Math. Ann. 201,283--300 (1973) O by Springer-Verlag 1973

Some Classes of Projective Boolean Algebras* S. Koppelberg

Denote by P the class of projective Boolean algebras and by K1 the class of those Boolean algebras which are isomorphic to a product of a finite set of algebras each of which is a coproduct of countable algebras. It is well-known that K 1 = P; answering a question of Halmos, we showed in [3] that Kl is a proper subclass of P. This result is improved in this paper in several ways. We define a notion of weak product of Boolean algebras and show that P is closed with respect to countable weak pro- ducts. Let K (resp. L) be the smallest class of Boolean algebras con- taining every countable algebra and being closed with respect to co- products and finite products (resp. coproducts, finite products and countable weak products). We shall show that K~ is a proper subclass of K, K is a proper subclass of L and L is a proper subclass of P. The algebra proving L :k P is the same as constructed for the counterexample in [3]; the algebras working for K1 4: K and K ~: L are much simpler examples proving K 1 # P.

After some preliminaries in Chapter 1, we show K 1 :~ K in Chapter 2. Weak products are introduced in Chapter 3. In Chapter 4, we prove that for every A e K there is a B ~ K1 such that A and B have isomorphic completions; this yields K + L . Finally, L~: P is derived in Chapter 5.

1. In this chapter, we shall list some definitions and propositions. The reader should consult [1] for 1.5, [2] for 1.3, 1.4, 1.6, [3] for 1.12, [4] for 1.5, 1.6, 1.10, 1.11, and [5] for t.7, 1.9, 1.10. Propositions which are not included in one of these references may be easily checked by the reader.

1.1. 09 and N o both denote the set of natural numbers. ,~ is the car- dinality of X. If {mille I} is a set of cardinals, supm, denotes its su-

i ~ l

premum and ~ m i its cardinal sum. If rn is an infinite cardinal, A a set, i ~ l

aeA and x~A', i.e. x:m--',A, then (a,x) denotes the element yeA" satisfying fa if i = 0

y(i)=lx(i-1) if l < i < N o (x(i) if N o < i < m .

* The research on these problems was sponsored by the Deutsche Forschungs- gemeinschaft.

284 S. Koppelberg:

1.2. If A, B are Boolean algebras and X, Y topological spaces, A _----- B and X - Y mean that A is isomorphic to B and X is homeomorphic to Y. A Boolean algebra has at least two elements. 2 is the two-element Boolean algebra.

1.3. We write ~ Xi for the sum of a family (X~[ i ~ I ) of topological

spaces and X o + ... + Xn i f / = {0, ..., n} is finite. We shall always suppose that Xic~X~ = fl if i,j E I s.t. i # j , and U Xi = ~. Xi. If X is a topological

i ~ l i ¢ I

space and m is a cardinal, m. X is the sum of m copies of X.

1.4. Let X be a topological space. If X is a Tychonoff space, fiX denotes the Cech-Stone compactification of X; if X is a locally compact Hausdorff space, ctX denotes the Alexandroff (one point) compactifica- tion of X.

1.5. A Boolean space is a non-void compact Hausdorff space which is 0-dimensional (i.e. the clopen subsets of X form a base of X). For every Boolean algebra A the dual (Boolean) space of A is denoted by S(A); for every Boolean space X the dual (Boolean) algebra is denoted by A(X); we assume that A(X) is the field of clopen subsets of X. If A, (X~) is a Boolean algebra (Boolean space) for every i ~ I, 1-] A, ( I ] X~]

f ¢ l ~ i e I ]

is the product algebra (product space) and L.~A~ the coproduct of the i ¢ I

family <Aili e / ) . We write .4 o ×. . . × . . . A , (Aou • uA, ) fo r a finite product It is known that S([-]A~)~-fl(~S[A,)] and (coproduct). S(UAi)

\ 7~-x l " -- " ~i¢! I \i~I I

= 1-I s ( a , ) .

1.6. If X is a topological space, w(X) (the weight of X) is the least cardinal m s.t. there exists a base of X of cardinality m. If x ~ X, w(x) (the weight of X at the point x) is the least cardinal ra s.t. there exists a base for the neighbourhood system of x of cardinality m. Clearly, w(x) < w(X). If M ~ X, w(M) _~ w(X). If x ~ M, w(x) in M is not greater than w(x) in X; i f M is an open subset of X, w(x) in M equals w(x) in X. If X is a Boolean space, w(X)= A'(~). - Suppose that, for every i t I, X i is a space whose topology is not indiscrete and that x = (xi{ i~ I ) ~ I~ Xi. Then, ~ ~ 1

w(,~Xl)=max[~,supw(X,)], ,,, /

and w(x) = max [~, sup w(x,) 1,

\ l~x /

provided that max ( t supw(X,)) resp. max {~, sup w(x,)t is infinite. \ te l /

Projective Boolean Algebras 285

1.7. The set D = {0, 1} is always given the discrete topology. Let m be an infinite cardinal number. Then D m (the Cantor space of weight m) is given the product topology. If n e o9, n. D m ~ D ~. N o • D" is not homeo- morphic to D ~, but D m contains a dense open subset homeomorphic to N O • D ~. If {rnll i e I} is a set of cardinals and m = ~ m~, H Din' ~ Din. if

i e l iE l

n~cO, N o < m 0 < ... < m, and X = DmO + ... + Dm~, then X × X ~_ X. If X is a Boolean space s.t. w ( X ) < N O < m, X ×/7" ~-/Y". If A is a countable

--~ A ,-~o~ atomless Boolean algebra, A = t~ j.

