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regular open algebra∗
CWoo†
2013-03-22 0:46:10
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 0 such that for any U , U 0 := U ⊥ , where U ⊥ is the
complement of the closure of U in X,
3. a binary operation ∧ such that for any U, V ∈ A, U ∧ V := U ∩ V , and
4. a binary operation ∨ such that for any U, V ∈ A, U ∨ V := (U ∪ V )⊥⊥ .
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 ∨ and ∧): Clearly, U ∧ U = U . Also, U ∨ U =
(U ∪ U )⊥⊥ = U ⊥⊥ = U , since U is regular open.
• Commutativity of the binary operations are obvious.
• The associativity of ∧ is also obvious. The associativity of ∨ goes as
follows: U ∨ (V ∨ W ) = (U ∪ (V ∪ W )⊥⊥ )⊥⊥ = U ⊥ ∩ (V ∪ W )⊥⊥⊥ =
U ⊥ ∩ (V ∪ W )⊥ , since V ∪ W is open (which implies that (V ∪ W )⊥
is regular open). The last expression is equal to U ⊥ ∩ (V ⊥ ∩ W ⊥ ).
Interchanging the roles of U and W , we obtain the equation W ∨
(V ∨ U ) = W ⊥ ∩ (V ⊥ ∩ U ⊥ ), which is just U ⊥ ∩ (V ⊥ ∩ W ⊥ ), or
U ∨ (V ∨ W ). The commutativity of ∨ completes the proof of the
associativity of ∨.
∗ hRegularOpenAlgebrai
created: h2013-03-2i by: hCWooi version: h40436i Privacy
setting: h1i hDefinitioni h06E99i
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• Finally, we verify the absorption laws. First, U ∧ (U ∨ V ) = U ∩ (U ∪
V )⊥⊥ = U ⊥⊥ ∩(U ∪V )⊥⊥ = (U ⊥ ∪(U ∪V )⊥ )⊥ = (U ⊥ ∪(U ⊥ ∩V ⊥ )⊥ =
(U ⊥ )⊥ = U . Second, U ∨ (U ∧ V ) = (U ∪ (U ∨ W )⊥⊥ = U ⊥⊥ = U .
2. A is complemented. First, it is easy to see that ∅ and X are the bottom
and top elements of A. Furthermore, for any U ∈ A, U ∧ U 0 = U ∩ U ⊥ =
U ∩ (X \ U ) ⊆ U ∩ (X \ U ) = ∅. Finally, U ∨ U 0 = (U ∪ U ⊥ )⊥⊥ =
(U ⊥ ∩ U ⊥⊥ )⊥ = (U ⊥ ∩ U )⊥ = ∅⊥ = X.
3. A is distributive. This can be easily proved once we show the following:
for any open sets U, V :
(∗)
U ⊥⊥ ∩ V ⊥⊥ = (U ∩ V )⊥⊥ .
To begin, note that since U ∩V ⊆ U , and ⊥ is order reversing, (U ∩V )⊥⊥ ⊆
U ⊥⊥ by applying ⊥ twice. Do the same with V and take the intersection,
we get one of the inclusions: (U ∩ V )⊥⊥ ⊆ U ⊥⊥ ∩ V ⊥⊥ . For the other
inclusion, we first observe that
U ∩V ⊆U ∩V.
If x ∈ LHS, then x ∈ U and for any open set W with x ∈ W , we have that
W ∩ V 6= ∅. In particular, U ∩ W is such an open set (for x ∈ U ∩ W ), so
that (U ∩W )∩V 6= ∅, or W ∩(U ∩V ) 6= ∅. Since W is arbitrary, x ∈ RHS.
Now, apply the set complement, we have (U ∩ V )⊥ ⊆ U { ∪ V ⊥ . Applying
⊥
next we get (U ∩ V )⊥⊥ for the LHS, and (U { ∪ V ⊥ )⊥ = U {−{ ∩ V ⊥⊥ =
{{
U ∩ V ⊥⊥ = U ∩ V ⊥⊥ for RHS, since U { is closed. As ⊥ reverses order,
the new inclusion is
(∗∗)
U ∩ V ⊥⊥ ⊆ (U ∩ V )⊥⊥ .
From this, a direct calculation shows U ⊥⊥ ∩ V ⊥⊥ ⊆ (U ⊥⊥ ∩ V )⊥⊥ ⊆
(U ∩ V )⊥⊥⊥⊥ = (U ∩ V )⊥⊥ , noticing that the first and second inclusions
use (∗∗) above (and the fact that ⊥⊥ preserves order), and the last equation
uses the fact that for any open set W , W ⊥ is regular open. This proves
the (∗).
Finally, to finish the proof, we only need to show one of two distributive
laws, say, U ∧ (V ∨ W ) = (U ∧ V ) ∨ (U ∧ W ), for the other one follows from
the use of the distributive inequalities. This we do be direct computation:
U ∧ (V ∨ W ) = U ∩ (V ∪ W )⊥⊥ = U ⊥⊥ ∩ (V ∪ W )⊥⊥ = (U ∩ (V ∪ W ))⊥⊥ =
((U ∩ V ) ∪ (U ∩ W ))⊥⊥ = ((U ∧ V ) ∪ (U ∧ W ))⊥⊥ = (U ∧ V ) ∨ (U ∧ W ).
Since a complemented distributive lattice is Boolean, the proof is complete.
Theorem 2. The subset B of all clopen sets in X forms a Boolean subalgebra
of A.
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Proof. Clearly, every clopen set is regular open. In addition, 1 ∈ B. If U is
clopen, so is the complement of its closure, and hence U 0 ∈ B. If U, V are clopen,
so is their intersection U ∧ V . Similarly, U ∪ V is clopen, so that U ∨ V = U ∪ V
is clopen also.
Theorem 3. In fact, A is a complete
T Boolean algebra. For an
S arbitrary subset
K of A, the meet and join of K are ( {U | U ∈ K})⊥⊥ and ( {U | U ∈ K})⊥⊥
respectively.
S
S
Proof. Let V = ( {U
| U ∈ K})⊥⊥ . For any U ∈ K, U ⊆ {U | U ∈ K} so
S
that U = U ⊥⊥ = ( {U | U ∈ K})⊥⊥ = V . This shows that V is an upper
bound
S of elements of K. If W is anotherSsuch upper bound, then U ⊆ W , so
that {U | U ∈ K} ⊆ W , whence V = ( {U | U ∈ K})⊥⊥ ⊆ W ⊥⊥ = W . The
infimum is proved similarly.
Theorem 4. A is the smallest complete Boolean subalgebra of P (X) extending
B.
More to come...
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