9. Boolean algebras and forcing orders

Transcription

1 9. Boolean algebras and forcing orders To introduce the apparatus of generic extensions and forcing in a clear fashion, it is necessary to go into a special set theoretic topic: Boolean algebras and their relation to certain orders. A Boolean algebra (BA) is a structure A, +,,, 0, 1 with two binary operations + and, a unary operation, and two distinguished elements 0 and 1 such that the following axioms hold for all x, y, z A: (A) x + (y + z) = (x + y) + z; (A ) x (y z) = (x y) z; (C) x + y = y + x; (C ) x y = y x; (L) x + (x y) = x; (L ) x (x + y) = x; (D) x (y + z) = (x y) + (x z); (D ) x + (y z) = (x + y) (x + z); (K) x + ( x) = 1; (K ) x ( x) = 0. The main example of a Boolean algebra is a field of sets: a set A of subsets of some set X, closed under union, intersection, and complementation with respect to X. The associated Boolean algebra is A,,, \, 0, X. Here \ is treated as a one-place operation, producing X\a for any a A. This example is really all-encompassing every BA is isomorphic to one of these. We will not prove this, or use it. As is usual in algebra, we usually denote a whole algebra A, +,,, 0, 1 just by mentioning its universe A, everything else being implicit. Some notations used in some treatments of Boolean algebras are: or for +; or for ; for. These notations might be confusing if discussing logic, or elementary set theory. Our notation might be confusing if discussing ordinary algebra. Now we give the elementary arithmetic of Boolean algebras. We recommend that the reader go through them, but then approach any arithmetic statement in the future from the point of view of seeing if it works in fields of sets; if so, it should be easy to derive from the axioms. First we have the duality principle, which we shall not formulate carefully; our particular uses of it will be clear. Namely, notice that the axioms come in pairs, obtained from each other by interchanging + and and 0 and 1. This means that also if we prove some arithmetic statement, the dual statement, obtained by this interchanging process, is also valid. Proposition 9.1. x + x = x and x x = x. Proof. the second statement follows by duality. x + x = x + x (x + x) by (L ) = x by (L); Proposition 9.2. x + y = y iff x y = x. Proof. Assume that x + y = y. Then, by (L ), x y = x (x + y) = x. 66

2 The converse follows by duality. In any BA we define x y iff x + y = y. Note that the dual of x y is y x, by 9.2 and commutativity. (The dual of a defined notion is obtained by dualizing the original notions.) Proposition 9.3. On any BA, is reflexive, transitive, and antisymmetric; that is, the following conditions hold: (i) x x; (ii) If x y and y z, then x z; (iii) If x y and y x, then x = y. Proof. x x means x + x = x, which was proved in 9.1. Assume the hypothesis of (ii). Then x + z = x + (y + z) = (x + y) + z = y + z = z, as desired. Finally, under the hypotheses of (iii), x = x + y = y + x = y. Note that Proposition 9.3 says that is a partial order on the BA A. There are some notions concerning partial orders which we need. An element z is an upper bound for a set Y of elements of X if y z for all y Y ; similarly for lower bounds. And z is a least upper bound for Y if it is an upper bound for Y and is any other upper bound for Y ; simlarly for greatest lower bounds. By antisymmetry, in any partial order least upper bounds and greatest lower bounds are unique if they exist. Proposition 9.4. x + y is the least upper bound of {x, y}, and x y is the greatest lower bound of {x, y}. Proof. We have x + (x + y) = (x + x) + y = x + y, and similarly y + (x + y) = y + (y + x) = (y + y) + x = y + x = x + y; so x + y is an upper bound for {x, y}. If z is any upper bound for {x, y}, then (x + y) + z = (x + (y + z) = x + z = z, as desired. The other part follows by duality(!). Proposition 9.5. (i) x + 0 = x and x 1 = x; (ii) x 0 = x and x + 1 = 1; (iii) 0 x 1. 67