1.8. Let A be a Boolean algebra (X a topological space). A (resp. X) satisfies the countable chain condition (c.c.c.) if every set of non-zero (non-void) pairwise disjoint elements of A (open subsets of X) is countable.

1.9. Let A be a Boolean algebra and a ~ A, a =~ 0. A I a = {x e A Ix < a} is (with respect to the partial ordering induced by A) a Boolean algebra which is called the relative algebra of A with respect to a. A is said to be homogeneous if A [ a -~ A for every a ~ A s.t. a :~ 0. If A is a complete algebra, a, b c A , O < a ~ b and A ~ A t a, then A ~- A t b.

1.10. Let A, B be Boolean algebras. A is a dense subalgebra of B if A is a subalgebra of B and, for every b e B s.t. 0 < b, there is a e A s.t. 0 < a < b. B is a completion of A if B is complete and there is a mono- morphism e : A ~ B s.t. e(A) is a dense subalgebra of B. If B, C are com- pletions of A, B ~ C. - Let X be a topological space. Then R ( X ) = { M ~_ X I M is a regular open subset of X} is a complete Boolean algebra with respect to set theoretical inclusion. IfA is a Boolean algebra, R(S(A)) is a completion of A. The completion of A (determined up to isomorphism) is denoted by A*.

1.11. A Boolean algebra P is said to be projective if it satisfies the following condition: if A and B are Boolean algebras, p : B - ~ A is an epimorphism and f : P ~ A is a homomorphism, then there is a homo- morphism i : P ~ B s.t. f = p o i . The following facts are well-known:

(a) P is projective iff it is a retract of a free algebra F (i.e. there exist homomorphisms i : P ~ F , p : F ~ P s.t. p o i = ide).

(b) Every countable Boolean algebra is projective. (c) If, for every i ~/ , At is projective, so is I I Ai.

i ~ l

(d) If n e co and, for 0 < i < n, Ai is projective, so is I-I Ai. O~i~n

(e) A projective algebra satisfies c.c.c, and does not include anyinfmite complete subalgebra.

A Boolean space X is called injective if A ( X ) is projective. Thus X is injective iff it is a retract of a Cantor space. The class of injective spaces contains every space with a countable base and is dosed with respect to products and finite sums. 19 Math, Ann. 201

286 S. K o p p e l b e r g :

1.12. Let K 1 be defined as above and let A be a Boolean algebra. Then, A ~K1 iff there exist n~o9 and Boolean algebras A 0 . . . . . A, s.t. A o is countable or free, AI, ..., A, are uncountable and free, Ao . . . . , A, are pairwise non-isomorphic and A ~ A o × ... x A,. If A is atomless, A e K~ iff A is isomorphic to a product of a finite set of pairwise non- isomorphic infinite free algebras.

2. 2.1. Definition. For every ordinal ~, a class K, of Boolean algebras is defined by: K o is the class of all countable Boolean algebras. Kat+~ is the class of those algebras which are isomorphic to a product of a finite number of algebras each of which is a coproduct of elements of Kat. If 2 is a limit number, K a = U K~. Finally, K = U {Kat Is an ordinal}.

at<),

It is clear that, for ~ < fl, Kat ~_ Ko ~_ K. By 1.11 (b)-(d), K _~ P. Evi- dently, K is the smallest class M of Boolean algebras s.t. K o =c M and M is closed with respect to coproducts, finite products and isomorphic copies.

2.2. Theorem. K 1 is a proper subclass o f K 2.

Proof. Let A be the two-element Boolean algebra, B the free Boolean algebra on N~ free generators. A x B e KI, by 1.12. Define C = [_J C.

rtl E al

where C. = A × B for every n e ~o. Clearly, C e K 2. We shall show that C ~ K r Consider Z = S(C). We may suppose that Z = (X + y)~o, where X = S(A) = {Xo} and Y = S(B) = D ~'. C is atomless, since Z has no iso- lated points. Suppose that C e K ~ . Using 1.12 and ~ = N 1 , we see that either C ~- A(D ~') or C _~ A(D ~°) x A(D~'), i.e. Z ~ D ~' or Z ~ D ~° + D ~'. But D ~' contains no point of weight N o and D~°+ D ~' contains 2 ~° points of weight N O (namely the points belonging to D~°), whereas Z has exactly one point of weight No (namely the point z o e Z s.t. zo(n)= x o for every n e ~o).

2.3. Corollary. KI is a proper subclass o f K.

3. Suppose that X~ is a Boolean space for every i e L Then ~ X i is i e l

a Tychonoff space and thus has a compactification, say X. Assuming that X~ is injective for every i e L we may ask whether X is injective, provided it is 0-dimensional.

If 1 is finite, ~ Xi is compact and X = ~ Xt, so X is injective. i E l i ~ l

Suppose I is uncountable. Since ~ X~ is locally compact, it may be

assumed to be an open subset of X. Thus {X~li e I} is an uncountable set of non-void open and pairwise disjoint subsets of X; X does not satisfy c.c.c, and hence cannot be injective.

