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Introduction to Boolean Algebras Solutions Manual Steven Givant

Contents Preface

v

Boolean Rings

1

Boolean Algebras

15

Boolean Algebras versus Rings

48

Duality

59

Fields of Sets

62

Elementary Relations

75

Order

102

Inﬁnite Operations

115

Topology

136

Regular Open Sets

160

Subalgebras

171

Homomorphisms

189

Extensions of Homomorphisms

217

Atoms

229

Finite Boolean Algebras

240 ii

Contents

iii

Atomless Boolean Algebras

251

Congruences and Quotients

262

Ideals and Filters

273

Lattices of Ideals

295

Maximal Ideals

304

Homomorphism and Isomorphism Theorems

324

The Representation Theorem

331

Canonical Extensions

337

Complete Homomorphisms and Complete Ideals

347

Completions

364

Products of Algebras

373

Isomorphisms of Factors

416

Free Algebras

441

Boolean σ-algebras

455

The Countable Chain Condition

484

Measure Algebras

494

Boolean Spaces

506

Continuous Functions

533

Boolean Algebras and Boolean Spaces

564

Duality for Ideals

590

Duality for Homomorphisms

616

Duality for Subalgebras

627

iv

Introduction to Boolean Algebras

Duality for Completeness

641

Boolean σ-spaces

648

The Representation of σ-algebras

657

Boolean Measure Spaces

664

Incomplete Algebras

670

Duality for Products

678

Sums of Algebras

708

Isomorphisms of Countable Factors

741

Preface This manual contains the solutions to all of the more than 800 exercises in the textbook Introduction to Boolean Algebras, by Steven Givant and Paul Halmos. An eﬀort has been made to present these solutions in the same prose style that is used in the textbook itself. The answers have been checked several times in order to eliminate errors, but some mistakes may not have been caught. I would be grateful if readers would bring any that they ﬁnd to my attention.

Steven Givant Department of Mathematics and Computer Science Mills College Oakland, California 94613 [email protected]

v

Boolean Rings 1. Verify that 2 satisﬁes ring axioms (1)–(9). Solution. As examples, we verify the associative law (2) for multiplication and the distributive law (8) for multiplication over addition. The proofs use the tables for addition and multiplication in the ring 2, and proceed by cases. Consequently, the arguments have a somewhat brute force ﬂavor. The following identities, which are clear from the table for multiplication in 2, will be used repeatedly: p·0=0·p=0

and

p·1=1·p=p

for all p in 2. To establish axiom (2), it must be shown that p · (q · r) = (p · q) · r for all p, q, and r in 2. If p = 0, then p · (q · r) = 0 · (q · r) = 0 = 0 · r = (0 · q) · r = (p · q) · r. If q = 0, then p · (q · r) = p · (0 · r) = p · 0 = 0 = 0 · r = (p · 0) · r = (p · q) · r. If r = 0, then p · (q · r) = p · (q · 0) = p · 0 = 0 = (p · q) · 0 = (p · q) · r. If p = q = r = 1, then p · (q · r) = 1 · (1 · 1) = 1 · 1 = (1 · 1) · 1 = (p · q) · r. To establish axiom (8), it must be shown that p · (q + r) = p · q + p · r for all p, q, and r in 2. If p = 0, then 1

2

Introduction to Boolean Algebras p · (q + r) = 0 · (q + r) = 0 = 0 + 0 = 0 · q + 0 · r = p · q + p · r. If p = 1, then p · (q + r) = 1 · (q + r) = q + r = 1 · q + 1 · r = p · q + p · r. 2. Verify that 23 satisﬁes ring axioms (1)–(9). Solution. As an example, we verify the associative law (1) for addition. The veriﬁcation of the other axioms is entirely analogous. It must be shown that p + (q + r) = (p + q) + r for all p, q, and r in 23 . If p = (p0 , p1 , p2 ),