3 Proof. By (K) and Proposition 9.4, 1 is the least upper bound of x and x; in particular it is an upper bound, so x 1. Everything else follows by duality, Proposition 9.2, and the definitions. Proposition 9.6. For any x and y, y = x iff x y = 0 and x + y = 1. Proof. holds by (K) and (K ). Now suppose that x y = 0 and x + y = 1. Then y = y 1 = y (x + x) = y x + y x = 0 + y x = y x; x = x 1 = x (x + y) = x x + x y = 0 + x y = x y = y. Proposition 9.7. (i) x = x; (ii) if x = y then x = y; (iii) 0 = 1 and 1 = 0; (iv) (DeMorgan s laws) (x + y) = x y and (x y) = x + y. Proof. If we apply Proposition 9.6 with x and y replaced respectively by x and x, we get x = x. Next, if x = y, then x = x = y = y. For (iii), by 9.5(iii), 0 1 = 0 and = 1, so by 9.6, 0 = 1. Then 1 = 0 by duality. For the first part of (iv), (x + y) x y = x x y + y x y = = 0, and (x + y) + x y = x (y + y) + y + x y = x y + x y + y + x y = y + x y + x y = y + y = 1, so that (x + y) = x y by Proposition 9.6. Finally, the second part of (iv) follows by duality. Proposition 9.8. x y iff y x. Proof. Assume that x y. Then x + y = y, so x y = y, i.e., y x. For the converse, use the implication just proved, plus 9.7(i). Proposition 9.9. If x x and y y, then x + y x + y and x y x y. Proof. Assume the hypothesis. Then (x + y) + (x + y ) = (x + x ) + (y + y ) = x + y, and so x + y x + y ; the second conclusion follows by duality. Proposition x y iff x y = 0. 68

4 Proof. If x y, then x = x y and so x y = 0. Conversely, if x y = 0, then so that x y. x = x (y + y) = x y + x y = x y, Elements x, y A are disjoint if x y = 0. For any x, y we define this is the symmetric difference of x and y. Proposition (i) x = y iff x y = 0; (ii) x (y z) = (x y) (x z); (iii) x (y z) = (x y) z. x y = x y + y x; Proof. For (i), is trivial. Now assume that x y = 0. Then x y = 0 = y x, so x y and y x, so x = y. For (ii), we have as desired. Finally, for (iii), x (y z) = x y z + x z y = (x y) (x z) + (x z) (x y) = (x y) (x z), x (y z) = x (y z + y z) + (y z + y z) x = x ( y + z) (y + z) + x y z + x y z = x y z + x y z + x y z + x y z; if we apply the same argument to z (y x) we get z (y x) = z y x + z y x + z y x + z y x, which is the same thing. So the obvious symmetry of gives the desired result. One further useful result is that axiom (D ) is redundant: Proposition (D ) is redundant. Proof. (x + y) (x + z) = ((x + y) x) + ((x + y) z) = (x (x + y)) + (z (x + y)) = x + ((z x) + (z y)) = x + ((x z) + (y z)) = (x + (x z)) + (y z) = x + (y z). 69

5 Complete Boolean algebras If M is a subset of a BA A, we denote by M its least upper bound (if it exists), and by M its greatest lower bound, if it exists. A is complete iff these always exist. Note that frequently people use M and M instead of M and M. Proposition Assume that A is a complete BA. (i) i I a i = i I a i. (ii) i I a i = i I a i. Proof. For (i), let a = i I a i; we show that a is the greatest lower bound of { a i : i I}. If i I, then a i a, and hence a a i ; thus a is a lower bound for the indicated set. Now suppose that x is any lower bound for this set. Then for any i I we have x a i, and so a i x. So x is an upper bound for {a i : i I}, and so a x. Hence x a, as desired. (ii) is proved similarly. The following (possibly infinite) distributive law is frequently useful. One should be aware of the fact that more general infinite distributive laws do not hold, in general. Since this will not enter into our treatment, we do not go into a counterexample or further discussion of really general distributive laws. Proposition If i I a i exists, then i I (b a i) exists and b a i = a i ). i I(b i I Proof. Let s = i I a i; we shall show that b s is the least upper bound of {b a i : i I}. If i I, then a i s and so b a i b s; so b s is an upper bound for the indicated set. Now suppose that x is any upper bound for this set. Then for any i I we have b a i x, hence b a i x = 0 and so a i (b x) = b + x; so b + x is an upper bound for {a i : i I}. It follows that s b + x, and hence s b x, as desired. Forcing orders A forcing order is a triple P = (P,, 1) such that is a reflexive and transitive relation on the nonempty set P, and p P(p 1). Note that we do not assume that is antisymmetric. Partial orders are special cases of forcing orders in which this is assumed (but we do not assume the existence of 1 in partial orders). Note that we assume that every forcing order has a largest element. Many set-theorists use partial order instead of forcing order. Frequently we use just P for a forcing order; and 1 are assumed. We say that elements p, q P are compatible iff there is an r p, q. We write p q to indicate that p and q are incompatible. A set A of elements of P is an antichain iff any two distinct members of A are incompatible. WARNING: sometimes antichain is used to mean pairwise incomparable, a much different notion. A subset Q of P is dense iff for every p P there is a q Q such that q p. 70

