regular open algebra

A regular open algebra is an algebraic system $𝒜$ 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 $U$,  $U^{\prime }:=U^{\perp }$, where $U^{\perp }$ is the complement of the closure of $U$ in $X$,

3. a binary operation $\wedge$ such that for any $U,V\in 𝒜$, $U\wedge V:=U\cap V$, and

4. a binary operation $\vee$ such that for any $U,V\in 𝒜$, $U\vee V:={(U\cup V)}^{\perp \perp }$.

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.

$𝒜$ is a Boolean algebra

Proof.

We break down the proof into steps:

1. $𝒜$ is a lattice. This amounts to verifying various laws on the operations:

   • (idempotency of $\vee$ and $\wedge$): Clearly, $U\wedge U=U$. Also, $U\vee U={(U\cup U)}^{\perp \perp }=U^{\perp \perp }=U$, 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: $U\vee (V\vee W)={(U\cup {(V\cup W)}^{\perp \perp })}^{\perp \perp }=U^{\perp }\cap {(V\cup W)}^{\perp \perp \perp }=U^{\perp }\cap {(V\cup W)}^{\perp }$, since $V\cup W$ is open (which implies that ${(V\cup W)}^{\perp }$ is regular open). The last expression is equal to $U^{\perp }\cap (V^{\perp }\cap W^{\perp })$. Interchanging the roles of $U$ and $W$, we obtain the equation $W\vee (V\vee U)=W^{\perp }\cap (V^{\perp }\cap U^{\perp })$, which is just $U^{\perp }\cap (V^{\perp }\cap W^{\perp })$, 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, $U\wedge (U\vee V)=U\cap {(U\cup V)}^{\perp \perp }=U^{\perp \perp }\cap {(U\cup V)}^{\perp \perp }={(U^{\perp }\cup {(U\cup V)}^{\perp })}^{\perp }=(U^{\perp }\cup {(U^{\perp }\cap V^{\perp })}^{\perp }={(U^{\perp })}^{\perp }=U$. Second, $U\vee (U\wedge V)=(U\cup {(U\vee W)}^{\perp \perp }=U^{\perp \perp }=U$.

2. $𝒜$ is complemented. First, it is easy to see that $\mathrm{\varnothing }$ and $X$ are the bottom and top elements of $𝒜$. Furthermore, for any $U\in 𝒜$, $U\wedge U^{\prime }=U\cap U^{\perp }=U\cap (X\setminus \overline{U})\subseteq \overline{U}\cap (X\setminus \overline{U})=\mathrm{\varnothing }$. Finally, $U\vee U^{\prime }={(U\cup U^{\perp })}^{\perp \perp }={(U^{\perp }\cap U^{\perp \perp })}^{\perp }={(U^{\perp }\cap U)}^{\perp }={\mathrm{\varnothing }}^{\perp }=X$.

3. $𝒜$ is distributive. This can be easily proved once we show the following: for any open sets $U,V$:

   $$(*)\mathit{\hspace{1em}\hspace{1em}}{U}^{\perp \perp }\cap V^{\perp \perp }={(U\cap V)}^{\perp \perp }.$$

   To begin, note that since $U\cap V\subseteq U$, and ${}^{\perp }$ is order reversing, ${(U\cap V)}^{\perp \perp }\subseteq U^{\perp \perp }$ by applying ${}^{\perp }$ twice. Do the same with $V$ and take the intersection, we get one of the inclusions: ${(U\cap V)}^{\perp \perp }\subseteq U^{\perp \perp }\cap V^{\perp \perp }$. 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 $W\cap V\ne \mathrm{\varnothing }$. In particular, $U\cap W$ is such an open set (for $x\in U\cap W$), so that $(U\cap W)\cap V\ne \mathrm{\varnothing }$, or $W\cap (U\cap V)\ne \mathrm{\varnothing }$. Since $W$ is arbitrary, $x\in$ RHS. Now, apply the set complement, we have ${(U\cap V)}^{\perp }\subseteq U^{\mathrm{\complement }}\cup V^{\perp }$. Applying ${}^{\perp }$ next we get ${(U\cap V)}^{\perp \perp }$ for the LHS, and ${(U^{\mathrm{\complement }}\cup V^{\perp })}^{\perp }=U^{\mathrm{\complement }-\mathrm{\complement }}\cap V^{\perp \perp }=U^{\mathrm{\complement }\mathrm{\complement }}\cap V^{\perp \perp }=U\cap V^{\perp \perp }$ for RHS, since $U^{\mathrm{\complement }}$ is closed. As ${}^{\perp }$ reverses order, the new inclusion is

   $$(**)\mathit{\hspace{1em}\hspace{1em}}U\cap V^{\perp \perp }\subseteq {(U\cap V)}^{\perp \perp }.$$

   From this, a direct calculation shows $U^{\perp \perp }\cap V^{\perp \perp }\subseteq {(U^{\perp \perp }\cap V)}^{\perp \perp }\subseteq {(U\cap V)}^{\perp \perp \perp \perp }={(U\cap V)}^{\perp \perp }$, noticing that the first and second inclusions use $(**)$ above (and the fact that ${}^{\perp \perp }$ preserves order), and the last equation uses the fact that for any open set $W$, $W^{\perp }$ 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: $U\wedge (V\vee W)=U\cap {(V\cup W)}^{\perp \perp }=U^{\perp \perp }\cap {(V\cup W)}^{\perp \perp }={(U\cap (V\cup W))}^{\perp \perp }={((U\cap V)\cup (U\cap W))}^{\perp \perp }={((U\wedge V)\cup (U\wedge W))}^{\perp \perp }=(U\wedge V)\vee (U\wedge W)$.

Since a complemented distributive lattice is Boolean, the proof is complete.∎

Theorem 2.

The subset $\mathrm{B}$ of all clopen sets in $X$ forms a Boolean subalgebra of $\mathrm{A}$.

Proof.

Clearly, every clopen set is regular open. In addition, $1\in ℬ$. If $U$ is clopen, so is the complement of its closure, and hence $U^{\prime }\in ℬ$. 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, $\mathrm{A}$ is a complete Boolean algebra. For an arbitrary subset $\mathrm{K}$ of $A$, the meet and join of $\mathrm{K}$ are ${\mathrm{(}\mathrm{\bigcap }\mathrm{\{}U\mathrm{\mid }U\mathrm{\in }\mathrm{K}\mathrm{\}}\mathrm{)}}^{\mathrm{\perp }\mathrm{\perp }}$ and ${\mathrm{(}\mathrm{\bigcup }\mathrm{\{}U\mathrm{\mid }U\mathrm{\in }\mathrm{K}\mathrm{\}}\mathrm{)}}^{\mathrm{\perp }\mathrm{\perp }}$ respectively.

Proof.

Let $V={(\bigcup \{U\mid U\in 𝒦\})}^{\perp \perp }$. For any $U\in 𝒦$, $U\subseteq \bigcup \{U\mid U\in 𝒦\}$ so that $U=U^{\perp \perp }={(\bigcup \{U\mid U\in 𝒦\})}^{\perp \perp }=V$. This shows that $V$ is an upper bound of elements of $𝒦$. If $W$ is another such upper bound, then $U\subseteq W$, so that $\bigcup \{U\mid U\in 𝒦\}\subseteq W$, whence $V={(\bigcup \{U\mid U\in 𝒦\})}^{\perp \perp }\subseteq W^{\perp \perp }=W$. The infimum is proved similarly.∎

Theorem 4.

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

More to come…