saharon shelah- superatomic boolean algebras: maximal rigidity

8/3/2019 Saharon Shelah- Superatomic Boolean Algebras: Maximal Rigidity

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SUPERATOMIC BOOLEAN ALGEBRAS: MAXIMAL RIGIDITY

Saharon Shelah

Abstract. We prove that for any superatomic Boolean Algebra of cardinality > 4there is an automorphism moving uncountably many atoms. Similarly for largercardinals. Any of those results are essentially best possible

Annotated Content

0 Introduction

1 Superatomic Boolean algebras have nontrivial automorphisms

[We prove that ifB is a superatomic Boolean Algebra, then it has a quitenontrivial automorphism; specifically ifB is of cardinality > 4() then Bhas an automorphism moving > atoms. We then discuss how much we

can weaken the superatomicity assumptions.]

2 Constructing counterexamples

[Under some assumptions we construct examples of superatomic BooleanAlgebras for which every automorphism moves few atoms.]

3 Sufficient conditions for the construction assumptions

[We deal with the assumptions of the construction in 2 deducing that inmany cases, even usually the bound in 1 is essentially best possible.]

Key words and phrases. Set Theory, Boolean Algebras, superatomic, rigid; pcf, MAD.

I would like to thank Alice Leonhardt for the beautiful typing.Latest Revision - 07/July/30sh704 paper

Typeset by AMS-TEX

1


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4 On Independence

[We show e.g. that 4() consistently can be improved.]

0 Introduction

We show that any superatomic Boolean Algebra has an automorphism movinguncountably many atoms if it is large enough, really > 4; similarly replacing 0by ; (an automorphism moves an atom if its image is not itself). We then showthat those results are essentially best possible. Recall that for many other naturalclasses of Boolean Algebras behave differently; there are arbitrarily large memberswith few automorphisms (and even endomorphisms). Of course, we can expressthose results in topological terms. See [M] and [M1] on Boolean Algebra.

Rubin and Koppleberg [RuKo01] have proved: if + + 2+

= ++, then there isa superatomic Boolean Algebra B of cardinality ++ with atoms and exactly +

automorphisms answering a question 80 of Monk [M2], i.e. in a preliminary versionasking for a consistent example.By [Sh 641, 1], provably in ZFC, there is a superatomic Boolean Algebra B suchthat |Aut(B)| < |End(B)| answering question 96 of Monk [M2, p.291].By [Sh 641, 2], provably in ZFC, there is a superatomic Boolean Algebra B suchthat |Aut(B)| < |B|, answering Problem 80 of [M2, p.291]. In fact, if is stronglimit, > cf() = 0 and = Min{ : 2 > 2}, then there is a Boolean AlgebraB with 2 atoms, 2 elements and every automorphism ofB moves < atoms so

|Aut(B)| 2

< 2

.

Notation

0.1 Definition. 1) For a Boolean Algebra B its operation are denoted by x y, x y, x y, x and 0B is its zero. Let us define the ideal id(B) by induction:

id0(B) = {0}

id(B) = {x1 . . . xn : n < and x B for = 1, . . . , n such that forsome < and for each {1, . . . , n} the element x/id(B) is an atom

ofB/id(B) or x id(B)}.

Hence for limit we have

id(B) =


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MOVING ATOMS 3

2) For x id(B) let rk(x,B) = Min{ : x id+1(B)}.3) B is superatomic ifB = id(B) and rk(B) is the ordinal such that B/id(B)is a finite Boolean Algebra (so B = id+1(B)).

4) For a Boolean Algebra B and x B let B x be B restricted to {y B : y B x},it is a Boolean Algebra.5) Define by induction on n = 1, 2, . . . :

1(< ) = 2


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Let I =: []


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MOVING ATOMS 5

z I) (z c)(y z I)], just close times recalling is regular.Now if y I then |y| < hence y c [c] 2(< ); let i x fori < (2(< ))

+ be pairwise distinct, let ai be a B-autonomous set of cardinality 2


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6 SAHARON SHELAH

{x/I : < }, so for some , x/I = x/I hence |x\x| < hence|x x| hence x x = hence x / J, a contradiction.]