6 Now we are going to describe how to embed a forcing order into a complete BA. We take the regular open algebra of a certain topological space. We assume a very little bit of topology. To avoid assuming any knowledge of topology we now give a minimalist introduction to topology. A topology on a set X is a collection O of subsets of X satisfying the following conditions: (1) X, O. (2) O is closed under arbitrary unions. (3) O is closed under finite intersections. The members of O are said to be open. The interior of a subset Y X is the union of all open sets contained in Y ; we denote it by int(y ). Proposition (i) int( ) =. (ii) int(x) = X. (iii) int(y ) Y. (iv) int(y Z) = int(y ) int(z). (v) int(int(y )) = int(y ). (vi) int(y ) = {x X : x U Y for some open set U}. Proof. (i) (iii), (v), and (vi) are obvious. For (iv), if U is an open set contained in Y Z, then it is contained in Y ; so int(y Z) int(y ). Similarly for Z, so holds. For, note that the right side is an open set contained in Y Z. (v) holds since int(y ) is open. A subset C of X is closed iff X\C is open. Proposition (i) and X are closed. (ii) The collection of all closed sets is closed under finite unions and intersections of any nonempty subcollection. For any Y X, the closure of Y, denoted by cl(y ), is the intersection of all closed sets containing Y. Proposition (i) cl(y ) = X\int(X\Y ). (ii) int(y ) = X\cl(X\Y ). (iii) cl( ) =. (iv) cl(x) = X. (v) Y cl(y ). (vi) cl(y Z) = cl(y ) cl(z). (vii) cl(cl(y )) = cl(y ). (viii) cl(y ) = {x X :for every open set U, if x U then U Y }. Proof. (i): int(x\y ) is an open set contained in X\Y, so Y is a subset of the closed set X\int(X\Y ). Hence cl(y ) X\int(X\Y ). Also. cl(y ) is a closed set containing 71

7 Y, so X\cl(Y ) is an open set contained in X\Y. Hence X\cl(Y ) int(x\y ). Hence X\int(X\Y cl(y ). This proves (i). (ii): Using (i), (iii) (v): clear. (vi): (vii): X\cl(X\Y ) = X\(X\int(X\(X\Y ))) = int(y ). cl(y Z) = X\int(X\(Y Z)) by (i) = X\int((X\Y ) (X\Z)) = X\(int(X\Y ) int(x\z)) by 9.15(iv) = [X\int(X\Y )] [X\int(X\Z)] = cl(y ) cl(z). cl(cl(y )) = cl(x\int(x\y )) = X\int(X\(X\int(X\Y ))) = X\int(int(X\Y )) = X\int(X\Y ) = cl(y ). (vii): First suppose that x cl(y ), and x U, U open. By (i) and Proposition 9.15(vi) we have U X\Y, i.e., U Y, as desired. Second, suppose that x / cl(y ). Then by (i) and 9.15(vi) there is an open U such that x U X\Y ; so U Y =, as desired. Now we go beyond this minimum amount of topology and work with the notion of a regular open set, which is not a standard part of topology courses. We say that Y is regular open iff Y = int(cl(y )). Proposition (i) If Y is open, then Y int(cl(y )). (ii) If U and V are regular open, then so is U V. (iii) int(cl(y )) is regular open. (iv) If U is open, then int(cl(u)) is the smallest regular open set containing U. (v) If U is open then U cl(y ) cl(u Y ). (vi) If U is open, then U int(cl(y )) int(cl(u Y )). (vii) If U and V are open and U V =, then int(cl(u)) V =. (viii) If U and V are open and U V =, then int(cl(u)) int(cl(v )) =. (ix) For any set M of regular open sets, int(cl( M) is the least regular open set containing each member of M. Proof. (i): Y cl(y ), and hence Y = int(y ) int(cl(y )). 72

8 (ii): U V is open, and so U V int(cl(u V )). For the other inclusion, int(cl(u V )) int(cl(u)) = U, and similarly for V, so the other inclusion holds. (iii): int(cl(x)) cl(x), so cl(int(cl(x))) cl(cl(x)) = cl(x); hence int(cl(int(cl(x)))) int(cl(x)); the other inclusion is clear. (iv): By (iii), int(cl(u)) is a regular open set containing U. If V is any regular open set containing U, then int(cl(u)) int(cl(v )) = V. (v): and (v) follows. (vi): U (X\(U Y )) X\Y, hence U int(x\(u Y )) = int(u) int(x\(u Y )) = int(u (X\(U Y ))) int(x\y ), hence X\int(X\Y ) X\(U int(x\(u Y ))) = (X\U) (X\int(X\(U Y ))), hence U (X\int(X\Y )) (X\int(X\(U Y ))), U int(cl(y )) = int(u) int(cl(y )) = int(u cl(y )) int(cl(u Y )) by (v). (vii): U X\V, hence cl(u) cl(x\v ) = X\V, hence cl(u) V =, and the conclusion of (vii) follows. (viii): Apply (vii) twice. (ix): If U M, then U M int(cl( M). Suppose that V is regular open and U V for all U M. Then M V, and so int(cl( M)) int(cl(v ) = V. We let RO(X) be the collection of all regular open sets in X. We define operations on RO(X) which will make it a Boolean algebra. For any Y, Z RO(X), let Y + Z = int(cl(y Z)); Y Z = Y Z; Y = int(x\y ). Theorem The structure RO(X), +,,,, X 73

9 is a complete BA. Moreover, the ordering coincides with. Proof. RO(X) is closed under + by Proposition 9.18(ix), and is closed under by Proposition 9.18(ii). Clearly it is closed under, and, X RO(X). Now we check the axioms. The following are completely obvious: (A ), (C ), (C). Now let unexplained variables range over RO(X). For (A), note by 9.18(i) that U U + V (U + V ) + W; and similarly V (U +V ) +W and W U +V (U +V ) +W. If U, V, W Z, then by 9.18(iv), U + V Z and hence (U + V ) + W Z. Thus (U + V ) + W is the least upper bound in RO(X) of U, V, W. This is true for all U, V, W. So U + (V + W) = (V + W) + U is also the least upper bound of them; so (A) holds. For (L): U + U V = int(cl(u (U V ))) = int(cl(u)) = U. (L ) holds by 9.18(i). For (D), first note that Y (Z + W) = Y int(cl(z W)) int(cl(y (Z W))) by 9.18(vi) = int(cl((y Z) (Y W))) = Y Z + Y W. On the other hand, (Y Z) (Y W) = Y (Z W) Y, Z W, and hence easily Y Z + Y W = int(cl((y Z) (Y W))) int(cl(y ) = Y and Y Z + Y W = int(cl((y Z) (Y W))) int(cl(z W) = Z + W; so the other inclusion follows, and (D) holds. (K): For any regular open Y, from Proposition 9.17(ii) we get Y = int(x\y ) = X\cl(X\(X\Y )) = X\cl(Y ). Hence X = cl(y ) (X\cl(Y )) cl(y ) cl((x\cl(y )) = cl(y (X\cl(Y ))), and hence X = Y + Y. (K ): Clearly = Y int(x\y ) = Y Y. Thus we have now proved that RO(X), +,,,, X is a BA. Since is the same as, is the same as. Hence by Proposition 9.18(ix), RO(X), +,,,, X is a complete BA. Now we return to our task of embedding a forcing order into a complete Boolean algebra. Let P be a given forcing order. For each p P let P p = {q : q p}. Now we define O P = {X P : (P p) X for every p X}. 74