Projective Boolean Algebras 287

Even if T=No, X need not be injective. Assume that X = f l ( ~ X i ) . \ 1 ~ 1 f

Then X is 0-dimensional and A(X)_~ 1--1 A(Xi). But H A(Xi) includes the i ~ l i e l

complete and infinite algebra I-I Bi (where BI = 2 for every ieI) as a sub- i e I

algebra and therefore cannot be projective. - One might thus expect that the Cech-Stone compactification is "too large" (in the usual partial

\

ordering of compactifications of ~ X/] for our purpose and that a smaller i e I I

compactification may work. Indeed, since ~ X i is locally compact, it

,E, ( Z x , ) has a smallest compactification, namely ~ ( Z Xi). Suppose that \ i E l / \ i E l /

= U {z} where z ¢ U X,. Y X, tis a compact Hausdorff space and i E l i a l \ i E l !

has a hase consisting of clopen sets. A subset M of ~ (~lXi) is clopen iff

(a) z ¢ M and there is an Io c= I s.t. I o is finite and M is a clopen subset of U Xi or (b) z ~ M and cz(i~IXi)\M satisfies (a).

iElo

The fact that ~ X i has a 0-dimensional one point compactification iE1

and the description of the clopen subsets of • ( ~ Xi] does, of course, not \ i E l /

depend on the assumption T = No.

3.1. Theorem. Suppose that, for every n ~ o2, X, is an injective Boolean space. Then ~ ( ~ X~) is injective.

Proof. Suppose X = ~ ( ~,X~ 1 = U X~w{z}. Since X is a Boolean \

space, we may assume that it is a closed subset of some Cantor space I5. Define, by induction, a sequence (Y~I n e co) of clopen subsets of Y s.t. X~= Yj~X and Y , n [ l ) Yk~ = 0 for every neco and put Y = Y \ U Y~.

\k~'. ] CO n~oJ

Since X, is injective and a closed subset of Y,, we may choose a con- tinuous function f , : Y ~ X , s.t. f ,(x)=x for every xeX~. Define f : Y ~ X by

f(y)={f. fy) if y e Y . if yeY,,,.

f is a well-defined function from Y to X s.t. f(x)= x for every x a X. To prove that f is continuous we show (since X is a Boolean space) that f - l (M) is clopen if M is a clopcn subset of X. We may assume that z ¢ M. Then there is an n e co s.t. M = ~ M i where M~ is a clopen

O ~ i ~ n

subset of Xt for 0 < i < n. Thus, f - l (M)= ~ fi-l(Mt), which is a 1 9 " O<=i~--n

288 S, Koppe lbe rg :

clopen subset of U Yt and hence of Y. X is therefore a retract of O<i<n

a Cantor space and injective. Given a family (A~lie I) of Boolean algebras, we define its weak

w

product 1~ At by i e l~v

I-I Ai = la ~ 1-I Ai I there is an I o ___ I s.t. 7 o < N o and i e l t i ~ l

(a i = 0 for every i e f "d o or

a~ = I for every i e/'X/o) } .

It is clear that I~A~ is a subalgebra of 1~.4~. Assuming A~=A(X~) for i ~ l i e l

every ie l , and X = c t ( ~ Xt) = UX~){z}, we have I-IAi_~A(X). An \ i e l I i e l i e l w

isomorphism q~ is given as follows: ifa ~ H A i and a~ = 0 for every i E 1",/o t e l

(where I o _~ I, 7 o < No) and a i = M i E A ( X t ) for i t I o, define ¢p(a)= 0 M;; i~lo

ifa~ = 1 for almost every i e I, define ~0(a) = X \ ~ o ( - a), where - a denotes the complement of a. - Theorem 3.1 may be slightly improved to:

3.2. Theorem. The weak product of a family (Aili e I) of Boolean algebras is projective iff A i is projective for every i ~ I and I is countable

w

Proof. Assume that 1-I Ai is projective, i.e. that o~(~Xil is injective, i e l \ i e l l

and that i~ I. We may assume that X i is a clopen subset of ~ ( ~ Xil. \ i e l ]

Let Xo be a fixed point of Xi. A continuous function f : a ( ~ . X i ) ~ Xi is /

defined by f ( x ) = { x if xEXt

xo if x e X i .

Thus Xi is a retract of the injective space a (~_Xit and hence a retract \ t e / [

of a Cantor space. - The necessarity of the assumption 7 < No and the sufficiency of both assumptions is clear by the remark at the beginning of this chapter and Theorem 3.1.

Considering the previous problem, it may be seen that a ( _ ~ X~) /

need not be the only injective 0-dimensional compactification of ~ Xn n E ~

(provided that X, is injective for every n e a~). It follows easily from the fact that fl (~,~ X~)is the largest and~ (,~,o Xn)the smallest 0-dimensional

compactification of ~ X~ that the 0-dimensional compactifications of n 6 ¢ ~

Projective Boolean Algebras 289

X/l correspond to those subalgebras of I-I An (where A/l = A(Xn)) / l E O ) / l ~ ( D

W

which include H An. If An is countable for every n ~ co, H An is count-

able and there are countable (and hence projective) subalgebras of

I-I A~ including I~I A, as a proper subalgebra.

3.3. Definition. L is the smallest class of Boolean algebras including K 0 which is closed with respect to finite products, countable weak products and arbitrary coproducts.

It is clear that K _c L ~ P.

4. Let K* = {A* I A e K} and K* = {A* I A E K 1 }. It is not hard to see that K + L follows from K * = K*. It suffices, by Lemma 4.3, to prove K~ = K*, and this is dear if we are able to show that {Aili ~ I} ~ K1 i m p l i e s ( ~ A i ) * ~ K * . I n t h e p r o o f o f t h i s p r o p o s i t i o n , w e h a v e t o c o n -

sider several cases which are handled in Lemmas 4.7, 4.8. The main tool for these lemmas is Lemma 4.5, a sort of isomorphism theorem.

