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[parent] regular open algebra (Definition)

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 $ '$ such that for any $ U$, $ U':=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 well-defined (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... ...\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'=U\cap U^\bot=U\cap (X\setminus \overline{U}) \subseteq \overline{U}\cap (X\setminus \overline{U})=\varnothing$. Finally, $ U\vee U'=(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$:
    $\displaystyle (*) \qquad U^{\bot\bot}\cap V^{\bot\bot}=(U\cap V)^{\bot\bot}.$
    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
    $\displaystyle U\cap \overline{V}\subseteq \overline{U\cap V}.$
    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 \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 \varnothing$, or $ W\cap (U\cap V)\ne \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
    $\displaystyle (**) \qquad U\cap V^{\bot\bot}\subseteq (U\cap V)^{\bot \bot}.$
    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)^{\b... ...{\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. $ \qedsymbol$
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'\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. $ \qedsymbol$
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 \lbrace U\mid U\in \mathcal{K}\rbrace)^{\bot\bot}$ and $ (\bigcup \lbrace U\mid U\in \mathcal{K}\rbrace)^{\bot\bot}$ respectively.
Proof. Let $ V=(\bigcup \lbrace U\mid U\in \mathcal{K}\rbrace)^{\bot\bot}$. For any $ U\in \mathcal{K}$, $ U\subseteq \bigcup \lbrace U\mid U\in \mathcal{K}\rbrace$ so that $ U=U^{\bot\bot}=(\bigcup \lbrace U\mid U\in \mathcal{K}\rbrace)^{\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 \lbrace U\mid U\in \mathcal{K}\rbrace \subseteq W$, whence $ V=(\bigcup \lbrace U\mid U\in \mathcal{K}\rbrace)^{\bot\bot}\subseteq W^{\bot\bot}=W$. The infimum is proved similarly. $ \qedsymbol$
Theorem 4   $ \mathcal{A}$ is the smallest complete Boolean subalgebra of $ P(X)$ extending $ \mathcal{B}$.

More to come...



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Cross-references: infimum, upper bound, join, meet, complete Boolean algebra, addition, Boolean subalgebra, clopen sets, subset, Boolean, distributive lattice, distributive inequalities, distributive laws, preserves, order, closed, inclusions, intersection, order reversing, open sets, distributive, top, bottom, complemented, absorption, completes, equation, expression, implies, open, associativity, obvious, commutativity, idempotency, lattice, Boolean algebra, regular open, well-defined, parent, binary operation, closure, complement, unary, operations, topological space, regular open sets, universe, algebraic system

This is version 13 of regular open algebra, born on 2008-03-22, modified 2008-04-15.
Object id is 10436, canonical name is RegularOpenAlgebra.
Accessed 161 times total.

Classification:
AMS MSC: 06E99 (Order, lattices, ordered algebraic structures :: Boolean algebras :: Miscellaneous)
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