10 We check that this gives a topology on P. Clearly P, O. To show that O is closed under arbitrary unions, suppose that X O. Take any p X. Choose X X such that p X. Then (P p) X X, as desired. If X, Y O P, suppose that p X Y. Then p X, so (P p) X. Similarly (P p) Y, so (P p) X Y. Thus X Y O P, finishing the proof that O P is a topology on P. We denote the complete BA of regular open sets in this topology by RO(P). Now for any p P we define e(p) = int(cl(p p)). Thus e maps P into RO(P). This is our desired embedding. Actually it is not really an embedding in general, but it has several useful properties, and for many forcing orders it really is an embedding. The useful properties mentioned are as follows. We say that a subset X of P is dense below p iff for every r p there is a q r such that q X. Theorem Let P be a forcing order. Suppose that p, q P, F is a finite subset of P, a, b RO(P), and N is a subset of RO(P) (i) e[p] is dense in RO(P), i.e., for any nonzero Y RO(P) there is a p P such that e(p) Y. (ii) If p q then e(p) e(q). (iii) p q iff e(p) e(q) =. (iv) If e(p) e(q), then p and q are compatible. (v) The following conditions are equivalent: (a) e(p) e(q). (b) {r : r p, q} is dense below p. (vi) The following conditions are equivalent, for F nonempty: (a) e(p) q F e(q). (b) {r : r q for all q F } is dense below p. (vii) The following conditions are equivalent: (a) e(p) ( q F e(q)) N. (b) {r : r q for all q F and e(r) s for some s N} is dense below p. (viii) e(p) a iff there is no q p such that e(q) a. (ix) e(p) a + b iff for all q p, if e(q) a then e(q) b. Proof. (i): Assume the hypothesis. By the definition of the topology and since Y is nonempty and open, there is a p P such that P p Y. Hence e(p) = int(cl(p p)) int(cl(y )) = Y. (ii): If p q, then P p P q, and so e(p) = int(cl(p p)) int(cl(p q) = e(q)). (iii): Assume that p q. Then (P p) (P q) =, and hence by Proposition 9.18(viii), e(p) e(q) =. Conversely, suppose that e(p) e(q) =. Then (P p) (P q) e(p) e(q) =, and so p q. (iv): If e(p) e(q), then e(p) e(q) = e(p), so p and q are compatible by (iii). (v): For (a) (b), suppose that e(p) e(q) and s p. Then e(s) e(p) e(q), so s and q are compatible by (iv); say r s, q. Then r s p, hence r p, q, as desired. 75

11 For (b) (a), suppose that e(p) e(q). Thus e(p) e(q) 0. Hence there is an s such that e(s) e(p) e(q). Hence e(s) e(q) =, so s q by (iii). Now e(s) e(p), so s and p are compatible by (iv); say t s, p. For any r t we have r s, and hence r q. So (b) fails. (vi): We proceed by induction on F. The case F = 1 is given by (v). Now assume the result for F, and suppose that t P \F. First suppose that e(p) q F e(q) e(t). Suppose that s p. Now e(p) q F e(q), so by the inductive hypothesis there is a u s such that u q for all q F. Thus e(u) e(s) e(p) e(t), so by (iv), u and t are compatible. Take any v u, t. then v q for any q F {t}, as desired. Second, suppose that (b) holds for F {t}. In particular, {r : r q for all q F } is dense below p, and so e(p) q F e(q) by the inductive hypothesis. But also clearly {r : r t} is dense below p, so e(p) e(t) too, as desired. (vii): First assume that e(p) ( q F e(q)) N, and suppose that u p. By (vi), there is a v u such that v q for each q F. Now e(v) e(u) e(p) N, so 0 e(v) = e(v) N = s N (e(v) e(s)). Hence there is an s N such that e(v) e(s) 0. Hence by (iii), v and s are compatible; say r v, s. Clearly r is in the set described in (b). Second, suppose that (b) holds. Clearly then {r : r q for all q F } is dense below p, and so e(p) q F e(q) by (vi). Now suppose that e(p) N. Then e(p) N 0, so there is a q such that e(q) e(p) N. By (iv), q and p are compatible; say s p, q. Then by (b) choose r s and t N such that e(r) t. Thus e(r) e(s) t e(p) t ( N) N = 0, contradiction. (viii) : Assume that e(p) a. Suppose that q p and e(q) a. Then e(q) a a = 0, contradiction. (viii) : Assume that e(p) a. Then e(p) a 0, so there is a q such that e(q) e(p) a. By (vii) there is an r p, q with e(r) a, as desired. (ix) : Assume that e(p) a+b, q p, and e(q) a. Then e(q) a ( a+b) b, as desired. (ix) : Assume the indicated condition, but suppose that e(p) a + b. Then e(p) a b 0, so there is a q such that e(q) e(p) a b. By (vii) with F = {p} and N = {a b} we get q such that q p and e(q) a b. So q p and e(q) a, so by our condition, e(q) b. But also e(q) b, contradiction. We now expand on the remarks above concerning when e really is an embedding. Note that if P is a simple ordering, then the closure of P p is P itself, and hence P has only two regular open subsets, namely the empty set and P itself. If the ordering on P is trivial, meaning that no two elements are comparable, then every subset of P is regular open. An important condition satisfied by many forcing orders is defined as follows. We say that P is separative iff it is a partial order (thus is an antisymmetric forcing order), and for any p, q P, if p q then there is an r p such that r q. Proposition Let P be a forcing order. (i) cl(p p) = {q : p and q are compatible}. (ii) e(p) = {q : for all r q, r and p are compatible}. (iii) The following conditions are equivalent: (a) P is separative. 76

