Introduction to Boolean Algebras (Instructor Solution Manual, Solutions)

755 Pages·2008·3.739 MB·English

by  Steven Givant, Paul Halmos

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
prosestylethatisusedinthetextbookitself. Theanswershavebeenchecked
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 multiplica-
tion 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 = (p ,p ,p ), q = (q ,q ,q ), r = (r ,r ,r ),
0 1 2 0 1 2 0 1 2
then
p+(q+r) = (p ,p ,p )+((q ,q ,q )+(r ,r ,r ))
0 1 2 0 1 2 0 1 2
= (p ,p ,p )+(q +r ,q +r ,q +r )
0 1 2 0 0 1 1 2 2
= (p +(q +r ),p +(q +r ),p +(q +r ))
0 0 0 1 1 1 2 2 2
= ((p +q )+r ,(p +q )+r ,(p +q )+r )
0 0 0 1 1 1 2 2 2
= (p +q ,p +q ,p +q )+(r ,r ,r )
0 0 1 1 2 2 0 1 2
= ((p ,p ,p )+(q ,q ,q ))+(r ,r ,r )
0 1 2 0 1 2 0 1 2
= 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 (coordi-
natewise) 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
for all p, q, and r in 2X. The right and left sides of this equation are 2-
valued 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
sumoftwofunctionsin2X, whilethethirdequalityusestheassociative
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
translationsofthecorrespondingoperationsinthering2. Forinstance,
¯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 ¯1
¯0 ¯0 ¯1 and ¯0 ¯0 ¯0 .
¯1 ¯1 ¯0 ¯1 ¯0 ¯1
Thefunctionfrom2to2X thatmaps0and1to¯0and¯1respectivelyisa
bijectionthatpreservestheoperationsofadditionandmultiplicationin
the sense that it translates the addition and multiplication tables for 2
into the addition and multiplication tables for 2X. For instance, the
entryintheadditiontableof2forthesum0+1is1. Thecorresponding
entryintheadditiontableof2X forthesum¯0+¯1isjustthetranslation
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 the inverse laws
p+−p = 0 and −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,
bytheinverselaw. Invokethecancellationlawtoconcludethatq = −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.Inaring, thecancellationlawsforadditionhold, andadditive
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 opera-
tion 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 (cid:2), 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
elementsofR,thensoarep⊕qandp(cid:2)q. Hereistherequiredcalculation
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
secondequalityusesthedistributive,associative,andcommutativelaws
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(cid:2)q
is similar but simpler: