L$ is the smallest dense sublocale of $L.$ A frame is fit if and only if each sublocale is the intersection of open sublocales, so in a fit frame $L$, the Boolean algebra $B\_L$ is the intersection of all open sublocales of $L$. If $S$ is a sublocale of a frame $L$, then the Heyting operation on $S$ is the same as the Heyting operation on $L$. In particular, if $S$ is a dense sublocale of $L$, then the pseudocomplement operation on $S$ coincides with the pseudocomplement operation on $S$. Therefore, if $S$ is a dense sublocale of a frame $L$, then $B\_S=B\_L$.