12 (b) e is one-one, and for all p, q P, p q iff e(p) e(q). Proof. (i) and (ii) are clear. For (iii), (a) (b), assume that P is separative. Take any p, q P. If p q, then e(p) e(q) by 9.20(ii). Suppose that p q. Choose r p such that r q. Then r e(p), while r / e(q) by (ii). Thus e(p) e(q). Now suppose that e(p) = e(q). Then p q p by what was just shown, so p = q since P is a partial order. For (iii), (b) (a), suppose that p q p. Then e(p) e(q) e(p), so e(p) = e(q), and hence p = q. So P is a partial order. Suppose that p q. Then e(p) e(q). Choose s e(p)\e(q). Since s / e(q), by (ii) we can choose t s such that t q. Since s e(p), it follows that t and p are compatible; choose r t, p. Clearly r q. Now we prove a theorem which says that the regular open algebra of a forcing order is unique up to isomorphism. Theorem Let P be a forcing order, A a complete BA, and j a function mapping P into A\{0} with the following properties: (i) j[p] is dense in A, i.e., for any nonzero a A there is a p P such that j(p) a. (ii) For all p, q P, if p q then j(p) j(q). (iii) For any p, q P, p q iff j(p) j(q) = 0. Then there is a unique isomorphism f from RO(P) onto A such that f e = j. That is, f is a bijection from RO(P) onto A, and for any x, y RO(P), x y iff f(x) f(y). Note that since the Boolean operations are easily expressible in terms of (as least upper bounds, etc.), the condition here implies that f preserves all of the Boolean operations too; this includes the infinite sums and products. Proof. Before beginning the proof, we introduce some notation in order to make the situation more symmetric. Let B 0 = RO(P), B 1 = A, k 0 = e, and k 1 = j. Then for each m < 2 the following conditions hold: (1) k m [P] is dense in B m. (2) For all p, q P, if p q then k m (p) k m (q). (3) For all p, q P, p q iff k m (p) k m (q) = 0. (4) For all p, q P, if k m (p) k m (q), then p and q are compatible. In fact, (1) (3) follow from 9.20 and the assumptions of the theorem. Condition (4) for m = 0, so that k m = e, follows from 9.20(iv). For m = 1, so that k m = j, it follows easily from (iii). Now we begin the proof. For each m < 2 we define, for any x B m, g m (x) = {k 1 m (p) : p P, k m (p) x}. The proof of the theorem now consists in checking the following, for each m 2: (5) If x, y B m and x y, then g m (x) g m (y). 77