4.1. Definition. Let X, Y be topological spaces. X and Y are said to be equivalent (X ~ Y) if R(X) ~ R(Y).

4.2. Lemma. Let, for every i ~ L Ai be a Boolean algebra. Then, i = and i = •

i e l

Proof. We may assume that A i is a dense subalgebra of A* and A i (A*) is a subalgebra °fllif, A'(iUI A*). It is then clear that ,~,I~ Ai i sa dense

subalgebra of the complete algebra I-I A*. Similarly, L_J Ai is a dense i ~ l i ~ I

subalgebra of [_~ A* and hence a dense subalgebra of the complete i ~ I

algebra (Recall that r l A, is incomplete unless > \ i e I ~ / i e l

finite, { ie l IA~>No} is 0 or {io} for some ioeI and, in the latter case, A~o is complete.)

4.3. Lemma. I f X is a topological space and Y a dense open subset of X, then X ,,, Y (where Y is given the topology induced by X).

Proof. For M e R(X), put o(M) = Mc~ Y. For N e R(Y), put ~p(N) = N hi (where h and i are the closure operator and the interior operator with respect to X). It is easily checked that q~ : R(X)--,R(Y), v2 : R ( Y ) ~ R ( X ) , q~ o ¥ = idRtrr Thus q~ is surjective. Since Y is a dense subset of X, q~ is injective. So, ~ and ~p are both bijective and monotonic and hence isomorphisms.

290 S. Koppelberg:

4.4. Lemma. Suppose that, for every i ~ I, X~ and Yt are Boolean spaces s.t. X, ,,- Y~. Then, l-I x , ~ I-I Y~ and ~ X, ,.~ ~, gi.

ie l ie l i¢~I iel

X *"~ A(X~)) *~- 4.2) (,.1) (,.

4.5. Lemma. Let X and Y be Boolean spaces without isolated points satisfying c.c.c. Suppose that (Bnln e o9} is a sequence of Boolean spaces s.t. for every n ~ o9,

(a) B, has a dense open subset homeomorphic to N o • Bn, (b) B, has a clopen subset homeomorphic to Bn+l and (c) for every non-void open subset U of X or Y, there is an n ~ co

s.t. U has a clopen subset homeomorphic to B,. Then X ,~ Y.

Proof. Let us first remark that, by (b) and (c), if U and V are non-void open subsets of X and Y and n o e 09, there is an nl e o9 s.t. no < nl and U, V both have clopen subsets homeomorphic to B,,. We shall construct by induction three sequences (kil l e co}, (Mil i e co} and (Nil i e 09) s.t.

(~) Mi(Ni) is a clopen subset of X(Y) , k i t co, and Mi ~- BR,~-Ni. (fl) i f i:~ j, M i n M j = N i c ~ N j = O . (~) ki < k~+l. (6) U M~ =k X and U Ni * Y for every n e o9.

i'~n i~_n For i = 0 , let X o (I1o) be a clopen subset of X (Y) s.t. 0 4 : X o ~ : X

( 0 . Yo4: Y). Choose ko~oJ and a clopen subset M o (No) of X o (Yo) s.t. Mo B o- No.

Suppose that, for every i< n, k i ~ co and Mi, Ni have been defined according (a)-(6). X \ U M, ( Y \ U N,] is a non-void clopen subset of

i~n k i_n /

X ( Y ) containing more than one point, since X ( Y ) has no isolated point. Let X,+,(Y,+,)bca clopen subset of X \ U X, , s,IY\, u Yit/s.t. ~* X,., 4:x\Uxilo4:Y,,+x. YNUYt] and choose k,,+,eog, M,,+,~_X,,+, (N.+I-~ Y.+l) s . t .k. < k.+ 1, M.+I(N,,+I) is a clopen subset of X,,+t(Y,,+I) and M . + I - B ~ + ~ - N . + I .

Suppose that k., M. and N. have been constructed for every n 6 co according (~)-(fi). By Zom's lemma, extend {M.I n ~ co} ({N.I n e co}) to a set ~ ( ~ ) of clopen subsets of X(Y) which is maximal with respect to the property: its memtmrs are pairwisc disjoint and each of them is homeomorphic to some Bt, (i ~ co). Since {kil i ~ co} is a cofinal subset of co, (c) implies that U ~ ( U 91) is a dense open subset of X(Y). X (Y ) satis-

Projective Boolean Algebras 291

lying c.c.c., ~lR(gl) is countable. By (a), choose for each M ~ ~ (N e 9/) a dense open subset CM o f M (D N of N) homeomorphic to N o • M (N O • N). Evidently, U C u ( U DNI is a dense open subset of X(Y) which is

M e ~ \ N ~ /

homeomorphic to ~ N o , Bk,. By 4.3, we have X -.~ U Cu ~ U Ds ~ Y. i e ~ Me~i~ N e ~

4.6. Lemma. Let X be a topological space s.t. every non-void re#ular open subset o f X has an open subset V satisfying V,,~ X. Then R ( X ) is homogeneous.

Proof. Suppose that U e R(X), U =4= 0. Let V be an open subset of U s.t. V ~ X. Define W = V h~. Then W e R(X) , W ~_ U and V is a dense open subset of W, hence W ~ V ~ X. By the isomorphism theorem of 1.9, U ~ X. This proves the lemma.

4.7. Lemma. I f m is an uncountable cardinal and <X.ln~e~> a se- quence o f Boolean spaces s.t. A(Xn) e K 1 and w(Xn) = m for every n ~ 09, then X = 1-I x . . .~ Dm.

n~o~

Proof. For every nero, by 1.12, X~ is, up to homeomorphism, de- composable in the following way:

X = X.,o + Dm~,. + ... + Dmk.,.

where k .eeg , ml , , . . . . . ink.,, are uncountable cardinals s.t. w(X.,o) < ml, . < ... < m k . , . = m and X..o is 0 or a Boolean space satisfying w(X..o) < No. It is clear that

Dm × X ~_ Dm x X,, o + ... + Dm x D ra

~ D m ,

If M is a non-void clopen subset of X~, D" x M is a non-void clopen subset of D" x X~ and thus D m x M ~ D m.

Now, let U be a non-void open subset of X. There is an n ~ e~ and, for every i < n, a non-void clopen subset M i of X i s.t. 11 Mix 1~ X~ ~_ U.

i < n n~_t

Define V = l--I M i x D "k-'- x l-I xi . Then V is a clopen subset of U and i < m n < i

V = H M, x D ' x I-I X , i < n n < i

_~ I-I M, x (Dm)"x(Om)"°x I - I x , i < n n < i

- I-I(Mi×DM)x l-I (x, x O") l < n n < i

~_ ( D')t%

"~ D m .

292 S. Koppelberg:

Define Y = I~ D'~"'* ~ Dm and B, = D m for every n ~ co. Now, Lemma 4.5

applies and yields X ~ Y, i.e. X ~ D =.

4.8. l emma . I f <X, ln~co> is a sequence of Boolean spaces s.t. A(X~) ~ K 1 for every n ~ 09 and b¢ o < w(Xo) < w(X 0 < w(X2) < ..., then X = H x...~ D= where m = ~ w(X~).

nE¢o n60~

Proof. X , is decomposable in the following way:

x . = X . , o + X.,l + ... + x.,k.

where k, e co. w(X,.o) < w(X,,. 1) < "'" < w(X,.k.) = w(X,); X,a . . . . . X,.k. are Cantor spaces and X,.o is a Cantor space or a Boolean space satisfying w(X,.o)<No. Define m,=w(X,) . By our assumptions, m , < m , + l for every n e o9.

Case (i). Suppose that X.. o = D~° for every n e o2. Put

B . = X o . o X ' . . x X . _ l , o x X . , k x I-Ix~. n<i

Then, B,, ~ D ' " x l ] Xi n<i

~(o ' - ) "+1 x I l x , n<i

- - (X o x D '" ) x -.. x (X, x O"") x l - I x / n < i

(D"")" + 1 x 1--[ Xi

~-o'" × I-I X~.

We first want to show, by 4.5, that X ,,~ B,o for every n o ~ to. Since D"- has a dense open subset homeomorphic to No" D"", B, ~ D "~ × I~ Xi has

a dense open subset homeomorphic to N0.B, , and the sequence (B.[ n ~ w> satisfies condition (a) of 4.5. Concerning condition (b), it is clear that

C n ~ " ~ X o , o X " ' " x X n - l . o × X . , k n × X n + l , k n + t X H X~ n+l <t

is a clopen subset of B, and

c.--_o' .*,× l-I x , n + l < i

"~B n + l "

We now check condition (c) for X (the proof for Y = B,o is similar). Let U be a non-void open subset of X. There is an n e o2 and non-void clopen subsets Mo . . . . . M,-1 of Xo . . . . . X , - i s.t. [ I M ~ x [ I X ~ C U . Then

t<n n~_i

C = l~ M~ × X,,k, × I ] Xi is a clopen subset of U and C - B,. i < n n < f

Projective Boolean Algebras 293

By 4.5, we have X .~ B. o for every n o e w and thus B. ~- B,. for m,n ~ co. By condit ion (6) of 4.5, 4.6 applies and R(X) is homogeneous. Fo r every n e o2, define

A.=Xo ,o× ... xX._x,o × 1-Ix, ~ O " ° x H X ,

n < i

~ H x , . n < i

AI* is a clopen subset of X and hence A. ~ X and X ~ I-[ Xi. Since X." + t _~ X., for every n e co, we have I* -< i

x = H x .~ I-I x; +' ~E¢,O I*EtO

~ H Ux , nEOJ n ~ i

~ X ~° by X ~ H X i and 4.4 I*<=i ~ I I x . ~o

1~ D'~ by 4.7 n ~ o J

D m .

Case (ii). We now assume that w(Xi*,o) <= No for every n~ co. Defining Y. = D ~° + XI*, 1 + "'" + X.,k., Y = H I1. and B. ~_ X as in ease (i), we may

apply ¢ 5 to X and Y and obtain X ~ Y ~ D" by case (i). Case (iii). Fo r the most general ease, without any restriction on X.,o,

define I o = { n e c o l k . = O } and l l = { n ~ w l k . > O and w(Xi*,o)>No}. If n e I o, X n = Dr". If n ¢ I o, define

JD~° + XI*,I + "" + X,,.k. if n s lt

Y"= IX . if n ¢ l o w l a .