q = (q0 , q1 , q2 ),

r = (r0 , r1 , r2 ),

then p + (q + r) = (p0 , p1 , p2 ) + ((q0 , q1 , q2 ) + (r0 , r1 , r2 )) = (p0 , p1 , p2 ) + (q0 + r0 , q1 + r1 , q2 + r2 ) = (p0 + (q0 + r0 ), p1 + (q1 + r1 ), p2 + (q2 + r2 )) = ((p0 + q0 ) + r0 , (p1 + q1 ) + r1 , (p2 + q2 ) + r2 ) = (p0 + q0 , p1 + q1 , p2 + q2 ) + (r0 , r1 , r2 ) = ((p0 , p1 , p2 ) + (q0 , q1 , q2 )) + (r0 , r1 , r2 ) = p + (q + r). The ﬁrst and last equalities use the assumption about the form of p, q, and r; the second, third, ﬁfth, and sixth equalities use the (coordinatewise) deﬁnition of addition in 23 ; and the fourth equality uses the associative law for addition in 2. 3. Verify that 2X satisﬁes ring axioms (1)–(9) for any set X. What ring do you get when X is the empty set? Solution. As an example, we verify the associative law (1) for addition. The veriﬁcation of the other axioms is entirely analogous. It must be shown that p + (q + r) = (p + q) + r 2X .

for all p, q, and r in The right and left sides of this equation are 2valued functions on the set X, and the two functions are equal just in case they agree at each argument. Suppose x is in X. Then

1 Boolean Rings

3 (p + (q + r))(x) = p(x) + (q + r)(x) = p(x) + (q(x) + r(x)) = (p(x) + q(x)) + r(x) = (p + q)(x) + r(x) = ((p + q) + r)(x).

The ﬁrst, second, fourth, and ﬁfth equalities use the deﬁnition of the sum of two functions in 2X , while the third equality uses the associative law for addition in 2. When the set X is empty, there is just one 2-valued function on X, namely the empty function. In this case, 2X is the one-element ring. 4. Essentially, what ring is 2X when X is a set consisting of just one element? Can you make this statement precise? Solution. If X is a one-element set, then 2X is essentially the ring 2. In more detail, if X = {x}, then the set 2X has two elements, namely the 2-valued functions ¯ 0 and ¯ 1 on X determined by ¯ 0(x) = 0

and

¯1(x) = 1.

The operations of addition and multiplication in the ring 2X are just translations of the corresponding operations in the ring 2. For instance, ¯ 0 + ¯1 = ¯1, because (¯ 0+¯ 1)(x) = ¯ 0(x) + ¯1(x) = 0 + 1 = 1 = ¯1(x). Analogous computations lead to the operation tables + ¯ 0 ¯ 1

¯ 0 ¯ 0 ¯ 1

¯ 1 ¯ 1 ¯ 0

and

· ¯0 ¯1

¯0 ¯0 ¯0

¯1 ¯0 ¯1

.

The function from 2 to 2X that maps 0 and 1 to ¯0 and ¯1 respectively is a bijection that preserves the operations of addition and multiplication in the sense that it translates the addition and multiplication tables for 2 into the addition and multiplication tables for 2X . For instance, the entry in the addition table of 2 for the sum 0+1 is 1. The corresponding entry in the addition table of 2X for the sum ¯0+ ¯1 is just the translation of 1, namely ¯ 1.

4

Introduction to Boolean Algebras 5. A group is a non-empty set, together with a binary operation + (on the set), a unary operation −, and a distinguished element 0, such that the associative law (1), the identity laws p+0=p

and

0 + p = p,

and

−p+p=0

and the inverse laws p + −p = 0

are all valid. Show that in a group the cancellation laws hold: if p+q =p+r

or

q + p = r + p,

then q = r. Conclude that in a group, the inverse element is unique: if p + q = 0, then q = −p. Solution. Assume p + q = p + r. Add −p to both sides and use the associative law, the additive inverse law, and the additive identity law, to obtain −p + (p + q) = −p + (p + r), (−p + p) + q = (−p + p) + r, 0 + q = 0 + r, q = r. The second cancellation law is demonstrated in a completely analogous fashion. To establish the uniqueness of inverses, assume p + q = 0. Then p + q = p + −p, by the inverse law. Invoke the cancellation law to conclude that q = −p. 6. Prove that in an arbitrary ring, p·0=0·p=0

and

p · (−q) = (−p) · q = −(p · q)

for all elements p and q. Solution. In a ring, the cancellation laws for addition hold, and additive inverses are unique, by Exercise 5. To prove that p · 0 = 0, use the identity law for addition, and the distributive law, to obtain p · 0 = p · (0 + 0) = p · 0 + p · 0. Apply the identity law again:

1 Boolean Rings

5 p · 0 + 0 = p · 0 = p · 0 + p · 0.