13 (6) g 1 m g m is the identity on B m. In fact, suppose that (5) and (6) have been proved. If x, y RO(P), then x y implies that g 0 (x) g 0 (y) by (5); g 0 (x) g 0 (y) implies that x = g 1 (g 0 (x)) g 1 (g 0 (y)) = y by (5) and (6). Also, (6) holding for both m = 0 and m = 1 implies that g 0 is a bijection from RO(P) onto A. So g 0 is the desired function f of the theorem. Now (5) is obvious from the definition. To prove (6), assume that m 2. We first prove (7) For any p P and any b B m, k m (p) b iff k 1 m (p) g m (b). To prove (7), first suppose that k m (p) b. Then obviously k 1 m (p) g m (b). Second, suppose that k 1 m (p) g m (b) but k m (p) b. Thus k m (p) b 0, so by the denseness of k m [P] in B m, choose q P such that k m (q) k m (p) b. Then p and q are compatible by (4), so let r P be such that r p, q. Hence k 1 m (r) k 1 m (p) g m (b) = {k 1 m (s) : s P, k m (s) b}. Hence k 1 m (r) = {k 1 m (s) k 1 m (r) : s P, k m (s) b}, so there is an s P such that k m (s) b and k 1 m (s) k 1 m (r) 0. Hence s and r are compatible; say t s, r. Hence k m (t) k m (r) k m (q) b, but also k m (t) k m (s) b, contradiction. This proves (7). Now take any b B m. Then g 1 m (g m (b)) = {k m (p) : p P, k 1 m (p) g m (b)} = {k m (p) : p P, k m (p) b} = b. Thus (6) holds. This proves the existence of f. Now suppose that g is also an isomorphism from RO(P) onto A such that g e = j, but suppose that f g. Then there is an X RO(P) such that f(x) g(x). By symmetry, say that f(x) g(x) 0. By (ii), choose p P such that j(p) f(x) g(x). So f(e(p)) = j(p) f(x), so e(p) X, and hence j(p) = g(e(p)) g(x). This contradicts j(p) g(x). EXERCISES E9.1. Let (A, +,,, 0, 1) be a Boolean algebra. Show that (A,,, 0, 1) is a ring with identity in which every element is idempotent. This means that x x = x for all x. E9.2. Let (A, +,, 0, 1) be a ring with identity in which every element is idempotent. Show that A is a commutative ring, and (A,,,, 0, 1) is a Boolean algebra, where for any x, y A, x y = x + y + xy and for any x A, x = 1 + x. Hint: expand (x + y) 2. 78

14 E9.3. Show that the processes described in exercises E9.1 and E9.2 are inverses of one another. E9.4. A filter in a BA A is a subset F of A with the following properties: (1) 1 F. (2) If a F and a b, then b F. (3) If a, b F, then a b F. An ultrafilter in A is a filter F such that 0 / F, and for any a A, a F or a F. Prove that a filter F is an ultrafilter iff F is maximal among the set of all filters G such that 0 / G. E9.5. (Continuing exercise E9.4) Prove that for any nonzero a A there is an ultrafilter F such that a F. E9.6. (Continuing exercise E9.4) Prove that any BA is isomorphic to a field of sets. (Stone s representation theorem) Hint: given a BA A, let X be the set of all ultrafilters on A and define f(a) = {F X : a F }. E9.7 (Continuing exercise E9.4) Suppose that F is an ultrafilter on a BA A. Let 2 be the two-element BA. (This is, up to isomorphism, the BA of all subsets of 1.) For any a A let { 1 if a F, f(a) = 0 if a / F. Show that f is a homomorphism of A into 2. This means that for any a, b A, the following conditions hold: f(a + b) = f(a) + f(b); f(a b) = f(a) f(b); f( a) = f(a); f(0) = 0; f(1) = 1. E9.8. (Lindenbaum-Tarski algebras; A knowledge of logic is assumed.) Suppose that L is a first-order language and T is a set of sentences of L. Define ϕ T ψ iff ϕ and ψ are sentences of L and T = ϕ ψ. Show that this is an equivalence relation on the set S of all sentences of L. Let A be the collection of all equivalence classes under this equivalence relation. Show that there are operations +,, on A such that for any sentences ϕ, ψ, [ϕ] + [ψ] = [ϕ ψ]; [ϕ] [ψ] = [ϕ ψ]; [ϕ] = [ ϕ]. Finally, show that (A, +,,, [ v 0 ( (v 0 = v 0 ))], [ v 0 (v 0 = v 0 )]) is a Boolean algebra. 79

