J. 1. A («-5 “if..L.g:‘;'. R! l (7.3, UK" " ' " ‘H W ‘;J LIBRARY Michigan State University PLAifin RETURN BOX to remove this checkout from your record. TO A “mom m due. ! DUE [ ll T MSU Is An Amrmmive Action/Equal Opportunity Institution CHARACTERIZATIONS OF THE SYMMETRIC DIFFERENCE AND THE STRUCTURE OF STONE ALGEBRAS By JACK GRESHAM ELLIOTT A THESIS Submitted to the School for Advanced Graduate Studies of Michigan State University of Agriculture and Applied Science 1n.part1al fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1958 . (,3: adv. v To Verda without Whose patience and faith.this thesis would not have been written. ACKI‘IOWLEDGEMEN TS The writer wishes to indicate his deep gratitude to Professor L. M. Kelly for his guidance, inspiration and friendly criticism during the studies leading to and the preparation of this thesis. The writer also wishes to express his appreciation of the interest and encouragement extended to him by Professor E. A. Nordhaus. Section 1. Section 2. Section 3. 360131011 1].. TABLE OF CONTENTS Page Introduction............................. 1 Characterizations of the symmetric difference operation.in a Boolean algebra.......................... 5 Structure of Stone algebras.............. 25 Characterization of certain Stone algebras as direct products............t. hl Section 1. Introduction Studies of Brouwerian algebras in which there is defined a binary operation analogous to the distance function of a metric space have been carried out by Nordhaus and Lapidus [ 11] and by Lapidus[ 8 ] . Their work generalized many of the earlier similar investigations of Ellis [5.6] and B1umenthal[ L; ] in the field of Boolean algebras. Ellis, in particular, observed that in a Boolean algebra the symmetric difference operation satisfied lattice relationships formally equivalent to the postulates defining a metric distance function, and showed that many purely geometric concepts could be carried over into this new setting. The first goal of this thesis is to show that in a Boolean algebra the symmetric difference is the only binary Operation Which satisfies the requirements of an.abstract metric and is simultaneously a group operation. One important difference between Boolean and Brouwerian algebras is thefact that the Boolean complement x' of an element x is disjoint from.x, while the Brouwerian ccmplement‘lx of an element x is not necessarily disjoint from x. However, in.many (but not all) Brouwerian algebras it is true that, given any element x, the elements 1x and'Tix are disjoint, where]]x denotes the Brouwerian complement of 'TX. M. H. Stone has asked, "What is the most general Brouwerian algebra in which, for every x, the elements 7x and 71x are disjoint?"1 ‘ The second goal of this thesis is to determine the basic structure of these Brouwerian algebras. In Section 2 the symmetric difference operation in a Boolean algebra is characterized as the only binary operation which.is at once an abstract metric and a group operation. By successive weakening or removal of some of the group and metric postulates generalizations of this result are obtained. Other characterizations of the symmetric differ- ence among the class of Boolean operations are found, and Section 2 is concluded with.further characterizations of the symmetric difference in a Boolean algebra as the only binary operation satisfying certain other side conditions. In Section 3 there is determined the basic structure of those Brouworian algebras in which, for every x, the elements 1x and 11x are disjoint. An interesting charac- terization of a wide sub-class of these special Brcuwerian algebras is presented in Section a. In the remainder of this section are presented fundamental definitions, concepts, and notation to be used throughout. 1This question appears as Problem.70 of Birkhoff [ 3 ], where it is phrased in the dual setting of pseudo- complemented lattices. -3- Definition: A partially ordered set P is a set of elements a, b, c, .4». together with a binary relation 9. 2 I) (read "a is over b", "a Contains b", or "b is under a") subject to the following postulates: . Pl: a _>_ a ‘ P2: Ifoa_>_b andb_>_a, thena=b P3 Definition: An £12222 £93313 of a subset x of P is an If aZb andb_>_c, thenaZc. .0 element a such that a _>_ 2: holds for every x in X. An element b is the M 11311:; M of x if b is an upper bound of x and if b _<_ a holds for every upper bound a of X. A 19193. Egan-9.. of X and the greatest $93.23 gains of X are defined similarly. ' I Definition: A partially ordered set P is a lattice if for each pair of elements a, b the greatest lower bound of a and b and the least upper bound of a and b exist. The greatest lower bound of a; and b is denoted by vb, or ab, and is referred to as the product, or lattice product, or 33933 of a and b; the least upper bound of a and b is written a + b and is called the gig, or lattice m, or 1&1}; of a and b. It is shown in Birkhoff [ 3 1 that the meet. and join Operations satisfy the following laws: Ll: (Idempotent law): a 4'- s w a and as II 9.. L2 (Commutative law): a + b = b + a and ab e3 be. L3 (Associative law): a + (b + c) = (a + b) + c and MM) == (ab)c. 11+ (Absorption law): a + ab = a and a(a + b) = a. -4- Definitiqn: A distributive lattice is a lattice in which for every triple of elements a, b, c the following relation- ships hold: L5 L6: a + be = (a +b)(a + c). a(b + c) =ab + ac. Section 2. Characterizations of the Symmetric Difference Operation in a Boolean Algebra Definition: A Boolean algebra is a distributive lattice with 0 and I in which for each element a there exists an element a' satisfying a + a' = I and aa' == 0. The element a' is referred to as the complement (or Boolean complement) of a. It can be shown that the complement a' of a is unique, and that complementation is ortho-complementation, i.e. that (a')' = a. 2 Definition: With each pair of elements a, b of an abstract set S let there be associated an element f(a, b) of a lattice L with an O. The binary function f is a metric function from S to L if the following three conditions hold: 151 f(a, b) 0 if, and only if, a = b, M2: f(a, b) f(b, a), M3 f(a, b) + f(b, c) _>_ f(a, c); and we say that "S is lattice-metrized by f". A metric function 1‘ from a lattice L to itself is called a metric operation, and in this case L is called an auto- metrized lattice. -6- The properties Ml, H2 and N3 are lattice-analogues of the familiar requirements of a distance function in a metric space. We carry the analogy further by referring to the lattice element f(a,b) as the "distance between.a and b", the elements f(a,b), f(b,c) and f(a,c) as "sides of the triangle whose vertices are a, b and c", and in general using geometric terminology wherever such usage is convenient and suggestive. It is particularly convenient to refer to M3 as the "triangle inequality". Definition: In a Boolean algebra the element ab! + a'b is the gymmotric difference of a and b. Theorem 2.1 [Ellis, 5] : The symmetric difference in a Boolean algebra is a metric operation. 2322:: Let d(a,b) denote the symmetric difference of a and b. First we observe that d(a,b) =laa' + a'a = O + O = 0. Next we show that d(a,b) = 0 implies a = b. d(a,b) = ab' + a'b = O can hold only if ab' = a'b = 0. To each side of the equation ab! = O we add ab, obtaining (2.1) ab! + ab = 0 + ab = ab. Then ab = ab! + ab = a(b' + b) = a1 = a, using the fact that a Boolean algebra is a distributive lattice. But ab = a means that a E;b° Similarly, from a'b =g0 we con- clude that b;g_a. Therefore a =‘b, and Ml holds. Since the expression ab' + a'b is symmetric in a and b, it follows that M2 holds. To prove M3, we will show that [d(a,b) + d(b,cfl'd(a,c) = d(a,c), which of course implies d(a,b) + a(b,c) 2;d(a,c). To this end, we write (2.2) [d(a,b) + d(b,c)] ' d(a,c) = [(ab' + a'b) + (bc' + b'c)]-(ac' + a'c) = ab'ac' + a'bac' + bc'ac' + b'cac' .+ ab'a'c + a'batc + bc'a'c + b'ca'c = ab'c' + abc' + a'bc + a'b'c = ac'(b' + b) + a'c(b + b!) = ac' + a'c = d(a,c) and the proof is complete. In the following theorem, we let ash denote a metric group operation in a Boolean algebra, and show that sub = ab' + a'b necessarily. Lemma 1: If x, y and z are the sides of a triangle in a Boolean.algebra, then x +‘y = x + z é'y + z. - 23333: Since x + y 2 z by M3, we add x to each side to get x+y2x+ z. Similarlyx+ zzybyMB, and addingxto each side yields x + z 2 x + y. This implies that x + y = x + z, and the proof for the other two cases is similar. Lemma 2: If a = bee, then ash = c and are = b. 23222; a =‘bwc implies as(b*c) = O by M1. The associative law then gives (ash) so = O, whence ash = c by M1. The proof for the other case is similar. Lemma : O-zt-a = a. M: Let can: 1:. By Lemma 2, am: = 0. Hence a = x by Ml. Lemma 1.1:: a::-I = a'. 33223: Let asa' = b, and consider the triangle 0, a, a', the sides of which are Ora = a, Oea' = a', asa' =‘b. Lemma lgivesusa+b=a+a'=I, anda'+b=a+a'=I. -8- Hence I = (a + b)(a' + b) = b. We conclude that awa‘ = I, and Lemma 2 gives us awI = a' and airI = a. Theorem 2.2: The only metric group operation in a Boolean algebra is the symmetric difference. nggf: Let xry = p. Consider the triangle I, x, y, the sides of which are I%x = x', Try = y', xwy = p by Lemma h. From Lemma 1 we conclude that x' + y' = x' + p and x' + y: = y' + p. Multiplying the first of these by x gives xy' = xp, and multiplying the second by y gives ny = yp. Adding, we obtain xy' + x'y = xp + yp = (x + y)p. From the triangle 0, x, y, whose sides by Lemma 3 are wa = x, Cry =‘y, and x-::~y = p, we obtain x + y 2 p by the triangle inequality. Hence (x + y)p = p, and xy' + x'y = p = xey, completing the proof. We now extend Theorem.2.2 by relaxing some of the group requirements. Definition: A semi-group is a system of elements together with an associative binary operation. Theorem 2.3: The only metric semi-group operation in a Boolean algebra is the symmetric difference. 2323;: The only group property used in the proof of Theorem 2.2 was the associative law. Definition: A binary operation % is weakly associative if a*(a*b) = (awa)*b. Theorem 2.h: The only metric weakly associative operation in a Boolean algebra is the symmetric difference. 23223: In the proof Of Theorem.2.2, the associative law was used OnTy to show that a = bsc implies b = ass and c = ash, i.e. in the proof Of Lemma 2. we will show that these relations follow from the weak associative law and the fact that the symmetric difference is a metric operation. Then the proof Of Theorem 2.1 suffices as a proof of this theorem. The fact that Osa = a follows from M1 and the weak associa- tive law, for as(as0) = (asa)so = oso = 0 implies a = aso. Now let a = bsc, x = ash, and y = asc. Then (2.3). x = hsa = hs(bsc) = (hsh)sc = osc = c (2.h) y = csa = cs(csh) = (csc)sh = Osh = h. Hence a =‘hsc implies h = asc and c = ash.) Theorems 2.3 and 2.h.were generalizations Of Theorem.2.2 obtained through.relaxations Of the group postulates. In Theorem.2.5, which.fellows shortly, the associativity is abandoned. Definition: A quasiggroup is a system consisting Of a set of elements, together with.a binary Operation which satisfies the law Of unique solution, i.e. if a = hsc and two of these are known, the third is uniquely determined. A.lggp is a quasi-group with a two-sided identity element. ‘ Definition:' The Ptolemaic inequality holds'for a quadri- lateral if the three products (meets) Of Opposite sides satisfy the triangle inequality (M3). Theorem.2.5: The only metric loop Operation in a Boolean algebra is the symmetric difference. Before proceeding with the proof, some lemmas will be -10- established. Lemma 1: The loop identity is 0. PM: Let e‘denote the loop identity. Since ese = e by definition and ass = e by m, we have a = 0. Lemma 2: asI = a' and _asa' = I. £329.93: By the law Of unique solution there exists y such that asy = a'. The sides of triangle 0, a, y are asy = a', Osa = a, and Osy = y. The triangle inequality implies (2.5) a+y_>_a' and a'+y?_a. Tlms (2.6) as! + a'y _>_ a' and aa' + ay _>_; a, or (2.7) a'y 2 a' and ay 2 a. But these imply that'y _>_; a' and y _>_ a. Hence y a I, or asI = a'. Consider the triangle 0, a, a' whose. sides are Osa = a, Osa' = a' and asa'. Again the triangle inequality implies (2.8) a + (asa') _>_; a' and a' + (asa') _>_ a. Multiplying the first Of these by a' and the second by a gives (2.9) a'(asa') _>_ a' and a(asa') _>_~ a. Fram these we conclude that asa' Z a' and asa' _>_ a; hence asa' = I. ‘ Lemma :2: The Ptolemaic inequality holds in any quadrilateral 0, I, a, b. M: In the quadrilateral O, I, a, b, the side Gsa is Isb, the side Osb is Opposite Isa, and the side OsI is -11- Opposite ash. We will show only that (2.10) (OsaHI-u-b) + (Osb)(Isa) 2 (OsI)(ash) or (2.11) ab' 4- a'b _>_ I 0 (ash) = ash; the proofs for the other two cases are similar. The triangle I, a, b has sides Isa = a', Ish =‘b' and ash by Lemma 2. The triangle inequality gives (2.12) a' + b' 3 ash. By Lemma 1, the sides Of the triangle 0, a, h are Osa = a, Osb =‘b and ash. The triangle inequality here yields (2.13) a + h 2 ash. Hence (2.114.) (a + b) (a' + b') _>_ ash or (2.15) ab' + a'b‘z ash, which is what we set out to show. Proof Of Theorem.2.5: Let ash =‘x. We knew from.Lemma 3 that ab' + a'b‘z'x. We will complete the proof by showing ab' + a'b;g x. Applying the Ptolemaic inequality to the quadrilateral O, I, a, b', we have (2.16) (0sa)(Ish') + ('Osb'HI-sa) _>_ (0sI)(asb') or (2.17) ab + a'b' _>_ I - (ash') = asb'. Since ab' 4- a'b _>_ x, we Obtain (2.18) (ab' + a'b)( ab + a'b') 2 x . (asb'). But (ab! + a'b)(ab + a'b') = 0, hence (2.19) x . (asb') = O. -12- The triangle a, b, b' has sides ash = x, asb', and bsh' = I by Lemma 2. Using the triangle inequality, we get (2.20) x + (ash') = I. Thus x is the complement of ash', i.e. (2.21) x' = asb'. A similar argument shows that (2.22) x' = a'sb. Using the identity usv 5 u + v, we have (2.23) x' 5 a + b' and x' _<_ a' + b. Hence (2.2h) x"5 (a + h')(a' + h) = ab + a'b'. By DeMorgan's laws, we get (2.25) x 2 ab' + a'b. This, together with the earlier result x‘s