Then, for every n e/1, we see that X . ' ~ D t" x Y., where li*= w(X.,o), and

x = YI D--× 1-I n'-× 1-[ Y.. n e l t n¢~lo

to I-I Y-" Since E m . = E mi* = m, n¢~ lo n¢~ lo n¢¢o

Z II*< Z m.<=m, X ~ D ' . If ~o",/o

n E l o

If m \ / o is infinite, case (ii) applies

I-[ Y. "~ D". Since ~ m. __< m and n ¢ l o n ¢ I o n ¢ l t nel~

is finite, choose no e o~ s.t. i < n o for every i e co\lo. Then we have

x ~ l-I x . × n ' - o × 1-I n ' - n < i * 0 I*O < I*

-- D""o x I~ D ' " n o < t *

D = .

294 S. Koppelberg:

4.9. Lemma. K 1 is closed with respect to finite coproducts.

Proof. Let A = A 1 × -.. x A,,, B = B 1 x ... x B, where A 1 . . . . , B n are coproducts of countable algebras. Since

(X 1 + ... + X,,) x (Y1 + "'" + Y,) -~ ~ Xi x Y; i , j

for a rb i t ra ry topological spaces Xa , . . . , Y,, it follows that

A ~ B ~ I-I (ait2Bj) ~ K t . i , j

Thus, A u B E K1 whenever A, B ~ K 1, which implies the lemma.

4.10. Lemma. Suppose 14= 0 and A i ~ K1 for every i ~ L Then

Proof. Let us r emark that, if X, Y and Z are Boolean spaces, X ~ Y x Z and R(Y) e K~, it is sufficient to prove R(Z) e K'~ to get R(X) ~ K'~. F o r supposing R(Y)~-B'~, R(Z)~-B~ where B1, B 2 ~KI , we have

R(X) ~ R(Y x Z)

_~ (A ( r ) t2 A (Z)) *

~ - (R(Y)uR(Z) )* by 4.2

= ( B I I-I B2)

(B 1 t2 B2)*

and B t t_l B 2 ~ K 1 by 4.9. Let X i = S(A£ A = 11Ai and X = S(A) = I-I Xt. If A t = 2 for every

i e I i e I

i e I, A = 2 E K~. Suppose now that I o = {i ~ I I,~i > 2} 4: 0. Then A ~ L,~ Ai, and we m a y assume I o = L ~ x0

If I is finite, A ~ K t by 4.9. We may thus assume that T > N o. Define I 1 = {i ~ I IA t < No}, 12 = IN/1. Since [_.j A t e K l, we have (by the previous

t6J'l remark) to show that ( I ~ AiI* ~ K~ and we m a y assume 12 = L

\ i~.I2 / Define, f o r j ~ I, raj = A) = ~(Xj), I~ = {k~I]m~ = rnj} and pu t /3 = {j~IlI)

is infinite}. I f j ~ h and / j = / i = S o , 1~ Xk ~ D ' j by 4.7; if l j>So, we k ~ l j

m a y split l j into lj subsets each of which has power N o and thus obta in I-I x a ~ D "J' 'J. Thus, I-I x j - ~ / y " where m = Z m~. By the previous

k e l / . tel3 remark , we have to show tha t ( [_] A,]* ~ Kr , J, ,3 and we assume tha t for

\ 1 ¢ I 3 / every j e I, I~ is finite. Since, for every j E I, L.J Ak is an element of K1 and

k e l j has cardinal i ty mj, we m a y collect {A~lk ~ I;} to [ l A, and then assume

tha t m~4:m~ for i, j e l s.t. i 4=j. ~

Projective Boolean Algebras 295

Define M = {mi[ i E I}. M is a set of cardinals and hence well-ordered with respect to the membership relation. Let ). + n be the order type of M where n ~ ~o and 2 is 0 or a limit ordinal. If q~ is an order iso- morphism from M to / t+n, define, for i~I , c(i)=q~(mi), c : I - . * 2 + n is a bijection. If n > 0 , we see that [ ~ Ai~K~. It suffices to prove

t l Ai]* ~ K*. Since I is infinite, 2 cannot be 0. So there is an ordinal t/ c(i)<~ /

s.t. 2 = o9. t/. For every ¢ < q, we get, by 4.8, II {Xilo9- ~ < c(i) < o9. (¢ + 1)} ~ D "~ where m~ = Z{mi[~o • ~ =< c(i) < o9. (~ + 1)}, and hence I-[ X~ ,-, D" where m = Y', m~. a0<~

4.11. Theorem. I r A ~ K, then A* ~ K*.

Proof. This is clear forA ~ K o, since K o _~ K I. Suppose that {Ail i ~ 1} is a subset of K and, for every i e I, there is B i ~ K~ s.t. A* ~ B*. Then,

(~[~i Ai)* ~ \ t ~ ([-j A*t*/= ~i~l ([~ B*f*/~([-j\i~I Bit*/ and, by 4.10, (iL~lB~)*~ K~.

It is proved similarly that (A1 × "" x A,)* s K~ if A*, ..., A* ~ K*. The class of those Boolean algebras A s.t. A* e K* includes K 0 and is closed with respect to coproducts and finite products and isomorphic copies and thus includes K.

4.12. Theorem. Let B be a complete Boolean algebra. B ~ K* iff there exist n ~ 09, an ordinal k < N O and infinite cardinals m~ < m 2 < -.- < m, s.t.

B ~- 2 k x R(D m~) x ... × R(D~").

I f B is atomless, B ~ K* iff

(,)

B ~- R(D m~) x ... x R(D'" ) .

Proof. Let C be a countable Boolean algebra. Suppose C is a sub- algebra of C*. Define X = {xe C l x is an atom}, k = X and s=supC*X. If s = 0 , X = 0 , C is atomless and hence C ~ - A ( D ~°) and C*~-R(D~°). If s = 1, C is atomic and hence C* ~ 2 k. If 0 < s < 1,

C* ~ C*ls x C*l - s (**) ~- 2 k x R(D~°) ,

since C* Is is atomic with k atoms and C ' l - s is atomless and has a countable dense and atomless subalgebra.

