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## Academic research paper on topic "Boolean algebras admitting a countable minimally acting group"

﻿Cent. Eur. J. Math. • 12(1) • 2014 •46-56 DOI: 10.2478/s11533-013-0325-6

VERS ITA

Central European Journal of Mathematics

Boolean algebras admitting a countable minimally acting group

Research Article

Aleksander Btaszczyk1*, AndrzejKucharski1^, Stawomir Turek2*

1 Institute of Mathematics, University of Silesia, Bankowa 14, 40-007 Katowice, Poland

2 Institute of Mathematics, Jan Kochanowski University, Swi^tokrzyska 15, 25-406 Kielce, Poland

Received SO November SOtt; accepted 19 March S013

Abstract: The aim of this paper is to show that every infinite Boolean algebra which admits a countable minimally acting group contains a dense projective subalgebra.

MSG: 54B35, 54A05, 52A01, 54D30

Keywords: Projective Boolean algebra • Dense subalgebra • Regular subalgebra • Cohen skeleton • Cohen algebra • Group of automorphisms

1. Regular and relatively complete subalgebras

AH Boolean algebras considered here are assumed to be Infinite. Boolean algebraic notions, excluding symbols for Boolean operations, follow the Koppelberg's monograph . In particular, if (B, A, V, — , 0,1) is a Boolean algebra, then B+ = B \ {0} denotes the set of all non-zero elements of B. A set A C B is a subalgebra of the Boolean algebra B, A < B for short, if 1, 0 e A and A is closed under Boolean operations or, equivalently, u — w e A for all w, u e A. We shall write A = B whenever A and B are isomorphic Boolean algebras. A non-empty set X C B+ is called a partition of B whenever x A y = 0 for distinct x, y e X and

V X = 1,

* E-mail: ablaszcz@ux2.math.us.edu.pl î E-mail: akuchar@ux2.math.us.edu.pl

* E-mail: sturek@ujk.edu.pl

Springer

I.e. the supremum of X in B exists and equals 1. Therefore, a partition Is just a maximal set consisting of non-zero palrwlse disjoint elements of a Boolean algebra. The set of all partitions of B will be denoted here by PartB. For a Boolean algebra B the symbol c(B) denotes the Souslin number of B, i.e.

c(B) = sup {|P| : P G PartB}.

A subalgebra A of B is called regular, A <reg B for short, whenever every partition of A is a partition of B; see e.g. Koppelberg [10, p. 123], and also Heindorf and Shapiro [8, p. 14]. Let us recall that a set X C B+ is dense in a Boolean algebra B if for every b G B+ there exists a G X such that a < b. The cardinal number

^(B) = min {|X| : X C B+ and X is dense in B}

denotes the density of B. For A < B we say that A is a dense subalgebra of B, A <d B for short, whenever A is a dense subset of B. It is easy to see that every dense subalgebra is regular, i.e. A <d B implies A <reg B .

Let us recall that a complete Boolean algebra Bc is the completion of a Boolean algebra B whenever B is a dense subalgebra in Bc. From the Sikorski Extension Theorem it easily follows that if A is isomorphic to a dense subalgebra of B then Ac = Bc; see e.g. Koppelberg . However, there exist Boolean algebras, say A and D, for which Ac = Dc but neither A is isomorphic to a dense subalgebra of D nor D is isomorphic to a dense subalgebra of A; see e.g. .

If A < B, then an element b G B+ is called A-regular in B whenever there exists an element q(b) G A+ which is minimal among all the elements of A which are greater than b, i.e.

q(b) = min {d G A : b < d};

see also Koppelberg  for an equivalent definition of q(b). It is clear that if A < B then every element of A is A-regular in B since q(b) = b for every b G A in that case. A Boolean algebra A is called a relatively complete subalgebra of a Boolean algebra B, A <rc B for short, provided that every element of B is A-regular. It is not difficult to show that every relatively complete subalgebra is regular;see Corollary 1.3 below. It is clear that if A <rc B and B is complete, then A is complete as well. Indeed, if X C A and u G B is the supremum of X in B, then q(u) is the supremum of X in A. From the definition we obtain immediately the following lemma.

Lemma 1.1.

If A < B and for some b, c G B+ there exist both q(b) and q(c) then the following conditions hold true:

(a) for every a G A+ there exist q(a A b) and q(a A b) = a A q(b),

(b) there exists q(b V c) and moreover q(b V c) = q(b) V q(c).

Some of the conditions in the next proposition were proved by Koppelberg ;see also Balcar, Jech and Zapletal  or Heindorf and Shapiro . For the sake of completeness we give its proof.

Proposition 1.2.

If A < B then the following conditions are equivalent:

(a) A <reg B,

(b) for every b G B+ there exists a G A+ such that whenever x G A+ and x < a, then x A b = 0,

(c) for every b G B+ there exists a G A+ such that A\(a A-b) = {0},

(d) the set of all non-zero A-regular elements of B is dense in B,

(e) for every b G B+ there exists a G A+ such that q(a A b) = a.

Proof. (a) => (b) Suppose that there exists b e B+ such that below every a e A+ there exists xa e A+ such that xa Ab = 0. The set X = {xa : a e A+} is dense in A. By the Kuratowski-Zorn lemma there exists a maximal disjoint set Y C X. Since X is dense in A, Y is a partition of A. On the other hand, since y A b = 0 for each y e Y, the set Y is not a partition of B. We have a contradiction.

(b) ^ (e) Let b e B+ be fixed. By condition (b) there exists a e A+ such that for each x e A+ we have the following implication:

x < a x A b = 0. (*)

In particular we have 0 < a A b < a. We shall show that a A b is an A-regular element of B. For this goal it is enough to show that

a = min {y e A : a A b < y}.

We set Y = {y e A : a A b < y}. Since a e Y, it remains to show that a is the lower bound of Y. Suppose that a — x = 0 for some x e Y. Since a — x < a, by condition (*), we have (a — x) A b = 0. On the other hand, we have a A b A—x = 0 because x e Y. Again we get a contradiction.

(d) ^ (a) Suppose X C A+ is a partition of A and there exists b e B+ such that x A b = 0 for every x e X. By condition (d) we can assume that b is A-regular. Clearly 0 < q(b) since 0 < b < q(b). Moreover, q(b) < —x for each x e X since b < —x for each x e X. Therefore, X cannot be a partition of A; a contradiction.

Since the equivalence (b) ^ (c) and the implication (e) ^ (d) are obvious, the proof is complete. □

Immediately from Proposition 1.2 we obtain the following corollary.

Corollary 1.3.

For each Boolean algebras A and B, A <rc B implies A <reg B.

A Boolean algebra B is countably generated over a subalgebra A < B if there exists a countable set X C B such that B = (A U X}, i.e. B is generated by the set A U X. If A is countably generated over C, then we shall write C <rcw A whenever C <rc A and we shall write C <regw A if C <reg A.

For an infinite set X the symbol FrX denotes the free Boolean algebra generated by the set X as the set of free generators. If \X\ = \Y\ then the Boolean algebras FrX and Fr Y are isomorphic and are denoted by Frk, where k = \X\ = \ Y\. Clearly, if X C Y then FrX is a subalgebra of Fr Y. In fact we have much more: if X C Y, then FrX <rc Fr Y.

A Boolean algebra is called projective if it is a retract of a free Boolean algebra. Therefore, a Boolean algebra B is projective whenever there exists a cardinal number k and homomorphisms f: B ^ Fr k and g: Fr k ^ B such that the composition g o f is the identity on B. All free Boolean algebras are obviously projective but the converse statement is not true. A first internal characterization of projective algebras was obtained in topological language by Haydon . In terms of relatively complete subalgebras it can be written as follows:

Theorem 1.4 (Haydon's Theorem).

An infinite Boolean algebra B is projective iff there exists a sequence {Aa : a < \B\} of subalgebras of B such that the following conditions hold true:

• Ao = {0,1},

• Aa <rcw Aa+i for all a < \B\,

• Aa = U {A^ : ¡3 < a} whenever a < \B\ is a limit ordinal,

• B = U{A¡ : ¡< \B\}.

An algebraic proof of Haydon's Theorem can be found In Koppelberg  and also In Helndorf and Shapiro .