Invoke the cancellation law to conclude that 0 = p · 0. Here is the argument that p · (−q) = −(p · q): 0 = p · 0 = p · (q + −q) = p · q + p · (−q). Additive inverses are unique, so p · (−q) must be the additive inverse of p · q. 7. Let A be the set of all idempotent elements in a commutative ring R with unit. Deﬁne the sum p ⊕ q of two elements p and q in A by p ⊕ q = p + q − 2pq, where the right-hand term is computed in R (and pq means p · q). The distinguished elements of A are the same as those of R, and operation of multiplication in A is just the restriction of the operation of multiplication in R to the elements of A. Show that A is a Boolean ring. Solution. The ﬁrst task is to demonstrate that A is closed under the deﬁned operations of addition ⊕ and multiplication  , and that A contains the distinguished elements of R. The distinguished elements 0 and 1 of R are obviously idempotent, since 0·0=0

and

1 · 1 = 1,

and therefore they both belong to A. To establish the closure of A under the operations, it must be shown that if p and q are idempotent elements of R, then so are p⊕q and pq. Here is the required calculation for p ⊕ q: (p ⊕ q)(p ⊕ q) = (p + q − 2pq)(p + q − 2pq) = pp + pq − 2ppq + pq + qq − 2pqq − 2ppq − 2pqq + 4ppqq = p + pq − 2pq + pq + q − 2pq − 2pq + 4pq = p + q − 2pq = p ⊕ q. The ﬁrst and last equalities use the deﬁnition of addition in A; the second equality uses the distributive, associative, and commutative laws for the ring R; the third equality uses the assumption that p and q are idempotent elements in R; and the fourth equality uses the associative, commutative, and additive inverse laws for R. The calculation for p  q is similar but simpler:

6

Introduction to Boolean Algebras (p  q)(p  q) = (pq)(pq) = ppqq = pq = p  q. The next step is to verify that the axioms for Boolean rings are valid in A. As examples, we verify the associative law (1), the identity law (5), the distributive law (9), and the special law (10). Let p, q, and r be elements in A — that is, idempotent elements of the ring R. A simple computation using the deﬁnition of the operation ⊕ , and various laws that hold in the ring R, shows that p ⊕ (q ⊕ r) = p ⊕ (q + r − 2qr) = p + (q + r − 2qr) − 2p(q + r − 2qr) = p + q + r − 2qr − 2pq − 2pr + 4pqr = p + q + r − 2pq − 2pr − 2qr + 4pqr. A similar computation shows that (p ⊕ q) ⊕ r = p + q + r − 2pq − 2pr − 2qr + 4pqr. The two computations together yield p ⊕ (q ⊕ r) = (p ⊕ q) ⊕ r. The veriﬁcation of the identity law uses, in addition, Exercise 6: p ⊕ 0 = p + 0 − 2 · p · 0 = p + 0 + 0 = p. The veriﬁcation of the distributive law is analogous in spirit to the veriﬁcation of the associative law: p  (q ⊕ r) = p  (q + r − 2qr) = p(q + r − 2qr) = pq + pr − 2pqr, by the deﬁnitions of ⊕ and  , and by the distributive law for the ring R. Similarly, (p  q) ⊕ (p  r) = (pq) ⊕ (pr) = pq + pr − 2pqpr = pq + pr − 2ppqr = pq + pr − 2pqr (the identity pp = p is used in the ﬁnal step), so that p  (q ⊕ r) = (p  q) ⊕ (p  r). Here is the veriﬁcation of (10): p ⊕ p = p + p − 2pp = p + p − 2p = 2p − 2p = 0.

1 Boolean Rings

7

8. A Boolean group is a group in which every element has order two (in other words, the law (10) is valid). Show that every Boolean group is commutative (that is, the commutative law (3) is valid). Solution. Let p and q be elements of a Boolean group. Every element is its own inverse, so p, q, and p + q are their own inverses. In particular, 0 = (p + q) + (p + q). Add p to the left side of both equations, and q to the right side, to obtain p + 0 + q = p + p + q + p + q + q = 0 + q + p + 0. Use the identity law to conclude that p + q = q + p. 9. A zero-divisor in a ring is a non-zero element p such that p · q = 0 for some non-zero element q. Prove that a Boolean ring (with or without a unit) with more than two elements has zero-divisors. Solution. Suppose A is a Boolean ring (with or without a unit) with more than two elements. There must then be distinct non-zero elements p and q in A. Notice that the element p + q is non-zero, since p = q. If p · q = 0, then both p and q are zero-divisors. If p · q = 0, then p · q and p + q are zero-divisors, since p · q · (p + q) = p · q · p + p · q · q = p · p · q + p · q · q = p · q + p · q = 0. 10. Prove that every Boolean ring without a unit can be extended to a Boolean ring with a unit. To what extent is this extension procedure unique? Solution. To motivate the construction, let A be a Boolean ring (with or without a unit), and consider a Boolean ring B that extends A in the sense that the operations of addition and multiplication in A are the restrictions of the corresponding operations of B, and both rings have the same zero. (The technical way of describing this situation is to say that A is a subring of B.) Assume that B has a unit 1 that does not belong to A. If p and q are distinct elements of A, then 1 + p and 1 + q are distinct elements of B, by the cancellation law, and neither one of them is in A. Indeed, if r = 1 + p were in A, then p + r would be in A, because of the closure of A under addition; it would then follow that 1 is in A, since