Now, if B ~ K*, we may suppose B -- A* where A e K 1. (,) then follows from 1.12, 4.2 and the representation (**) for the countable factor C of A (if there is one).

296 S. Koppelberg:

If, conversely, (,) holds, define A = A k x A(D'O × ..- x A(D""), where

= / 2k, if k < N o

Ak [the finite -cofinite algebra of co, if k = N o.

Then, A e K 1 and A* ~ B.

The second part of the theorem follows immediately from the first one.

Trivially, the algebra 2 is homogeneous. If m is an infinite cardinal, R(D m) is homogeneous by 4.6. Thus every algebra B E K* is decompos- able into homogeneous factors.

4.13. Theorem. K is a proper subclass of L.

Proof. Let, for every n e co, A, = A(D ~) and define A = f i A,. Clearly, t ' l e m

.4 E L Since .4 is a dense subalgebra of I~ A,, n E o J

kn~¢o / / I ~ ¢.o ? I ~ £ o

Let now l, m be infinite cardinals s.t. l < m. Then R(D ~) has a dense subalgebra of cardinality l, namely A(DZ). Suppose that R(Dt)~ R(Dm); thus, R(D") has a dense subalgebra of cardinality l, say B. Choose two functions f : A(D') ~ B and 9 : B-~ A(D') s.t. M ~ A(D"), M 4:0 implies fl ~= f (M) ~ M and M ~ B, M ~= 0 implies 0 # 9(M) c= M. The algebra C generated by {g(f(M))IMeA(O")} is a dense subalgebra of A(D") of cardinality not greater than I. A(D") is a free algeb=ra; suppose that {v~lie m} is a set of free generators of A(D'). Since C < l < m, there is a subset I o of m s.t. To< l and the subalgebra generated by {vilielo} includes C. Choose i o ~ m \ l o. Since v~o 4:0 and C is a dense subalgebra of A(D"), there exists some c e C s.t. 0 < c < rio. But e is generated by free generators distinct from vjo which is impossible. We have thus reached a contradiction; R(D") is not isomorphic to R(D~). Suppose that A* ~ K*, e.g. (by 4.12) A*_~R(Dm)x ... xR(D""). If aeA* s.t. a # 0 , there is a non-void subset I of { 1, ..., n} s.t. A* l a - 1-I R(O"); so there are at most

i ~ l

2 " - 1 pairwise non-isomorphic relative algebras of A*. But

A* -~ I-I R(D"")

shows that {R(D~")] n e o9} is an infinite set of pairwise non-isomorphic relative algebras of A*. Thus A is not an element of K.

4.14. Remark. The complicated proof given here for K #: L can be considerably simplified. It may be shown that, for every A ~ K, the set

Projective Boolean Algebras 297

{A tal a e A, a # 0}'~co is finite (the main step of the proof is similar to the proof of 4.10 without assuming several technically complicated lemmas). This implies K + L as in 4.13. We have choosen the proof presented above because we consider the fact that every algebra in K* is de- composable into homogeneous factors in a very simple way and that K * = K* to be of some interest in its own.

5. 5.1. Definition. If X is a topological space, define

z (X) = {x e X l w(x) = ~ , } .

5.2. Definition. A topological space X is said to have property (Z) if w(X)~ N~o and z(X) is a non-void and nowhere dense subset of X.

We shall show L ~ P by proving in 5.3 that, if A e L, S(A) does not satisfy (Z) and giving in 5.4 an example of an injective space with property (Z).

5.3. Lemma. I f A E L, then S(A) does not have property (Z).

Proof. It suffices to prove that the class of Boolean spaces not satisfying (Z) contains every Boolean space having a countable base and is closed with respect to finite sums, one point compactifications of countable sums and arbitrary products.

If X has a countable base, w(X) < N o and hence z(X) = 0.

SupposenowthatX=o~(,~oX,),e.g.X= ,~o,U X ,u{y}and tha tXhas

property (Z). Let x ~ z(X). Since w(y) = N 0, x + y. So there exists no e s.t. x ~ X,0, Being a clopen subspace of X, X.o satisfies w(X.o) < w(X) ~ N~,. Since w(x) = N~, in X,o, too, z(X.o) 4: 0. z(X.o) is nowhere dense in X.o , for otherwise there would exist a non-void open set U s.t. U ~_ z(X.o) h c= z(X) h, and z(X) would not be nowhere dense in X. Thus, X,o satisfies (Z). - The same argument shows that, if X1, . . . ,X. do not satisfy (Z), then X1 + ... + X. does not satisfy (Z).

