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regular open algebra
A regular open algebra is an algebraic system $\mathcal{A}$ whose universe is the set of all regular open sets in a topological space $X$, and whose operations are given by
1. a constant $1$ such that $1:=X$,
2. a unary operation ${}^{{\prime}}$ such that for any $U$,β $U^{{\prime}}:=U^{\bot}$, where $U^{\bot}$ is the complement of the closure of $U$ in $X$,
3. a binary operation $\wedge$ such that for any $U,V\in\mathcal{A}$, $U\wedge V:=U\cap V$, and
4. a binary operation $\vee$ such that for any $U,V\in\mathcal{A}$, $U\vee V:=(U\cup V)^{{\bot\bot}}$.
From the parent entry, all of the operations above are welldefined (that the result sets are regular open). Also, we have the following:
Theorem 1.
$\mathcal{A}$ is a Boolean algebra
Proof.
We break down the proof into steps:
1. $\mathcal{A}$ is a lattice. This amounts to verifying various laws on the operations:

(idempotency of $\vee$ and $\wedge$): Clearly, $U\wedge U=U$. Also, $U\vee U=(U\cup U)^{{\bot\bot}}=U^{{\bot\bot}}=U$, since $U$ is regular open.

Commutativity of the binary operations are obvious.

The associativity of $\wedge$ is also obvious. The associativity of $\vee$ goes as follows: $U\vee(V\vee W)=(U\cup(V\cup W)^{{\bot\bot}})^{{\bot\bot}}=U^{\bot}\cap(V\cup W% )^{{\bot\bot\bot}}=U^{\bot}\cap(V\cup W)^{\bot}$, since $V\cup W$ is open (which implies that $(V\cup W)^{\bot}$ is regular open). The last expression is equal to $U^{\bot}\cap(V^{\bot}\cap W^{\bot})$. Interchanging the roles of $U$ and $W$, we obtain the equation $W\vee(V\vee U)=W^{\bot}\cap(V^{\bot}\cap U^{\bot})$, which is just $U^{\bot}\cap(V^{\bot}\cap W^{\bot})$, or $U\vee(V\vee W)$. The commutativity of $\vee$ completes the proof of the associativity of $\vee$.

Finally, we verify the absorption laws. First, $U\wedge(U\vee V)=U\cap(U\cup V)^{{\bot\bot}}=U^{{\bot\bot}}\cap(U\cup V)^{{% \bot\bot}}=(U^{\bot}\cup(U\cup V)^{\bot})^{\bot}=(U^{\bot}\cup(U^{\bot}\cap V^% {\bot})^{\bot}=(U^{\bot})^{\bot}=U$. Second, $U\vee(U\wedge V)=(U\cup(U\vee W)^{{\bot\bot}}=U^{{\bot\bot}}=U$.

2. $\mathcal{A}$ is complemented. First, it is easy to see that $\varnothing$ and $X$ are the bottom and top elements of $\mathcal{A}$. Furthermore, for any $U\in\mathcal{A}$, $U\wedge U^{{\prime}}=U\cap U^{\bot}=U\cap(X\setminus\overline{U})\subseteq% \overline{U}\cap(X\setminus\overline{U})=\varnothing$. Finally, $U\vee U^{{\prime}}=(U\cup U^{\bot})^{{\bot\bot}}=(U^{\bot}\cap U^{{\bot\bot}})% ^{\bot}=(U^{\bot}\cap U)^{\bot}=\varnothing^{\bot}=X$.
3. $\mathcal{A}$ is distributive. This can be easily proved once we show the following: for any open sets $U,V$:
$(*)\qquad U^{{\bot\bot}}\cap V^{{\bot\bot}}=(U\cap V)^{{\bot\bot}}.$ To begin, note that since $U\cap V\subseteq U$, and ${}^{\bot}$ is order reversing, $(U\cap V)^{{\bot\bot}}\subseteq U^{{\bot\bot}}$ by applying ${}^{\bot}$ twice. Do the same with $V$ and take the intersection, we get one of the inclusions: $(U\cap V)^{{\bot\bot}}\subseteq U^{{\bot\bot}}\cap V^{{\bot\bot}}$. For the other inclusion, we first observe that
$U\cap\overline{V}\subseteq\overline{U\cap V}.$ If $x\in$ LHS, then $x\in U$ and for any open set $W$ with $x\in W$, we have that $W\cap V\neq\varnothing$. In particular, $U\cap W$ is such an open set (for $x\in U\cap W$), so that $(U\cap W)\cap V\neq\varnothing$, or $W\cap(U\cap V)\neq\varnothing$. Since $W$ is arbitrary, $x\in$ RHS. Now, apply the set complement, we have $(U\cap V)^{\bot}\subseteq U^{\complement}\cup V^{\bot}$. Applying ${}^{\bot}$ next we get $(U\cap V)^{{\bot\bot}}$ for the LHS, and $(U^{\complement}\cup V^{\bot})^{\bot}=U^{{\complement\complement}}\cap V^{{% \bot\bot}}=U^{{\complement\complement}}\cap V^{{\bot\bot}}=U\cap V^{{\bot\bot}}$ for RHS, since $U^{\complement}$ is closed. As ${}^{\bot}$ reverses order, the new inclusion is
$(**)\qquad U\cap V^{{\bot\bot}}\subseteq(U\cap V)^{{\bot\bot}}.$ From this, a direct calculation shows $U^{{\bot\bot}}\cap V^{{\bot\bot}}\subseteq(U^{{\bot\bot}}\cap V)^{{\bot\bot}}% \subseteq(U\cap V)^{{\bot\bot\bot\bot}}=(U\cap V)^{{\bot\bot}}$, noticing that the first and second inclusions use $(**)$ above (and the fact that ${}^{{\bot\bot}}$ preserves order), and the last equation uses the fact that for any open set $W$, $W^{\bot}$ is regular open. This proves the $(*)$.
Finally, to finish the proof, we only need to show one of two distributive laws, say, $U\wedge(V\vee W)=(U\wedge V)\vee(U\wedge W)$, for the other one follows from the use of the distributive inequalities. This we do be direct computation: $U\wedge(V\vee W)=U\cap(V\cup W)^{{\bot\bot}}=U^{{\bot\bot}}\cap(V\cup W)^{{% \bot\bot}}=(U\cap(V\cup W))^{{\bot\bot}}=((U\cap V)\cup(U\cap W))^{{\bot\bot}}% =((U\wedge V)\cup(U\wedge W))^{{\bot\bot}}=(U\wedge V)\vee(U\wedge W)$.
Since a complemented distributive lattice is Boolean, the proof is complete. β
Theorem 2.
The subset $\mathcal{B}$ of all clopen sets in $X$ forms a Boolean subalgebra of $\mathcal{A}$.
Proof.
Clearly, every clopen set is regular open. In addition, $1\in\mathcal{B}$. If $U$ is clopen, so is the complement of its closure, and hence $U^{{\prime}}\in\mathcal{B}$. If $U,V$ are clopen, so is their intersection $U\wedge V$. Similarly, $U\cup V$ is clopen, so that $U\vee V=U\cup V$ is clopen also. β
Theorem 3.
In fact, $\mathcal{A}$ is a complete Boolean algebra. For an arbitrary subset $\mathcal{K}$ of $A$, the meet and join of $\mathcal{K}$ are $(\bigcap\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}$ and $(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}$ respectively.
Proof.
Let $V=(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}$. For any $U\in\mathcal{K}$, $U\subseteq\bigcup\{U\mid U\in\mathcal{K}\}$ so that $U=U^{{\bot\bot}}=(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}=V$. This shows that $V$ is an upper bound of elements of $\mathcal{K}$. If $W$ is another such upper bound, then $U\subseteq W$, so that $\bigcup\{U\mid U\in\mathcal{K}\}\subseteq W$, whence $V=(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}\subseteq W^{{\bot\bot}}=W$. The infimum is proved similarly. β
Theorem 4.
$\mathcal{A}$ is the smallest complete Boolean subalgebra of $P(X)$ extending $\mathcal{B}$.
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