Scholarly article on topic 'Boolean algebras admitting a countable minimally acting group'

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

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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

© Versita Sp. z o.o.

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 [9]. 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 [9]. 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. [5].

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 [10] 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 [10];see also Balcar, Jech and Zapletal [3] or Heindorf and Shapiro [8]. 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 [7]. 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 [11] and also In Helndorf and Shapiro [8].

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 [11], motivated by the Cohen forcing. A topological theorem proved by Shapiro [12] 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 [10]. However, another topological result obtained by Shapiro [13] 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 [11]. 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. [1]. 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 [15] 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 [12] 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 [1] and next it was strongly improved by Balcar and Franek [2]. 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 [2] 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 [4], Turek [14], Geschke [6]).

Acknowledgements

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

References

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