8

Introduction to Boolean Algebras p + r = p + 1 + p = p + p + 1 = 0 + 1 = 1, and this would contradict the assumption that 1 is not in A. Easy computations show how to add and multiply elements of the form 1 + p (for p in A) with other elements of this form and with elements of A: (1 + p) + (1 + q) = 1 + 1 + p + q = 0 + p + q = p + q, (1 + p) + q = 1 + (p + q) = p + (1 + q), (1 + p) · (1 + q) = 1 + p + q + p · q, (1 + p) · q = q + p · q

and

p · (1 + q) = p + p · q.

We turn now to the construction of the desired ring extension. As before, A is an arbitrary Boolean ring with addition and multiplication operations + and · , and with zero element 0. For each element p in A, introduce a new element p¯ not in A, and chosen so that p¯ and q¯ are distinct when p and q are distinct elements of A. (The element p¯ is the analogue of the element 1 + p in the preceding paragraph.) Take B to be the set of elements in A together with the set of new elements: B = A ∪ {¯ p : p ∈ A}. Deﬁne binary operations ⊕ and  on B by p ⊕ q = p¯ ⊕ q¯ = p + q

and p¯ ⊕ q = p ⊕ q¯ = p + q

and p  q = p · q,

p¯  q¯ = p + q + p · q,

p¯  q = q + p · q,

p  q¯ = p + p · q,

for any elements p and q in A. (These deﬁnitions are motivated by the computations carried out at the end of the preceding paragraph.) It is not diﬃcult to show that B, under the operations ⊕ and  , with zero 0 and unit ¯ 0, is a Boolean ring with unit. Notice that the restrictions of ⊕ and  to A are just the addition and multiplication operations of A, so B is an extension of the ring A. A slightly diﬀerent way of proceeding is to let B be the Cartesian product of A and the two-element Boolean ring 2. In other words, let B be the set of ordered pairs (p, q) with p in A and q in 2. Addition in B is deﬁned coordinate-wise: (p1 , q1 ) ⊕ (p2 , q2 ) = (p1 + p2 , q1 + q2 ).

1 Boolean Rings

9

(The operations on the right side are performed in the rings A and 2 respectively.) Multiplication in B is deﬁned by the equation (p1 , q1 ) ⊗ (p2 , q2 ) = (p1 · p2 + q1 · p2 + q2 · p1 , q1 · q2 ), where the products p1 ·p2 and q1 ·q2 are formed in A and in 2 respectively, and where the product q1 · p2 is interpreted as p2 when q1 = 1, and as 0 when q1 = 0 (and analogously for q2 · p1 ). The veriﬁcation that B is a Boolean ring with unit (0, 1) under these operations is straightforward, but tedious. To prove that B is essentially an extension of A, identify each element p in A with the pair (p, 0) in B. (The element p¯ in the preceding paragraph corresponds to the pair (p, 1) in B.) The discussion in the ﬁrst two paragraphs shows that if A is a Boolean subring of an arbitrary Boolean ring C with unit 1, and if that unit is not in A, then the mapping that takes each element p in A to itself, and each element p¯ to 1 + p is an embedding of the extension B (constructed in the preceding paragraph) into C. This implies that B is, up to isomorphic copies, the minimal proper extension of A that has a unit. A Boolean ring A without a unit may have many extensions that are Boolean rings with units, but there is essentially just one minimal extension. 11. Does every ﬁnite Boolean ring have a unit? Solution. The positive solution to this exercise uses the notion of an (elementary) symmetric function in a ring. The symmetric functions of the n arguments v1 , v2 , . . . , vn are s1 = v1 + v2 + · · · + vn = s2 =



n 

vi ,

i=1

vi · vj ,

i