A Boolean algebra C is called a Cohen algebra if the completion of C is isomorphic to the completion of the product of countably many free Boolean algebras. The notion of a Cohen algebra is due to Koppelberg , motivated by the Cohen forcing. A topological theorem proved by Shapiro  says that every subalgebra of a free Boolean algebra is a Cohen algebra. In particular, a subalgebra of a projective Boolean algebra is a Cohen algebra;see e.g. [8, p. 133].

On the other hand, dense subalgebra of a projective Boolean algebra need not be projective;see e.g. Koppelberg . However, another topological result obtained by Shapiro  implies that every Cohen algebra has to contain a dense projective subalgebra. In much simpler way the same theorem follows from a nice characterization of Cohen algebras given by Koppelberg . For the sake of this characterization Koppelberg introduced the notion of the Cohen skeleton. A collection S of subalgebras of a Boolean algebra B is called a Cohen skeleton if it satisfies the following conditions:

• A <reg B for every A e S,

• an element of S contains a dense countable subalgebra,

• the union of every nonempty chain in S is a dense subset of a member of S,

• for every A e S and every countable set X C B there exists C e S such that A U X C C and a dense subalgebra of C is countably generated over A.

Then the Koppelberg characterization reads as follows.

Theorem 1.5 (Koppelberg's Theorem).

If a Boolean algebra B satisfies the countable chain condition, then the following conditions are equivalent:

• B is a Cohen algebra,

• B has a Cohen skeleton,

• B contains a dense projective subalgebra.

2. Automorphisms group acting on a Boolean algebra

We say that a group H of automorphisms of a Boolean algebra B acts minimally on B if for each b e B+ there exist h1,... ,hn e H such that

Mb) v-v hn(b) = 1. (1)

see e.g. . Clearly, if B is a homogeneous Boolean algebra then the group of all automorphisms acts minimally on B. In particular, for every k > t the group of all automorphisms of Fr k, the free Boolean algebra of size k, acts minimally on Fr k. Since the Boolean algebra Fr k is homogeneous, the size of this group is 2k. However, by homogeneity, there is a group H of automorphisms of algebra Fr k of size k which acts minimally on Fr k. Moreover, Turek  has shown that there exists an infinite cyclic group of automorphisms acting minimally on Fr2o.

Lemma 2.1.

If a group H of automorphisms acts minimally on a Boolean algebra B, then c(B) < \H\.

Proof. Suppose P e PartB is of cardinality greater than \H\ and choose an ultrafilter p on B. For every h e H the set {h(u) : u e P} is a partition of B. Hence, there exists at most one element uh e P such that h(uh) e p. Choose an arbitrary w e P \ {uh : h e H}. Then h(w) e p for every h e H. On the other hand, since H acts minimally on B, there exist h1,..., hn e H such that

1 = hi (w) V ••• V hn(w).

Since p is an ultrafilter, there exists i < n such that hi(w) e p; we get a contradiction. □

If H is a group of automorphisms of a Boolean algebra B, then a subalgebra A < B is called to be an H-proper subalgebra of B, shortly A <H-prop B, whenever for every a e A and every h e H there exists X C A such that

h(") = Vb X■

Clearly, if h |"A is an automorphism of A for every h e H, then A is an H-proper subalgebra of B. We get the following proposition.

Proposition 2.2.

If H is a group of automorphisms of B such that H acts minimally on B and A <H-prop B, then A <reg B.

Proof. Suppose T e Part A and there exists b e B such that

b A t = 0 (2)

for every t e T. There exist hi,..., hn e H with (1). Since A <H-prop B, for every k < n and every t e T there exists a set Xtk C A such that

hk (t) = VB Xtk. (3)

Now, for every k < n we set Xk = |J {Xtk : t e T}. We claim that

VA Xk = 1 (4)

for every k < n. For this goal we fix a e A+ and k < n. Since h—1 e H and A is an H-proper subalgebra of B, there exists z e A+ such that z < h—1(a). Since T e Part A, there exists t e T such that z A t = 0. Therefore hk(z) A hk(t) = 0 and hence a A hk(t) = 0. Then, by condition (3) we get a Ax = 0 for some x e Xtk. This completes the proof of condition (4). By this condition, for every k < n we can choose xk e Xk in such a way that

xi A • • • A Xn = 0. (5)

On the other hand, by conditions (2) and (3), we get xA hk(b) = 0 for every k < n and every x e Xk. Hence, by condition (1), we obtain

xi A-A xn = xi A-A xn A (hi (b) V •• •V hn(b)) < (xi A hi(b)) V • • •V (xn A hn (b)) = 0, which leads to a contradiction with (5). □

We shall also use the following property of Boolean algebras with a group of automorphisms which acts minimally.