ab! + a'b, shows that (2.26) x = ab! + a'b and completes the proof Of Theorem.2.5. It might be conjectured that a metric quasi-group Operation is a Boolean algebra is necessarily the symmetric difference. The following example shows that this is not the case. In the Boolean algebra Of four elements 0, a, a' and I define "distances" as shown in Table 1. Table 1 C3 C) a a' I 0 0 a' a I a a' O I a a' a I O a' I I a a' o -13- Sinoe each.element appears once and only once in each row and column Of Table l, the law of unique solution holds. That Ml holds '15 shown by the fact that the elements on the main diagonal, and only those elements, are 0. The symmetry about the main diagonal implies that M2 holds. It is easily seen that the sides of any non-degenerate triangle are a, a' and I, hence M3 holds. This shows that 69 is indeed a metric quasi-group Operation. However, 0 ® a = a', while Osa == Oa' + O'a = a. Hence 6.9 is not the symmetric difference. Bernstein [1,2] characterized the possible group operations in a Boolean algebra among the class Of Boolean Operations. The author is indebted to Professor B. M. Stewart for pertinent Observations which.led to the following theorem. This theorem is similar to those in[ l ] . Definition: An Operation s is a Boolean operation in a Boolean algebra if (2.27) xsy = Axy + Bxy' + ery + Dx'y', where A, B, C and D are fixed elements Of the Boolean algebra. Theorem 2.6: Anvaoolean group operation in a Boolean algebra is an abelian group Operation, and is of the form (2.28) xsy = e(xy + xry') + e'(xy| + x'y) where e is the group identity. 2323;: The proof consists Of evaluating the "constants" A, B, C and D under the assumption that s is a group operation. Repeatedly using (2.27), we write (2.29) OsD = AOD + BOD! + CID .+ DID! = CD, (2.30) OsC' = AOC' + ECO + CIC' 4- DIC = DC. -m- IBy the law Of unique solution, this implies D = C'. Now (2.31) Dso = ADO + BDI + CD'O + DD'I = BD, (2.32) B'so = AB'O + BB'I + CBO + DBI = DB. Again by the law of unique solution, D = B'. Next (2.33) AsD = AAD + BAD' + OA'D + DA'D' = AD + AB = AD + AD‘ = A implies that D = e by definition Of the group identity. Hence (2.27) can be written (2.3h) xsy = Axy + eixy' + e'x'y + exry'. Now (2.35) e = ese = Aee + e'ee' + e'e'e + ee'e' = A0. Since e = 3', this gives B' = AB'. Next we Observe that (2.36) A's-B = AA'B + e'A'B' + e'AB + eAB‘ == AB + AB' = A, (2.37) BsB = ABB + e'BB' + e'B'B + eB'B' = AB + B: = AB + AB' = A. By the law Of unique solution, we get A' = B. Collecting results, we can write (2.38) e = D = B' = C' = A and e' = D' = B = C = A'. Hence (2.27) becomes finally (2.39) xsy = exy + e'xy' + e'x'y + ex'y' = 6(xy + x'y') + e'(xy' + X'y). The fact that s is an abelian Operation follows from.the symmetry in x and y of the right side of (2.39). -15- Corollary 1: In a Boolean algebra, the only Boolean group Operation with O as the group identity is the symmetric difference. Corollarng: In a Boolean algebra, the only Boolean group Operation such.that OsO = O is the symmetric-difference. 'ggggf: Using (2.28), we write (2.hO) O = OsO = e(O-O + II) + e'(OI + IO) = e, and Corollary 2 follows from Corollary 1. We notice that, in the proof Of Theorem 2.6, nO use was made of the associative law. Thus Theorem 2.6 may be generalized to get Theorem.217: Any Boolean loop Operation in a Boolean algebra is an abelian group Operation, and is of the form (2.hl) xsy = e(xy + x'y') + e'(xy' + x'y), where e is the loop identity. £3223: Exactly as in the proof of Theorem.2.6, it can be shown that s is an abelian Operation of the form cited in the theorem.statement. We will now show that the associative law holds, in particular that (2.h2) zs(xsy) and x-y'z + xiylz + xlyzl + xylzl (2.h3) (zsx)sy - xyz + x'y'z + x'yz' + xytzt. -16- In what follows, DeMOrgan's laws are used repeatedly. (2.111;) ' zs(xsy) = e [z(xsy) + z'(xsy)'] + e' [z(>s!-y)' + z'(x-=:-y)] ' = (62 + e'z')(x*y) + (62' 4- e'2)(x-='rv)' = (62 + e'z') [a(ryhir X'y') + e'(xy' + X'fl] + (625' + 6'2) [div + X'Y')‘+ e'(x:v" + X'y)]' == ez(xy f" x'y') +~ e'z'(xy' + x'y) + (92' + e'z) [e(xy + x'y')]' [e'(xy' +‘x'y)]: ez(xy + x'y') +,e'z'(xy' + X'Y) + (ez' + e'z).[e' + (xy + x'y')‘] [afl- (xy' +3030!) ez(xy + x'y') + e'z'(xy' + x'y) (I + (ez' + e'z) [e'(xy' + x'y)' + e(xy + x'y')‘ + (xy + x'y')'*(xy' + x'y)'] = ez(xy + my) + e'z'(xy' +x'y) + e'z(xy' +ix'y)' + ez'(xy + x'y')' '+ (62' + e'ZHXY)'(x'y')'(xy')'(x'v)' .ezmr + X‘y') *9 e'z'(xy' + K'y) + 6'2(ry')'(x'y) ' l+._rezt(i:q))(xty')! + (ez' + e'z)(x' + y|)(x + y)(x' + y)(x + y') ez(xy + x'y') + e'z'(xy' + x'y) + e'z(x' + y) (x + y')++ ez'(x' + y')(_x + y’) 4- (ez' + e'sz'y + xy'Hx'v' + xy) = ez(xy + x'y') + e'z'(xy' + x'y) + e'z(xy + x'y') + ez'(xy' + x'y) = emz + ex'y'z + e'xy'z' + e'x'yz' + e'xyz + e'x'y'z + exy'z' + ex'yz'. Collecting terms, we obtain 8 (2.11.5) zs(xsy) == xyz + x'y'z + x'yz' + xy'z'. -17- TO find (zsx)*y, we use the fact that s is abelian tO write (2.1;6) (zsx)sy = ys(zsx). Replacing 2 by y, x.by z, and y by x in (2.hS), we get (2.14.?) (zsx)sy = zxy + z'x'y + z'xy' + zx'y'. Hence (2.h8) (zsx)sy = xyz + x'y'z + xiyz' + xyiz'. The right sides or (2.16) and (2.1m) are identical, which proves that the associative law holds. Since an associative lOOp is a group, the theorem.follows. Corollary 1: The only Boolean loop Operation with O as the loop identity in a Boolean algebra is the symmetric difference. Corollary 2: The only Boolean loop Operation such that OsO = O in a Boolean algebra is the symmetric difference. .zgggg: Since («Z-#9) my = abs + x'y') + e'(xy' + x'y) we can write that (2.50) O = OsO = e(OO + II) + e'(OI + IO) = 6. Then Corollary 2 follows from Corollary 1. It is interesting that the requirement that s be a Boolean Operation allowed us to remove the associative law from the assumptions needed to characterize the symmetric difference among the class of Boolean operations. Itn will be shown next that a similar phenomenon occurs with respect to the triangle inequality. Definition: A binary Operation is called semi-metric if it satisfies M1 and M2. -18- Theorem 2.8: The only Boolean semi-metric Operation in a Boolean algebra is the symmetric difference. 