Let us finally consider a product space X = ~ X~ and suppose that X

satisfies (Z). We may assume that ~i > 2 for every i e L X~ being homeo- morphic to a closed subspace of X, w(Xi)~ w(X)< No,. Furthermore, I < N,o, for otherwise, w(x) > No, would hold, by 1.6, for every x ~ X and z(X) = X would not be a nowhere dense subset of X. - We next show that there is an i e I s.t. z(X~) ~ ~. Suppose not; then w(xi) < No, for every i ~ 1 and x~ ~ Xi. I f T < N0, w(x) < No, for every x e X and thus z(X) = ~. So we have No _-< I < N,0.- We shall now establish a contradiction showing that z(X) is a dense subset of X. Let x = (xili~ I ) ~ z(X). By 1.6, supw(x0 = N,~, but w(x~)< N~, for every i t L Hence there is a sequence t e l

298 s. Koppelberg:

( i . In e co) in I s.t. w(Xio ) < w(xn)< ..- and sup w(x, . )= No. Suppose that

B is any basic cube in X, e.g. B = I-I U~ x 1-I X~ where I0 __c I is finite i e l o i(EIo

and U~ is a non-void open subset of X~ for every i e Io. Let n o be a natural number satisfying {i n I n ~ to, no ~ n} c~ I o = O. Let y be an element of X s.t. y e B and, for every n > n o in to, y~. = x~. It is then clear that y ~ B c~ z(X).

We thus know that there is an ioe I s.t. z(X~o) # 0. Suppose that z(X~o ) is not nowhere dense in X~o. Choose a non-void open set U~o g_ z(X~o) h in X ~ and define W = Uio x I ] Xi, M = z(Xio) x I-I x~. w is a non-void

i * io i~ io open subset of X, W = M h and M c= z(X), so W c= z(X) h contradicting property (Z) of X. Thus z(Xio ) is nowhere dense in Xio and X~o has property (Z).

5.4. Theorem. L is a proper subclass of P.

Proof. Throughout the proof, we shall abbreviate D ~- by D, for every n ~ co.

In [3], a Boolean algebra A is constructed which is projective, but not in Kt. A is the direct limit of a system ( ( A . I n ~ c o ) , ( g . l n ~ c o ) ) where A n = A(Do) x ,.. x A(Dn) and g,, : A,,~A,,+~ is a suitable mono- morphism. Let X = S(A), X,, = S(A,,) = D o + ... + D n and ~p. : X. + 1 ~ X. be the mapping dual to g.. It is then clear that X is the inverse limit of the system ( ( X . I n e c o ) , (~o.ln e to)). We shall describe the mappings ¢p., examine X and apply 5.3.

Assume n ~ co. We want to define

q~n: X,+l = Do + "'" + Dn+I--} X , = D o + "'" + Dn.

s.t. for O_<k_<n + 1

tpnlDk:Dk_.¢{Dk if O<k<_n Dk_ t if k = n + t .

More exactly, we define, if 0_< k < n + 1 and x e Dk,

x if O < k < n

gon(x)= (1, x ) if k = n

[(O, x l N , ) if k = n + l

where x lN. is the restriction of x to lg,. Obviously, tpn is a continuous function from X~+ 1 to Xn. I f f E I-I Xn,

we write f = ( f . [ n ~ o ) where f , e X. for every n ~ to. The inverse limit X

Projective Boolean Algebras 299

of the system ( (X . ln e o9>, (q~. In e o9>> is then

X= {fe~oX.lq~.(f .+l)=f. for every nero}.

We now show that X has property (Z). Since w(X.) = N., for every n e co, w ( . ~ X . ) = ~o, and w(X)<~,. Define

X~ = { f e X If . ~ O. for every n E co}.

Let f e l--I x . . It is clear that f e X~ iff fo:N 0 --*D is s.t. fo(/) = 0 for every n q o ,

/ eNo and, for every ne 09, f .eD,, is the restriction of f.+~ to N.. Thus, X. is homeomorphic to

W= ~fe]-[D.lfo(l)=O for every/~N0 and f,,~-f.+l for every n¢co / J

where Iq D. is given the product topology and W the topology induced n E O ,

by I I D.. W is homeomorphic to n 6 t J )

V = { g ~ D ~ I g ( / ) = 0 for every l~N0} ;

a homeomorphism h from W to F is established by h(f)= U f,, for

f e I4{. But V is homeomorphic to D ~\~° and thus to D ~. - I f f e X z, the weight o f f in X~ is No,, in X ~ No, and, by w(X) ~ No,, w(f) = N,o in X.

Assume now that g e X'xX~. Then there is some k e 09 s.t.

# e Yk = { f e X I k + 1 is the least number n s . t . f . ~ D.}.

We have

Y k = { f e X I f . e D . f o r e v e r y n < k a n d f . eDk for k<n}

and similar to X~ ~ D ~ it may be proved that Yk-~ Dk" Obviously, Yk = X n U where

U= { f e ~o,X.IfkeDk and fk+l CDk+l}.

U being a clopen subset of I-[ X., Yk is a clopen subset of X. Thus, O has

the same weight in X and Yk, namely Nk. This shows that z(X)= X.. It is easily seen that U Yk is a dense open subset of X. Thus z(X)

k e o ,

is a non-void nowhere dense subset of X. X satisfies (Z) and A ~ A(X) is projective but (by 5,3) not an element of L.

300 S. Koppelberg: Projective Boolean Algebras

References

1. Dwinger, Ph.: Remarks on the field representations of Boolean algebras. Ind. Math. 22, 213--217 (1960).

2. Engelking, R.: Outline of general topology. Amsterdam: North Holland 1968. 3. G6memann, S. : A problem of Halmos on projective Boolean algebras. To appear in

Colloquium Mathematicum. 4. Halmos, P. R.: Lectures on Boolean algebras Princeton, N.J.: Van Nostrand 1963. 5. Sikorski, R.: Boolean algebras, 2nd edition. Berlin-Heidelberg-New York: Springer

1964.

Dr. Sabine Koppelberg Mathematisches Institut der Universit~it D-5300 Bonn Wegelerstr. 10 Federal Republic of Germany

(Received January 26, 1972}