Lemma 2.3.

If a group of automorphisms H acts minimally on a Boolean algebra B, then ^(B) = n(B|"u) for every u e B+.

Proof. If there exists a dense set X C B|"u of size k > w, then for every h e H there exists in B h(u) a dense set of the same size k. By the assumptions, there exist hi,..., hn e H such that hi(u) V • • • Vhn(u) = 1. Hence B admits a dense set of size k. □

Using Propositions 2.2 and 1.2 we can modify a bit the definition of H-proper subalgebras.

Lemma 2.4.

Assume that a group H of automorphisms of a Boolean algebra B acts minimally on B. Then A <H-prop B iff for every h G H and every a G A there exists a set X Ç A+ of pairwise disjoint elements such that

Proof. Assume that X Ç A+ and h(a) = \/bX. Let Y Ç A+ be a maximal In A+ disjoint set such that every element of Y is below some element of X. Then \/B X = \/B Y. Otherwise one can choose u G B+ such that u < \/B X and u A y = 0 for every y G Y. By Propositions 1.2 and 2.2, we can assume that u is A-regular in B and u < x for some x G X. Since u < —y for every y G Y, we have q(u) < — y for every y G Y. We get a contradiction with the maximality of Y since q(u) G A+ and q(u) < x. The opposite implication is trivial. □

3. The main result

The next lemma gives a rather technical but very useful property of regular subalgebras.

Lemma 3.1.

Assume that A <reg B and n(B \b) > n(A) for each b e B+. Then for every b e B+ there exists an A-regular element c e B+ such that c < b and x A b = 0 implies x A (b - c) = 0 for all x e A+.

Proof. Since n(B tb) > n(A), the set {x A b : x e A+} cannot be dense in Bfb. Hence, there exists c e (Bfb)+ such that x A b A-c = 0 whenever x e A+ and x A b = 0. By Proposition 1.2 we can assume that c is A-regular since A is a regular subalgebra of B. □

The last lemma can be extended as follows.

Lemma 3.2.

Assume A <reg B and n(B\b) > n(A) for every b e B+. Then for every b e B+ there exists an infinite P e Part(B tb) which consists of A-regular elements and the following condition holds true: if R C P is finite, then

xab = 0 x a (b - y R) = 0 (6)

for each x e A+.

Proof. Let us consider the family Z of all sets S C (Bfb)+ of disjoint A-regular elements of B which satisfy the following property:

(**) for every x e A+ and every finite subfamily R C S we have (6).

By Lemma 3.1, there exists an A-regular element c e B such that the set S = {c} fulfills the condition (**). Hence the family Z is non-empty. By the Kuratowski-Zorn lemma there exists a maximal family P e Z. It remains to show that Vb P = b. Suppose that there exists d e (B tb)+ such that d Ap = 0 for each p e P. Again, by Lemma 3.1, we obtain an A-regular element c < d such that x A d = 0 implies x A (d — c) = 0 for every x e A+.

To get a contradiction it is enough to show that P U {c} e Z. For this goal assume that R C P is finite and x Ab = 0. We shall show that

x A (b - y (R U {c})j = 0. We have two cases. If x A d = 0, then x < -c since c < d. Since P e Z, we get

0 < x A b A-y R < x A b A-y R A-c = x A b A-(Y R v ^ = x A b A-y (R U {c}).

If x A d = 0, then x A d A-c > 0. Since d < b and d A\J R = 0, we have d < b A-\J R. Therefore we get

0 < x A d A-c < x A b A—\J R A-c = x A b A—\J (R U {c}),

which completes the proof. □

If Ri, R2 e Part B then we say that R2 is a refinement of Ri, R1 — R2 for short, if for every u e R2 there exists v e Ri such that u < v.