23922: According to Bernstein[ 1 1, a Boolean Operation has the form . (2.51) my = (lea-1m + (I-x-o) xv + (o-x-I) x'y + (cz-omyi. Thus (2.52) . IsI by M1, and u 9 8 u o (2.53) Ozz-I = IsO by MZ. Let OH = z. Then (2.51) yields (2.5h) Isz = zIz' 4 zI'z = O*+ O = O, and z = I by M1. Thus (2.55) xsy = xy' + x'y. Frink [ 7 ] has characterized the symmetric differ- ence as the only Boolean group Operat ion over which the meet distributes. A In what follows, however, we will not restrict ourselves to Boolean Operations. Theorem 2.9: The only semi-metric group Operation in a Boolean algebra over which the meet distributes is the symmetric difference. . M: The group identity is 0. Let a, b and c be the sides Of the triangle 1, m,..n. Using the associative law, M1 and M2, it is seen that (2.55) ash i= (lsm)s(m-::-n) = 1s[ms(m-::-n)] = 1s (m-::-m)-::-r£l = 1s(0sn) = 1sn -19- Similarly it can be shown that (2.56) asc = b i and (2.57) obit-c = a. New, using (2.55) - (2.57) and the distributivity assumption, (2.58) [(asb) + (hsc)] (as-c) = (c + a)(asc) [(9. + c)a]s[(a + c)c] ='- a'X‘O. Recall that the lattice relation (x + y)z = z implies x + y _>_ 2. Hence (2.58) yields (2.59) (ash) + (hsc) _>_ asc. Similarly it can be shown that (2.60) (ash) + (asc) 2 hsc (2.61) (hsc) + (asc) 2 ash. Thus M3 holds, and s is a metric group Operation. Then s is the symmetric difference by Theorem 2.2. A It might be conjectured that the meet necessarily distributes over every semi-metric group Operation in a Boolean calgebra. That this is not the case is shown by the following example. In the‘BOOlean algebra of eight' elements, define an operation 69 by the following table: Table 2 £13 0 a c a' b' c' I O O a b c a' b' c' I a a O b' c' I b c a' b b b O- a' c a I c' c c ' a' O-' b I a h' a' a' I c b O“ c' b' a b' b! b a I c' O a' c c' c' c I a b' a' O b I I a' c' b' a c b O -20.. New 0 appears only on the main diagonal, and the table is symmetric about the main diagonal, so MI and MZ hold. Clearly O is the group identity, and inverses are unique (each element is self-inverse). It has been verified that the associative law holds. Thus 69 is a semi-metric group Operation. However (2.62) a'(c®a) = 8'0' = b. while (2.63) (a'c) -::- (a'a) = GO 0 = c. which shows that the meet does not distribute over 69. Theorem 2.10: The only semi-metric semi-group operation in a Boolean algebra over which the meet distributes is the symmetric difference. 6 M: Ml guarantees that ass = 0. Thus if 0 is an identity element, then each element Of the Boolean algebra is its own inverse. But the associative law and M1 give us (2.66) (0-::-a)sa == 0:3(asa) == Oso = O whence Osa = 21, again by Ml, and O is an identity element. If e is any element such that esa = a holds. for all a, then ese = e. But ese = O by M1, so 0 is a unique identity. Thus s is a group Operation and Theorem 2.10 now follows from Theorem 2.9. Theorem 2.11:. In a Boolean algebra, the only semi-metric weakly associative operation over which the meet distributes is the symmetric difference. £13932: Using Ml and weak associativity, (2.611.) (Osa)sa == os-(asa) = OsO = O -21- implies_; (2.65) Osa = a. ' By the distributivity assumption and M2 (2.66) ab'(asb) = (ab'a)s(ab'b) = (ab')*0 ab' yields (2.67) . ab' 5 ash. Similarly (2.68) a'b |/\ ash, hence (2.69) ab' + a'b _<_ ash. Bow , (2.70) ab(asb) == (aba)s(abb) = (ah)s(ab) == 0, and therefore (2.71) [ab(ash)]' '= .I. By DeMorganfls laws (2.72) (ab)' + (asb)' = I. Then (2.73) (ab) [(abw + (ea-MI = (ab) gives (2.74) (ab)(as-b)l- =4 (ab) which.implies (2.75) ab 5 (9:314?) '. Next we Observe that (2.76) (a + h)(asb) = [(a + b)a]s[(a + b)b] '== ash, or (2.77) a + b _>_ ash. -22- Hence- (2.78) (a + h)' 5 (ash)' or (2.79) a'b' _<, (ash) '- Thus (2.75) and(2.79) yield (2.80) ab + a'b' _<_ (asb)' or (2.81) (ab + a'b')‘ 3 ash. Again applying DeMorgan's laws, we get (2.82) ab' + a'b Z ash. But (2.69) and (2.82) together imply (2.83) ab' + a'b = asb' and the theorem.is proved. Corollary: In a Boolean algebra, the only semi-metric operation s such that Osa = a for every a and such that the meet distributes over s is the symmetric difference. 23322: In the proof Of Theorem 2.11 the weak associativity preperty was used only to show that Osa = a for every a in the Boolean algebra. Definition: As usual, let s denote the symmetric difference. A binary Operation O is called quasi-analytical (Marczewski [10 ]) when (2.814.) (aob)chOd) 5 (ass) 4- (hsd) for all quadruples a, b, c, d Of a Boolean algebra. Theorem 2.12: (Marczewski): The only quasi-analytical group operation in a Boolean algebra with O as the group identity is the symmetric difference. -23- 3292.3 Marczewski showed in his proof that the Operation 0 is Boolean. . It then follows. from Corollary 1 to Theorem 2.6 or from Bernstein's results I: 1] that O is a metric Operation, whereupon the theorem follows from Theorem 2.2. Following is an independent proof Of Marczewski's theorem. First we note that a = a'l, for (2.85) a = asO = (aoO)s(aoa'1) _<_ (as-a) + (Osa'l) = a‘1 and 6 (2.86) a'1 = a'lso = (a‘loo)s(a'1oa) _-_<_ (a’lsa'l) + (Osa) = a give us respectively a 5 a"1 and a"1 5 a. . Since a ='a"1, we have aoa = 0. Let aob = 0. But aoa = 0, hence a = b by the law Of unique solution and M1 holds. To prove M2, we write (2.87) (aob)o(boa) = so [bO(boa)] ao [(boh)oa] ao(Ooa) II (I aoa ll 0. Thus 'aOb = boa by Ml. Let a, b and c be sides of a triangle 1, m, n, with a = lom, b = men and c = lon. Then (2.88) aOb == (lom)o(mon) = lon = c (2.89) sec = (lom)o(lon) == (mol)o(lon) == men s b (2.90) boc == (mon)o(lon) = (mon)o(nol) = mOl = a Now (2.91) .0 = aob (OOO)%:-(aob) _<_ (Osa) + (Osb) = a + b (2.92) b = aoc = (DOOM-(ace) < (Osa) + (Os-c) = a + c (2.93) a = boc (OOO)s(boc) _<_ (Osb) + (Osc) = b + c proves M3. Hence 0 is the symmetric difference by Theorem 2.2. ’ Section 3. Structure of Stone Algebras pgfinition: A Brouwerian algebra is a lattice L in which for every pair Of elements a, b there exists an element x such that (3.1) b + x Z a and (3:2) b + y _>_ (1 implies y 2 x. In other words x is the "smallest" element such that b + x _>_; a. The element x is the difference of a andb. and is denoted by a -' b. It may be verified (see McKinsey and Tarski, [9] ) that (3.3) a-bsc ifandonlyifa_<_b+c. Examples of Brouwerian algebras are numerous; among the Brouwerian algebras are all Boolean algebras, all chains with c, all finite distributive lattices, all distributive lattices in which descending chains are finite, and all complete and completely distributive lattices. Theorem 3.1: A Brouwerian algebra is a distributive lattice. Proof: We will show that (B-M) a + ylyg = (a + V1)(a + V2). -26- Let (3.5) b = (a + y1)(a + are). Then (3.6) a+y1_>_banda+y22b implies ‘ (3.7) ylzb-aand yZ‘Zb-a by (3.3). This gives (3.3) V1372 _>_: b - a. We can now write (3.9). e+y1y22a+ (b - a) _>_b. whore the last inequality follows from (3.1). Having (3.10) a + V132?- (a + y1)(a + y2 ), it remains to show that the Oreverse inequality also holds. But in any lattice (3.11) a g a + yl, and y1y2 5 a + yl implies (3.12) a + ylyg 5 a + yl. Similarly a + ylya _<_ a 4- ya. Hence (3.13) a + hire _<_ (a + y1)(a + V2): This shows that (3.14.) holds. Also valid is the dual of (3.14.), i.e.' the expression (3.1M a(yl + 3'2) = “3’1 + aye obtained from (3.1() by interchanging "4-" and ".". Definition: If the Brouwerian algebra has a greatest element I, the element I - a is the Brouwerian gomplement of a, and is denoted by “Is. Similarly I -"|a ="|']a, I ma =ma. and on. In what follows, we restrict ourselves to Brouwerian algebras having an O and an I. It is shown in [9] and [13] that (a) a Sb implies ]a Z'Ib (b) Tia _<_ a (3.15) (0) THa =1a ((1) Met) =']a +11. (e) ‘l(e + b) =‘l‘l(1s]b). M. H. Stone has asked the question: "What is the most general Brouwerian algebra B in which. IaJIa = 0 holds for every element a in B?". This problem.appears in its dual form as "Problem 70" of Birkhoff [3]. A simple example of a Brouwerian algebra in which this prOperty does p92 hold is the lattice whose five elements are c, ab, a, b, a + b = I, for in this lattice ‘Is = b, .1b =‘f1a = a, but 1a11a = ba‘# 0. On the other hand, this prOperty holds in every Boolean algebra, and in every chain with an O and an I. Definition: A S2933 algebra is a Brouwerian algebra in which.']a]1a ll 0 identically. Let B denote a Brouwerian algebra with O and I, and let X denote the set Of elements of B satisfying ']x]]x = 0. If x and y are in X, then (3.16) 1(x + 38TH): + y) =71(1x1y)(11x +T|v> _<_ IX‘M‘TIX +TIF) =‘lx'ly11x +"|x'|yT|y = o + o = O -28- and (3.17) 1(m)'|'|(xy) (1x + ly)'|'|(xy) (‘lx + ‘IyYI'K‘I'Ime S. (11 + 'IYYHXTW = WXTbITIy + “ly'l‘leiy = o + o = 0 show that X is a sub-lattice Of B, since X is partially ordered by the partially ordering of B. Further, using the relationship 1(a - b) =1a +‘Hbl, we see that (3.18) )(x - y)'l‘|(x - y) (1:: +11y)[))(11x1y.)] 5. (1x +‘I'Iy)T|x')y IXTIx'Iv +"I‘IyTlx‘ly =O+O=O That is, if x and y are in X, then so are x + y, xy and x - y. This proves the Theorem 3.2: In a Brouwerian algebra B with O and I, the collection 'Of elements x satisfying ijlx == 0 is a Stone sub-algebra. Birkhoff [3]has shown that in any Brouwerian algebra B the subset R of elements satisfying 'T)r ='|r is a Boolean algebra under the Operations a + b and a® b = Tush). In a Stone algebra, however, the subset R is a Boolean sub-algebra Of B, i.e. R is a Boolean algebra under the Operations a + b and ab which hold in B. This is, in fact, a characterization Of Stone algebras, as is shown by the following theorem.‘ 1This result is shown in[8]. Theorem 3.3: A Brouwerian algebra B is a Stone algebra if and only if R is a Boolean sub-algebra of B. Before proceeding with the proof Of this theorem, some lemmas will be established which not only facilitate the proof but also add some insight into the structure of Stone algebras. Let Q denote the set of all elements of B satisfying a‘|a = 0. Lemma 1: Q is a subset Of R. m: If a is in Q, then a]a = 0 implies (3.19) 118. = m + a‘la = (Tla + a)(TIa + 1a) = a I = a, and a is in R. Lemma 2: Q is a sub-lattice Of B. 23392: Let a and h be in Q. Then (3.20) (a + h)‘|(a'+ b) _<_ (a + b)('|a‘|b) = ala'lb + bla'lb == 0 + 0 == 0 and (3.21) (ab)'|(ab) = ab('la + ‘|b) == ab‘la + ab'lb = 0-'~+ O = 0 show that a + b and ab are in-Q. M 3: B is a Stone algebra if and only if Q = R. M: Let Q = R. Recalling that "Is =Tn a, it follows that 1a is an element of R, for every a in B. Since Q = R, we have that 'laTla = O, and B is a Stone algebra. Conversely. let 1x112: = 0 hold for every x in B. If x is in R, then 'I'lx = x implies 0 = ‘ flx. Hence R is a subset of Q. Using Lemma 1 we conclude that R = Q. -30- Proof Of Theorem 3.3: Let B be a Stone algebra. Then R = Q'by Lemma 3: hence if x is in R then xflx ==0. Since x +']x = I identically, it is seen that ']x is a Boolean complement Of x, and is unique since B is a distributive lattice. By Lemma 2, R = Q is a sub-lattice of B. Hence R is itself a complemented distributive lattice under the operations of B, 1.6. R is a Boolean sub-algebra Of B.’ Conversely, assume that R is a Boolean sub-algebra Of B. Then if x is in R, there exists an element x' in R satisfying x + x' = I and xx' = 0. Since x +’]x = I, we have (x +']x)(x + x') = x + x11.x = I. This, together with. x(x'1x) = 0, implies that x' = x'flx since B is a distributive lattice. Thus x' $11!. But x' satisfies 2: + x' _== I, hence ‘Ix 5 x' by definition Of the Operation 1. This shows that x' = ‘Ix, and hence xx' = £11: '5' O. From this it follows that R is contained in Q. Applying Lemma 1, we have that R = Q. Then B is a Stone algebra by Lemma 3, and the proof is complete. This. theorem suggests that Stone algebras may, in a sense, he built up from.Boolean algebras. This is indeed the case, and in the remainder Of this section we present a characterization theorem which gives some insight: into the general structure Of Stone algebras. Definition: An ideal J in a lattice K is a subset of K having the prOperties (3.22) x and y in J implies x + y is in J. (3.33) x in J and y;5 x implies y is in J. Let L be a distributive lattice with o and I. R be a Boolean sub-algebra Of L containing 0 and I, and T be an ideal in L having the properties W (3.21:) (a) The only element in L (common to both R and T is O. (b) T is a Brouwerian sub-algebra Of L. Remark: t1 4- t2 = I holds for no pair Of elements t1, t2 Of T. Proof: If t1+ t2 = I for some pair Of elements t1, t2 of T, then the fact that T is an ideal would imply that I is in T. This is impossible by (3.2ha). W: The relationship t 2 r )5 0 holds for no elements t in T and r in R. 2.29.9.2: Assume t 3 r. Since T is an ideal, it follo‘rs that r is'in T. Then, by (3.21)), r ='o. Let B denote the direct sum R a T of R and T, i.e.‘the set of elements of L Of the form r + t, where r is in R and t is in T. Theorem 34*: B is a lattice. M: Let r1 + tliand r2 4- t2 be elements of B. T1161: (3.25) (r1 + t1) + (ra + t2) = (r1 + r2) + (t1 + t2) 4 = r3 4- t3, where r3 = r._L + r2 is in R since R is a Boolean sub-algebra OfLandt3=t1+taisinTsinceTisanidealefL. Since L is a distributive lattice, we Observe that (3.26) (r1 + t1)(r2 + t2) = r1r2 + (r1t2 + rgtl + tlta) = :03 + 153, where r3 =-r1r2 is in.R since R is a Boolean sub-algebra of L and t3 = rlta + r2t1 + tlta is in T since T is an ideal of L. Lemma 1: r - (r1 + t1) = rrl'. 923225: ‘Using the fact that L is a distributive lattice, we write (3.27) (r1 + t1) + rrl' u r1 + rrli + t1 = (r1 + r)(r1 + r1|) + t1 = (r1 + r) I + t1 = r1 + r + t1' 3 r. Thus rr1' satisfies the first part (3.1) Of the definition.. or the difference or r and (r1 + t1). We show next that if (r1 + t1) + x _>_ r then x _>_ rlrli. Let x be any element of B, say x = r2 + t2, and assume (3.28) (r1 + t1) + (r2 + t2) _>_ r. Then (3.29) (T1 + r2) + (t1 + t2) _>_ r. Since R is a Boolean sub-algebra of L, there exists an element (r1 + r2)' in R such.that (r1 + r2)(r1 + r2)! = 0. Hence (3.30) (r1 + r2)(r1 + r2)| + (t1 + t2)(r1 + r2)! _>_ r(r1 4- r2)' or _ ' (3.31) (t1 + ta)(r1 + r2)! 2 r(r1 + 1'2)" -33- The left side of (3.25) is in T, since T is an ideal of L, and the right side is in R since R is a sub-algebra of B. But in an earlier remark we showed that t Z r )5 O is impossible. Hence (3.32) r(r1 + r2)' = 0. Since R is a Boolean algebra, DeMorgan's laws hold. Hence (3.33) r' + r1 + T2 = I. Multiplying both sides by rrl', we get (rI‘1')r2 == (rrli) which.in turn implies that (3.31;) r2 _>_ rrl'. Hence (3.35) 1-2 + .132. _>_ 1-23 rrlt and the proof Of Lemma 1 is complete. Lemma 2: t - (r1 4» t1) = t - [t(r1 4- t1)]. 