# regular open algebra

1. 1.

a constant $1$ such that $1:=X$,

2. 2.

a unary operation ${}^{\prime}$ such that for any $U$,  $U^{\prime}:=U^{\bot}$, where $U^{\bot}$ is the complement  of the closure   of $U$ in $X$,

3. 3.

a binary operation  $\wedge$ such that for any $U,V\in\mathcal{A}$, $U\wedge V:=U\cap V$, and

4. 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 well-defined (that the result sets are regular open). Also, we have the following:

###### Theorem 1.

$\mathcal{A}$

###### Proof.

We break down the proof into steps:

1. 1.

$\mathcal{A}$ is a lattice  . This amounts to verifying various laws on the operations:

2. 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. 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: $U\wedge(V\vee W)=U\cap(V\cup W)^{\bot\bot}=U^{\bot\bot}\cap(V\cup W)^{\bot\bot% }=(U\cap(V\cup W))^{\bot\bot}=((U\cap V)\cup(U\cap W))^{\bot\bot}=((U\wedge V)% \cup(U\wedge W))^{\bot\bot}=(U\wedge V)\vee(U\wedge W)$.

###### 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. ∎

###### 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. ∎

###### Theorem 4.

$\mathcal{A}$ is the smallest complete Boolean subalgebra of $P(X)$ extending $\mathcal{B}$.

More to come…

Title regular open algebra RegularOpenAlgebra 2013-03-22 17:56:21 2013-03-22 17:56:21 CWoo (3771) CWoo (3771) 16 CWoo (3771) Definition msc 06E99