15 E9.9. (A knowledge of logic is assumed.) Show that every Boolean algebra is isomorphic to one obtained as in exercise E9.8. Hint: Let A be a Boolean algebra. Let L be the first-order language which has a unary relation symbol R a for each a A. Let T be the following set of sentences of L : x y(x = y); x[r a (x) R a (x)] for each a A; x[r a b (x) R a (x) R b (x)] for all a, b A; xr 1 (x). E9.10. Let A be the collection of all subsets X of Y def = {r Q : 0 r} such that there exist an m ω and a, b m (Y { }) such that a 0 < b 0 < a 1 < b 1 < < a m 1 < b m 1 and X = [a 0, b 0 ) [a 1, b 1 )... [a m 1, b m 1 ). Note that A by taking m = 0, and Y A since Y = [0, ). (i) Show that if X is as above, c, d Y { } with c < d, c a 0, then X [c, d) A, and c is the first element of X [c, d). (ii) Show that if X is as above and c, d Y { } with c < d, then X [c, d) A. (iii) Show that (A,,, \,, Y ) is a Boolean algebra. E9.11. (Continuing exercise E9.10.) For each n ω let x n = [n, n + 1), an interval in Q. Show that n ω x 2n does not exist in A. E9.12. Let A be the Boolean algebra of all subsets of some nonempty set X, under the natural set-theoretic operations. Show that if a i : i I is a system of elements of A, then i + a i ) = 1 = i I(a a ε(i) i, ε I 2 i I where for any y, y 1 = y and y 0 = y. E9.13. Let M be the set of all finite functions f ω 2. For each f M let U f = {g ω 2 : f g}. Let A consist of all finite unions of sets U f. (i) Show that A is a Boolean algebra under the set-theoretic operations. (ii) For each i ω, let x i = U {(i,1)}. Show that ω 2 = (x i + x i ) i ω while ε ω 2 i ω x ε(i) i =, 80

16 where for any y, y 1 = y and y 0 = y. This is an example of an infinite distributive law that holds in some BAs (by exercise E9.12), but does not hold in all BAs. E9.14. Suppose that (P,, 1) is a forcing order. Define p q iff p, q P, p q, and q p. Show that is an equivalence relation, and if Q is the collection of all -classes, then there is a relation on Q such that for all p, q P, [p] [q] iff p q. Finally, show that (Q, ) is a partial order, i.e., is reflexive on Q, transitive, and antisymmetric (q 1 q 2 q 1 implies that q 1 = q 2 ); moreover, q [1] for all q Q. E9.15. We say that (P, <) is a partial order in the second sense iff < is transitive and irreflexive. (Irreflexive means that for all p P, p p.) Show that if (P, <) is a partial order in the second sense and if we define by p q iff (p, q P and p < q or p = q), then A (P, <) def = (P, ) is a partial order. Furthermore, show that if (P, ) is a partial order, and we define p q by p q iff (p, q P, p q, and p q), then B(P, ) def = (P, ) is a partial order in the second sense. Also prove that A and B are inverses of one another. E9.16. Show that if (P,, 1) is a forcing order and we define by p q iff (p, q P, p q and q p), then (P, ) is a partial order in the second sense. Give an example where this partial order is not isomorphic to the one derived from (P,, 1) by the procedure of exercise E9.14. E9.17. Prove that if (P,, 1) is a forcing order such that the mapping e from P into RO(P) is one-one, then (P, ) is a partial order. Give an example of a forcing order such that e is not one-one. Give an example of an infinite forcing order Q such that e is not one-one, while for any p, q Q, p q iff e(p) e(q). E9.18. (Continuing E9.14.) Let P = (P,, 1) be a forcing order, and let Q = (Q,, [1]) be as in exercise E9.14. Show that there is an isomorphism f of RO(P) onto RO(Q) such that f e P = e Q π, where π : P Q is defined by π(p) = [p] for all p P. References Koppelberg, S. General theory of Boolean algebras. Volume 1 of Handbook of Boolean Algebras, edited by J. D. Monk and R. Bonnet, North-Holland 1989, 312pp. Kunen, K. Set Theory. 81