Proposition 3.3.

Assume c(B) = w and A <reg B and n(B \'b) > n(A) for every b e B+. Then for every finite collection R1,..., Rn e PartB, there exists R e PartB consisting of A-regular elements such that

(a) Rt < R for every i < n,

(b) for every w e Ri U • • • U Rn the set {u e R : u < w} is a countable infinite partition of B \w,

(c) for every w e R1 U • • • U Rn and for every finite set P C {u e R : u < w} and every x e A+ we have

xA w = 0 xA (w - Y ^ = 0.

Proof. Assume R1.....R„ e Part B. Let Q = {w1 A w2 A---A wn : wt e R, i < ri} \ {0}. It is easy to check that

Q e PartB and Ri — Q for every i < n. Clearly, for every w e Rj the set

{ wi A • • • A w— A w A w+i A^ • • A wn : wi e Ri, i < n, i = j] \ {0} C Q

is a partition of B|"w. By Lemma 3.2, for every b e Q we obtain a partition Rb of B \b consisting of A-regular elements such that whenever P C Rb is finite and x e A+ then x A b = 0 implies x A (b - V P) = 0. By Lemma 3.2, |Rb| < w since c(B) < w. To complete the proof it is enough to set R = U {Rb : b e Q}. □

We are ready to prove our main result. Assume H is a countable group of automorphisms acting minimally on an infinite Boolean algebra B. We shall show that the collection

{A C B : A <H-prop B}

of subalgebras of B has got some properties which are similar to those of the Cohen skeleton. It appears that these properties determine that a Boolean algebra B is a Cohen algebra.

Theorem 3.4.

Assume H is a countable group of automorphisms acting minimally on an infinite Boolean algebra B. Then for every Boolean algebra A such that n(A) < n(B) and A <H-prop B and for every b e B+ there exists a Boolean algebra C such that

A <rcw C <H-prop B

and c < b for some c e C+.

Proof. We can assume that H = {hn : n = 1,2,... } and h1 is the identity on B. By Proposition 2.2 we have A <reg B. Since H acts minimally on B and \H\ < u>, by Lemma 2.1 we have c(B) < w and by Lemma 2.3 we also have n(B) = n(B fu) for every u e B+.

Let us consider the set Seq = U {n w : n < w} of all functions from n = {0,1,..., n — 1} into w, n < w. Let

Seq = {sn : n e w},

where s0 is the empty function. For g, f e Seq, we say that f extends g whenever g C f. Hence, f extends g iff dom g C dom f and f \ dom g = g. If g e n w and k e w then the symbol g"k denotes the sequence of length n + 1 that extends g and whose last term is k, i.e. g"k is the function f: n +1 —> w such that f\n = g and f(n) = k. We shall construct a sequence {Pn : n < w} of partitions of B. Every Pn consists of A-regular elements of B and are indexed by finite sequences of the length n, i.e. Pn = {ug : g e nw}, and the following conditions hold true:

(i) P0 = {u0}, where u0 = 1 and {b, -b} — P1,

(ii) V{ug-i : i < w} = ug for every g e nw,

(iii) ug^i A ug~j = 0 whenever g e nw and i = j,

(iv) for every i e {1,..., n} and every u e Pn-1 there exists an infinite family P C Pn such that hf(u) = \/ P,

(v) for every u e Pn-1, every finite subfamily P C Pn n B\u and every a e A+, a A u = 0 implies a A (u — \/ P) = 0.

To obtain P1 we consider the family R1 = {b, —b}. From Proposition 3.3 we get a countable infinite partition R of the algebra B consisting of A-regular elements such that the following conditions hold true:

• Ri R,

• if w e R1 the set {u e R : u < w} is a countable infinite partition of B |"w,

• if w e R1 and P C {u e R : u < w} is a finite set and x e A+ then x A w = 0 implies x A (w - V P) = 0.

