Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 • 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. 2 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... 3