23222: The right side exists since T is itself a Brouwerian algebra. Let y = t - [t(r1 + t1)]. Then (3.36) (r1 4- t1) + y = (r1 + t1) 4- [t - t(r1 + t1)] ‘2 t(r1 + t1) + [t - t(r1 + tlfl Z’t by definition of the difference Operation. If x in B satisfies (3.37) (r1 + 1:1) + x _>_ t than (3.38) t(r1 + t1) + tx Z t. Appealing to the second part (3.2) Of the definition Of the difference Operation, we see that (3039) 153‘ _>_ y: i.e. y = t - t(r1 + t1) is by definition the least element satisfying t(r1 4- t1) 4- y 2 t. Hence (3.1).0) x _>_ (31 2 .y, which completes the proof Of Lemma 2. Theorem 3,5: 3 is a Brouwerian algebra. Proof: Let (r + t) and (r1 4. t1) be any two elements Of B. We will show that (r + t) - (r1 + t1) exists in s, in particular that (3.1a) (r + t) - (r1 + t1) = [r - (1.1+ t1)] +[t - (r1 + t1)] Let (3.1.2). x=r-(r1+t1) and y=t-(r1+t1). The existence of x and y is guaranteed by Lemmas 1 and 2. .Further, . (3.1)-3) x + (r:L 4- t1) = [r - (r1 + 121)] + (r1 +t1) 2 r and f (3.142).) y 4- (r1 4- t1) = [t - (r1 + t1)] + (r1 4: t1) 2 t, by the definitions Of x and y. Combining (3.113) and (3.111).), we get (3.11.5) x+y+(r1+t1)2r+t. We will complete the proof by showing that if an element 2 OfB satisfies 2 4» (r1 + t1) 2r+ t then z_>_x+y. New (3.1i6) z+(r1+t1)3r+tzr3r-(r1+t1)ax gives 2 3 x by definition Of the difference operation, and similarly (3.1)?) 2+(r1+t1)_>_r+t2tgt-(r1+t1)=y yields z _>_ y. Hence 2 Z x + y and the proof is complete. Theorem.3.6: B is a Stone algebra. ‘gggggz Let r1 + t1 denote an arbitrary element of B. ‘We will use the relationship (3.1).8) r - (r1 + (:1) --- rrl' _ of Lemma 1 to obtain ‘I(r1 + t1) and‘11(r1 + t1). By definition Of the Operation '1 , we have that (3.h9) '1(r1 + t1) = I - (r1 + t1) = Irl' = r1' and, (3.50) '11(r1 + t1) =.I -_)(r1 + t1) = I - r1' = (rl')'= r1. Since R is itself a Boolean algebra, we have (3.51) )(r1 + t1)11(r1 + t1) = rl'rl = 0. Thus B is a Stone algebra. Structure Theorem: If B is the direct sum Of R and T, where R is'a Boolean sub-algebra (with least element 0 and greatest element I) Of a distributive lattice L with O and I and T is an ideal Of L such that g (a) the only element of L common to hoth.R and T is O (b) T is a Brouwerian sub-algebra of L, then.B is a Stone algebra. Further, every Stone algebra may be so described. The first part of the Structure Theorem.has already been proved. The remainder of this section, except for some remarks at the end, will be used to prove the last part Of the theorem. Definition: Let T denote the set Of elements of B.satisfying ‘]x = I and, as before, let R denote the collection of -36- elements Of'B satisfying 'Tlx =.x. Theorem 3.7: R is a Boolean sub-algebra of B. ‘23223: This has already been proved in Theorem.3.2. Theorem_3.8: T is an ideal of B. 2339;: Let a and b be elements of T. Then ’Ia db = I. and (3.52) 1(a + b) -.= “()aIb) = TII = I, by (3.15s). Hence a + b is in T. If a is in.T, and 0:5 a, then lo Z‘la "2 I by (3.15s). Thus 'IO = I, c is in T, and T is an ideal of B. Theoremg3.9: The only element of B common to both R and T is 0. 23232: Assume that an element a is in both.R and T. Then 1s = I, and O = ala = a1 = a. Theorem 3.10: IT is a Brouwerian sub-algebra Of B. ‘ggggg: In the proof Of Theorem.3.8 we showed that a + b is in T whenever a and b are in T. Now (3.53) 1(ab)=1a +‘Ib = I + I = I by (3.15d). Hence ab is in T, and T is a sub-lattice of B. It remains to prove that a - b is in T ifya and b are in T. But a- h_<_aby (3.3). Hence a - b is in T since T is an ideal Of B. Theorem 3,11: Every element b of B can be written in the form ‘b = r + t, where r is in R and t is in T. £2223: Since B is a distributive lattice, we may write (3.51)) 'Hb + b1b= (le + b)(le +_'|b). But (3.55) le + b =_b by (3.1%). and (3.56) 'I‘lb + 'lb = I by definition of the difference operation. Hence (3.57) le + b‘lb =1le + b)(T|b + 1b) '= bl = b . holds for every element b in B. Now 11b is in R, since (3.58) 11mm =lm1b) =‘|flb) =11}. by (3.150). Further, b-I‘b is in T, for (3.59) 'I(b'lb) ='|b +le = I by (3.15d). Thus we may set ‘le = r and b‘lb = t, and the desired representation is obtained. Theorems 3.7 through 3.11 complete the proof of the Structure Theorem. More insight into the make-up of Stone algebras may be obtained by interpreting the preceding work in terms of set theory. Definitigs A .gigg _o_f_ _s_e_t_s_ is a collection 0 of sets A, B, 0,." such that if A and B belong to 0 so does the set sum AUB and the set product AnB. A Boolean ring 3: 391:3 is a ring of sets which contains with any member A the set complement A' of A. Definition: Given two members A and B of a. ring of sets 8. A a- B denotes the smallest set of all sets X in 6 satisfying BUX 3A whenever this smallest set exists. 8 is a Brouwerian ring _o_I_‘_ sets if, for every pair of‘members A, B’ A E B 61513158 in C. An example of a Brouwerian ring of sets which is not a Boolean ring of sets is the collection 4% of all closed subsets of the plane. In W , A 7 B is the inter- section of A and the closure of the complement of B. The collection 0 of all Open subsets of the plane is a ring of sets which is not a Brouwerian ring'of sets. For, let A ' and B be Open sets, neith‘er containing the other, such that A03 is not empty. The smallest set satisfying BUXjA is An B', which is not in C9 . It is easily seen that there is no smallest open set containing AnB', hence A v5; B does not in general exist in 0’ . Let C be- a ring of sets containing the null set ¢ and a greatest set I, and let R be a Boolean sub-ring of C’ which also contains ¢ and I. Let 7 be a Brouwerian sub-ring of CD. which is an ideal and which has in common with I? only the null set ¢ . Finally, let 6 = i297 denote the collection of all sets of the form RUT, where Risin RsndTlsinfl’. Set-Theoretic Structure Theorem: Every ring of sets 6= HQ 37/, where if and 31’ satisfy the conditions laid down in the preceding paragraph, is a Stone algebra, and every Stone. algebra can be so described. M: The first part of the theorem follows from Theorem 3.5 and 3.6. Let B denote an arbitrary Stone algebra. Then B = R ® T where R is the set of elements of B satisfying 'I‘Ir = r and T is the set of elements of B satisfying ‘I t = I. Since any distributive lattice is isomorphic with.a ring of sets [cf. Birkhoff, p. lhq] we knew that B is isomorphic with a ring dg’of sets. The Boolean sub-ring fl? and the Brouwerian sub-ring.j7 are the respective images, under the isomorphism, of R and T. A direct application of Theorems 3.7 through 3.11 oan.now .be made to complete the proof of this theorem. This set-theoretic representation furnishes a method of constructing Stone algebras. Lot 21 denote an algebra of sets, Tyfan arbitrary collection of elements of W together with their complements, and W the collection of elements of Z/Atogether with their pairwise sums and products. If R1 and R2 are in /6), it is clear that RfLJRZ and RlnR2 are also in if . Further, if R is in I? , then R' is in 68 . For, if R is in Vthen R' is in ywhich is contained in fl . If R is not in ‘f, then either R = ViLJVé or R = V1f\V2, where V1 and V2 are elements of V. In the first case R' = Vl'flVZ' and in the second case R' = Vi'LJV2'. Since V1' and V2' are" elements'of 7V”, it follows that in either case R' is in i9. Hence fl is a Boolean ring of sets. From among the members of ZZnot already in fl choose a sub-collection.27 in such.a way that: (a) If T is in 17’, the set complement T' is not in ;?H. (b) If T1 and T2 are in Jar: so is TlLJTa. (c) If T is in 57’, so are all sets of'?