Let P1 = R. We enumerate all elements of the family P1 by elements of the set 1w. Assume that we defined the partitions P1,..., Pn. Applying again Proposition 3.3 for partitions

Ri = {hi(u) : u e Pn},

where i e {1,..., n}, we obtain a partition R e PartB+ which consists of A-regular elements and satisfies the following conditions:

(a) R i - R for every i < n ,

(b) for every w e R1 U • • • U Rn the set {u e R : u < w} is a countable infinite partition of B \ w,

(c) for every w e R1 U • • • U Rn and every finite set P C {u e R : u < w} and every x e A+, x A w = 0 implies xA (w - V P) = 0.

We set Pn+1 = R. By condition (b), the partition Pn+1 can be indexed by elements of n+1 w in such way that for every

g e nw,

{v e R : v < ug} = {ug~n : n < w}.

Let us observe that Pn+1 satisfies conditions (ii)-(v). In fact, since h1 is the identity, R1 = Pn and hence Pn — Pn+1. Condition (b) implies (ii) and (iv) and condition (c) implies (v). The induction is complete.

For any f,g e Seq we denote f ± g whenever neither g C f nor f C g. By conditions (ii) and (iii), for every g,h e Seq we get the following:

(vi) g C h and g = h imply uh < ug,

(vii) g .L f implies uf A ug = 0.

Let us recall that {sn : n e w} is the fixed enumeration of the set Seq of all finite sequences of natural numbers. Now we consider a sequence of algebras {An : n e w} where An is the subalgebra of B generated by AU{uf : f C sh i < n}.

Finally we set C = U{An : n e w}. By (vi) and (vii) we have the following claim.

Claim 1. Every element of C+ is a finite sum of elements of the form a A uf A-ug1 A • • • A-ugp, where a e A and gi L gj for distinct i,j < p and f C gt for all i < p and g1,... ,gp e {g : g C sh i < n} for some n e w.

By condition (i) there exists an element c e C+ such that c < b. It is easy to see that C is countably generated over A. We shall prove that A <rc C. For this goal let us fix an element x e C+. There exists some n e w such that x e An. We have to prove that there exists

q(x) = min {d e A : x < d}. By Lemma 1.1 (b), and Claim 1 we can assume that

x = a A Uf A—ug1 A^ • • A-ugp,

where a e A, gi ± g¡ for all distinct i,j < p and f C gi for all i < p. Now, by Lemma 1.1 (a), it suffices to prove that there exists

q(Uf A- Ug 1 Ugp).

Since for each m e w the partition Pm consists of A-regular elements, there exists q(uf). We shall show that q[uf A — ug1 A-^A- ugp) = q(uf). Clearly we have q(uf) e A and uf A — ug1 A^ • • A-ug < q(uf). Suppose that there exists some y e A+ such that q(uf) — y = 0. Then we have 0 = —y A uf and by condition (v), we get

0 < -y A (Uf - Y {UgI dom f+1 : i < p}},

since for every i < p we have f C gi and thus {ug¡idomf+1 : i < p} is a finite subfamily of Pdomf+1. Since ugi < ug^domf+1 we obtain

0 < -y A (Uf - Y {Ugi : i < p}} = Uf A- Ug1 A-^A-Ugp A-y.

Hence the element y cannot be an upper bound of Uf A-Ug1 A • • • A -Ugp. To complete the proof of the theorem it remains to show that C <H-prop B. Let c e C+ and hi e H be fixed. We shall show that there exists a set T C C such that hi(c) = \/B T. For this goal we fix some e e B+ such that e < hi(c). There exists n e w such that c e An. By Claim 1 we can assume that

c = a A Uf A-Ug1 A^ • • A-ugp,

where a e A+, and f, g1,..., gp e {f : f C si, i < n} and gi ± g¡ for all distinct i, j < p and f C gi for all i < p. We shall need the following claim.

Claim 2. If R1, R2 e Part B and R1 -< R2, then for every v e R1 and every b e B+ such that -v A b = 0 there exists u e R2 such that 0 = u A b < - v A b.

In fact, since R1 is a partition and - (b < v), there exists v' e R1 such that v A v' = 0 and b A v' = 0. Since v' = \/ {x e R2 : x < v'}, there exists u e R2 such that u < v' and u A b = 0. This completes the proof of the claim.

Now we return to the proof that C <H-prop B. Since e > 0 and e < hi(c), we have

Uf A-Ug1 A-^A-Ugp A a A h-1(e) = 0.

Hence, by Claim 2 there exists m e w and an element uk e Pm such that dom k > i and uk < uf A-ug1 A • • • A-ugp and

(viii) 0 = uk A a A h-(e) < uf A-ug1 A^ •• A-ug A a A h-1(e).

Since A <H-prop B, by condition (lv) there exist familles TUk, Ta C A+ of disjoint elements such that \/B Ta = ^¡(o) and Vb TUk = hi(uk). By distributivity lows we have \JB Ta A Vb TUk = Vb T, where T = {x A y : x G Ta, y G TUk}. Since Uf A—Ugi A - ■ ■ A — Ugp A a A h — (e) = c A a A h—1 (e), by condition (viii) we have

0 = hi(Uk) A h^a) A e < h(c) A h(a) A e < h(c) A e.

There exists t e T such that 0 = t A e < hi(uk A a) A e. Since e e B+ was chosen arbitrarily so that e < ht(c), we get

h(c) = Vb THence we get C <H-prop B, which completes the proof. □ As a immediate consequence of the last theorem we obtain the following theorem.

Theorem 3.5.

If a countable group of automorphisms acts minimally on a Boolean algebra B, then B contains a dense projective algebra of size ^(B).

Proof. Let t = ^(B) and let {ba : a < t} be a dense subset of B+. Since countable Boolean algebras are projective, we can assume that t > w. By transfinite induction we define a sequence of Boolean algebras {Aa : a < t} such that Aa <H-prop B for every a < t and the following conditions hold true:

(a) Ao = {0,1},

(b) Aa <rcw Aa+1 for all a < t,

(c) Aa = U {A^ : ¡3 < a} whenever a is a limit ordinal,

(d) for every a < t there exists a e A++1 such that a < ba.

If the Boolean algebras {Aa : a < y} satisfying conditions (a)-(d) have been constructed for some y < t and y is a limit ordinal we set Ay = U {Aa : a < y}. Definition of the H-proper subalgebras easily implies that Aa <H-prop B.

Assume that y is a successor ordinal, e.g. y = ^ + 1 and the conditions (a)-(c) are fulfilled for all 3 < y. It is clear that n(Ap) < + w < n(B). Then, by Theorem 3.4 there exists a Boolean algebra C such that

Ay <rcw C <H—prop B

and a < by for some a G C+. Then we set Ay = C. Now, by conditions (a)-(c) and Haydon's Theorem (Theorem 1.4), we conclude that

D = y {Aa : a <t}

is a projective Boolean algebra. From condition (d) it follows that D is a dense subalgebra of B since the set {ba : a < t} is dense in B. □

Remark 3.6.

The above theorem can also be proved by the use of Koppelberg's characterization of Cohen algebras;see Theorem 1.5. For this purpose one has to show that the family of all those subalgebras of B which are invariant with respect to a countable group of automorphisms constitute a Cohen skeleton.

In case of complete Boolean algebras we obtain the following corollary.

Corollary 3.7.

If B is a complete Boolean algebra and there exists a countable group of automorphisms acting minimally on B, then B = (Fr k)c, where k = ^(B).

Proof. It is an immediate consequence of Theorem 3.5. Indeed, let a projective Boolean algebra A be a dense subalgebra of B. Then B = Ac and, by Lemma 2.1, n(A) = n(A|"u) for every u e A+. From a theorem of Shapiro  it follows that if A is a projective Boolean algebra and n(A|U) is the same for every u e A+, then Ac is isomorphic to the completion of a free Boolean algebra;see also [8, p.116]. This completes the proof since k = n(B). □

Remark 3.8.

The above theorem was proved for the first time in  and next it was strongly improved by Balcar and Franek . They proved that if B(S) is the clopen algebra of the phase space of the universal minimal dynamical system over a semigroup S (see  for definitions) and B(S) is atomless and G is either concellative or has a minimal left ideal or is commutative, then B(S) is a Cohen algebra. In particular, if S is a countable group, then B(S) is a complete Boolean algebra which admits a countable group of automorphisms acting minimally on it (see also Bandlow , Turek , Geschke ).

Acknowledgements

The authors are deeply indebted to Professor Sabine Koppelberg for valuable comments to the early version of the paper.

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