‘ contained in T. -1LO- (d) The collections frand l?have in common only the null set. It is seen from (b) and (c) that fie a ring of sets, and (a) implies that ,7 is not -a Boolean ring of sets. If T1 and T2 are in j, the set T1 35(- T2 exists in ’M since W is an algebra of sets. But it is clear that T1 ? TZC T1, so that '1‘1 72 T2 = T1): T2 exists in 57 and 5/" is a Brouwerian ring of sets. Intuitively, the Brouwerian ring 3/ of sets serves to "fill out" the Boolean "skeleton" fl? . The desired Stone algebra fl is now obtained by forming the direct sum 5 = £6 57/. .41- Section.h. Characterization of Certain Stone Algebras In.this section we characterize a wide sub-class of Stone algebras. These Stone algebras are shown to be factorable into a direct product of Brouwerian algebras of a rather special kind called T-algebras. Definition: An element a.of a lattice L is Join-irreducible if x + y = a implies x = a or -y = a. Definition: A Brouwerian algebra with.I is a T-algebra if I is join-irreducible. T-algebras may be constructed in the following manner. To any Brouwerian algebra L adjoin a new element J in such a way that J is preperly over every element of L. Let'i denote the resulting lattice. It is seen that the adjoining of J to L leaves unchanged all the original differences a - b of elements of L. If x is an element of L, then J - x = J since for no y‘fi J can the relationship x + y = J hold. (Recall that J is preperly over every element of L, and that x + y is an element of L). This shows that there exists inlf the difference of any two elements, i.e. that‘L is a Brouwerian algebra. One of the results proved in this section is that the direct product of T-algebras is a Stone algebra. Thus a large collection of Stone algebras can be constructed by taking an arbitrary collection of'arbitrary Brouwerian algebras, converting each Brouwerian algebra into a T—algebra by adjoining an element J, and forming the direct product of the resulting T-algebras. Important concepts used throughout the rest of this section are presented in the following definitions. Definition: Let the set C be the indexing set for a collection of Join-irreducible elements apb’ec. The collection a1} is a representation 2;: _I. if (LI-01) I =\/a»‘. 650 The representation is irredundant if Kffi implies ”‘3" O. Definition: A lattice L is complete if every subset of L has a greatest lower bound and a least upper bound. Definition: A lattice L‘is completely distributive if arbitrary sums distribute over arbitrary products, and dually. Remark: Let D be the indexing set for an arbitrary subset of L, and let a;', 66D, denote the Boolean complementof as. For our purposes the full power of the complete 'distributive law is not needed; instead, it suffices that (ll-.2) AU“ + ag') = V[5/\a8(1)] . (1) GD where @335”) denotes a product formed by choosing, for (1)] ’ each 8613, either as or aS', “d(1)V[56/})a6 denotes the union of all such products. ”The following example illustrates the notation; the complete distributive law we require is the generalization of the following law: .43.. (ll.3) (a + a')(b + b').(c + c') = abo 4- abc' a-‘ab'o + abi'c' + a'bc + a'bc' + a'b'c + a'b'c'. Definition: Let A and B denote two algebraic systems having the same operations. The 3% product A XB of A and B is the set whose elements are pairs (a,b),. aeA and bEB, and whose operations are performed component-wise: - uni) r[(a1,b1).(a2.b2)] = [martian f(b1,b2)]. The direct product of an arbitrary number of algebraic systems, all having the same operations, is defined similarly. A Lemma 1: The direct product of an arbitrary collection of Stone algebras is itself a Stone algebra. £39.32: Let A be the indexing set for a collection of Stone algebras S ,v «EA. Let (ll-0.5) S F’ILS“ denote the direct product of the Stone algebras 88. An element x of S has components x“, where ages“. Then the element ‘11: = I - x of S has components 'Ix,‘= I.) - x4, and 111563 has components Tlx“ = 1,. - 'qu, since the difference operation is performed componentwise. Since the product Operation is also performed componentwise, the element 1x11xes has components 'lea‘lxd. But each 3,. is a Stone algebra, hence 'lxd‘l'lx.‘ = 0... Thus the components of ‘lelx are all 0, and S is a Stone algebra. Lemma 2: Every T-algebra is a Stone algebra. .up. 33.9.92: . Let x 75 I. Then 11: + x.=‘I implies that 1:: = I, since I is joinpirreducible. Hence “Tlx = 0, and ”1anx = ID = 0 holds for every x‘# I. The proof is completed by noting that 'TITTI = OI = O. The principal result of this section is presented in.the next two theorems. Theorem {p.l: If B is a complete Stone algebra, and if I has a representation as an irredundant Join of Join-irreduc- ible elements, then B is isomorphic with a direct product of T-algebras. Theorem.h,2: If B is a complete and completely distributive Stone algebra, then I can.be represented as an irredundant join of Join-irreducible elements. Proof of Theorem.h.1: Let C be the indexing set for the set of Join-irreducible elements az;fec, making up the representation of I, so that (LI-o6) I a: %ao’o Let AU denote the set of elements xEB satisfying 3: 5 at; and let D denote the direct product of the sets AK‘ The proof consists of three parts. In the first part it is shown that AF is a T-algebra. A one-to-one correspondence is established between B and D in the second part of the proof, and in the third part this correspondence is shown to be an isomorphism. AF is clearly a sub-lattice of B. If u and v are in Ar, then the fact that u - v _<_ 11 means that u - v is also in AU’ so that AU’ is itself a Brouwerian algebra. -15.. The element a, (which plays the role of I in A1,) is Join-irreducible, hence A, is a T-algebra, by definition. Let d be an element of D having components dav, where dreAr The correspondent x in B of “d in D is defined as (ll-o7) X = deo The sum exists since each do’ is in B, and B is complete. Let 3 denote another element of D, having components 35, and assume that ((+08) var = V315: {EC 1. e. assume that d and 5. map into' the same element X of B. If aflflEC, is one of the elements making up the representation if I, then from (2.}.8) we may write that (1;..9) 9.er = eggs? Using the infinite distributive law, which holds in B since B is complete, the above eXpression becomes (law) Mend.) = $451,521,). The fact that the representation is irredundant implies that the elements ab. are pairwise disjoint. Since at _<_ aa— implies 9;de 5 sea, = O forfl #1), expression (1)..10) reduces to (ll-011) 939% == afi'dp. But dc, g afl, and 'dp 5. eff, hence dp :36, Since p was an arbitrary member of the indexing set 0, this shows that d = d, i.e. that the correspondence defined in (1).?) is a -ue- one-to-one mapping of D into B. We will complete the second part of the proof of Theorem l)..1 by showing that every element in B is the image of an element of D. Let y be an element of B. Then yap. is in AU, and the element dy’ whose components. are yaT, is in D. The image of dyis ((+012) Xééyaz) == yxecaa. = yI = '3': again using the infinite distributive law. That this one-to-one correSpondence is operation- preserving follows from the fact that if d and d are, elements of D. satisfying d 5 3, then the components d3 of d and 35 of d indivually satisfy dis- Hr Hence (L13) Md. 5. ye). and the correspondence is order-preserving. But all the operations in B are defined in terms of the order relation; hence the correspondence is an isomorphism and the proof of Theorem l)..1 is complete. Proof of Theorem L2: If B is a T-algebra the theorem .is trivial. If not, the set R of elements of B satisfying _\')r = r contains elements other than 0 and I. For, if B is not a T-algebra, then there exists elements x and y, both different from I, such that x + y = I. This implies that -I(x + y) = O and T-Hx + y) = I. If-lx = I, then (milk) I =T|_ deArdU) '. Let y = Vr (1)'. Since y + rd)” = I holds for every 0’s in A, we have (($018) I = @(y + 13(1)) = y + ard(i) = y + x. By definition of 1x, this means y 3“) x, 1.6. . (#49) 1x 5 MIR”) '. From (h..17) and (4.19), We have (h.20) 'Ix = Mrguw. It is clear that, for every 0( in A, xrdU)‘ == 0. Hence .43- (ll-.21) O = Mnd(i)' = XMI'JH)‘ := 3(1):, ‘This shows that x is in Q and hence in R by Lemma 1 to Theorem 3.3, page 29. ‘ Not every term of (1+.16) is 0, since the sum of the terms is I. After discarding from.(h.16) those terms which are 0, the remaining terms may be relabelled so that (l)..16) becomes (”'22) I = sway It will be shown next that D is the indexing set for the atoms of R, i.e. those elements 8‘6 of R such that .0 5 r é 8‘8 holds for no element 'r in R. After that we (will show that the representation (L22) is irredundant, and the proof will be completed by showing each element a 8 is Join-irreducible. . Suppose that an element r of R satisfied (1)423) Osrgas, Ofir, aflér for some 6 in D. By the manner in which a6 was obtained, we observe that the expression for 8‘8 contains the letter 1', with or without a prime. If r appears as one of the members of the expression for 38’ then r _>_ as, which violates ((4.33). On the other hand, if r' appears as one of the members of the eXpression for as, then ra8= O, which also contradicts (L23). We conclude that no element r of R can satisfy (h.23), i.e. that 9‘6 is an atom of R. We remark parenthetically that a 5 may not be an atom of B. 4+9- Again using the fact that ‘each element r,< of R, with or without a prime, appears as a member of the expression for as, we see that 61 75 82 implies 8‘51 and a52 are different, i.e. at least one of the elements is primed in one term and not in the other. It follows that (11.21).) aslas2 = O for 61 75 62, which shows that the representation (1922) is irredundant. Assume that there exists elements x, y of B, each different from O and from as, which satisfy (14425) 05.35%: Osygag, x+y=a8 for some 6 in D. Recalling that 11x _<_ x, that ‘I'Ix is in. R, and that a6 is an atom of R, we have “Fix = 0. Hence 'lx s”: I, and, similarly,‘|y = I. Hence (L26) ‘(as =‘|(x + y) =')‘|(‘|x')y) =‘I'|(II) =T|I But if 183‘ I, then 0 ='|'|a6 = as since as is in R. I. This is a contradiction, for in the construction of (11.32) only the terms of (L916) different from 0 were retained. Thus (4.25) is impossible, and a5 is join-irreducible. This completes the proof of Theorem Ll..2. Corollary 1: Every complete and completely distributive Stone algebra is isomorphic with a direct product of T-algebras. Proof: This is a direct consequence of Theorems h.l and h.2. Corollary 2: Any finite Stone algebra is isomorphic with -50- a direct product of T-algebras. 23932: Any finite Stone algebra is complete and- completely ‘ distributive. Noticing that the essence of the proof of Theorem Li.2 was the discovery of the atoms of R, we are led to Theorem li.3: A Stone algebra B is isomorphic with a direct product of T-algebras if every descending chain in R is finite. W: Since R is a Boolean algebra in which descending chains are finite, we know that R‘ itself is finite (cf. Birkhoff[ 3 ], p. 159). Hence the atoms of R can be determined; it can be shown as in Theorem l)..2 that I is an irredundant join of the atoms of R and that the atoms of’ R are Join-irreducible elements of B. The proof is completed by applying Theorem 192. One further extension of Theorem l)..2 is obtained by noticing that the use of the infinite distributive law in the proof was confined to elements of R. Theorem huh: If B is a Stone algebra in which the Boolean sub-algebra R is complete and completely distributive, then B is isomorphic with a direct product of T-algebras. m: Exactly as in Theorems 1+.l and L)..2. An example of a Stone algebra which is not factorable is the "measure algebra" H (see Birkhoff[ 3 ], p. 1814.). This algebra may be constructed as follows. Let M denote the set of Lebesgue measurable subsets of the unit interval. Divide M into equivalence classes by placing in the same -51- class any two subsets whose symmetric difference is a set of measure zero. The equivalence classes are ordered by ' set inclusion. It is known that the resulting algebra H is a complete Boolean algebra without atoms. Since M is a Boolean algebra, it follows that T is a Stone algebra. However, H cannot be factored into a direct product of T-algebras since it has no join-irreducible elements. It might be conjectured that all Stone algebras are direct products, the factors being either T-algebras or Boolean algebras without atoms.. That this is not the case is shown by the following example, due to L. M. Kelly. First consider a non-atomic Boolean algebra, and consider its representation as a Boolean ring W of sets. ”may be regarded as embedded in an algebra of sets which.of course contains points. Let T be one of these points, and let the set ,57 consist of T together with.the null set. It is easily verified that the conditions of the Set-Theoretic Structure Theorem.(p. 38) are satisfied, hence 6 = £69? is a Stonealgebra. Since 43 contains only one join-irreducible element, namely T, the only possible factorization of (t; of the conjectured type is fi= ”XV. In the direct product flxyj the four elements (R,T), (R,D), (RQT) and (R50) are distinct, Where R denotes some member of fl different from O and I. But the point T lies in either R or R', hence in the direct sum flC—B 7the four elements R + T, R + O, R' + T and R! + O are not distinct. Thus no one-to-one -52- correspondence can be set up between £67} and flxj , ‘ 6V i.e. the Stone algebra Web/l cannot be factored in the conjectured manner. 1. 9. 10. 11. 12. 13. 11),. -53- BIBLIOGRAPHY B. A. Bernstein, Operations with respect to which the elements of a Boolean algpbra form a group, Trans. Amer. Math. Soc., 26(192h), 171-175. , On the existence of fields in Boolean algebras, Trans. Amer. Math. 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