Define an equivalence relation E on B : y1Ey2 iff y1 x = y2 x. Clearly Ehas 2|x| equivalence classes and 2|x| 4(< ); also y1Ey2 y1\y2 J, infact y1Ey2 (y1y2 J) (see Js definition). Choose a set of representatives{y : < } for E so || 4(< ) and let B be the subalgebra ofB which{y : <

} generates. So |B| 4(< ) and, being superatomic, the numberof ultrafilters ofB is also 4(< ). Next B is generated by J B as for y Bthere is such that yEy and y B

, y y J, y y J hence y J B.

For D an ultrafilter ofB let ZD = { < : (y B)( y y D)}.Clearly

5 for every \x there is a unique ultrafilter D = D[] on B such that

ZD (and the number of such ultrafilters is 4(< )).Now

6 4(< ).[Why? Assume that not. By 4 for each i < we can find a B-autonomousai such that |ai| 2


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MOVING ATOMS 7

(b) there is an ideal J I containing 2


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[Why? Let b Ik 0}, it is a subset ofZ, now for eachD Db we have Zb D| 1 in fact |Z D| 1. So if |Zb| then |Zb| > |Db|so for some c Zb we have (D Db)(c / D), but this contradicts the choice of

Zb. So |Zb| < , so {D : b D and D Dc for some c Zb} has cardinality

cZb

|Dc| < and is a subset of h(b).]

()3 h maps Ik


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[Why? If k = 1 we can find X of cardinality 2


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2 Constructing counterexamples

We would like to show that the bound 4(< ) from 1.1 is essentially bestpossible. The construction (in 2.1) is closely related to the proof in 1, but we needvarious assumptions. So in particular here corresponds to sup{|B a| : a Y} 2(< ) there, here corresponds to |Y| 3(< ) there, here corresponds to|atom(B)| 4(< ) there. We shall deal with them later.

2.1 Lemma. Assume

(a) and = cf() 1

(b) there is anA []0 almost disjoint (i.e. A = B A |A B| < 0)such that (A [])(B A)(B A) and |A| =

(c) B = B : <

(d) B is a superatomic Boolean Algebra with atoms such that any auto-morphism of B moves < atoms and |B| ; moreover if c1, c2 I(see below) and f is an isomorphism fromB (1 c1) onto B (1 c2)then > |{x atom(B) : x B c1 or f(x) = x}|

(e) I = {b B : |{x atom(B) : x b}| < } is a maximal ideal ofB

(f) there is an infinite set {an : n < } of pairwise distinct atoms ofB suchthat for every a I the set {n < : an a} is finite

(g) if = then for no a I, a I do we have

B (1B a) = B (1B a)(h) B is a superatomic Boolean Algebra

(i) B has atoms

(j) B has elements2

(k) if > then we have ,A, I satisfying:

() A []0 is a MAD family of cardinality

() I is an ideal ofB containing id1(B), included in idrk(B)(B

) suhcthat the Boolean algebraB/I is isomorphic to {a : a is finite orco-finite}; so |B| = follows

(gamma) if is a partial3 permutation of , Dom() = \Z1, Rang() = \Z2where Z = Z1 Z2 []


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Then we can find B such that:

() B is a superatomic Boolean Algebra() B has atoms and elements

() every automorphism g of B moves < atoms; i.e.|{x atom(B) : g(x) = x}| < .

Proof. Without loss of generality B is a Boolean Algebra of subsets of {w1, : , let Y B be such that |Y| = and {y/I :

y Y} is the set of atoms ofB/I with no repetitions; without loss of generality

0 for every y Y for some ,y/id(B) is an atom of B/id(B

) and(z)[z B y (z id(B

) z I)].

[Why is this possible? For each y B\I let = (y) =: Min{rkB(y x) : x I} and choose x0y exemplifying it, so (y x

0y)/id(B

) is the union offinitely many atoms of B/id(B

), say y1/id(B), . . . , yn/id(B

) where n 1and without loss of generality y B y. So {y1, . . . , yn} cannot be all in I and

there cannot be two y B\I, so there is a unique = () such that y / I, letxy = (1 y()) x

0y. Now {y x

y : y Y

} is as required.]Let Y+ be such that Y+ B, y/ idrk(y,B)(B

) : y Y+ list with no repeti-tions {y/ idrk(y,B) : y/idrk(y,B)(B

) an atom ofB/idrk(y,B)(B)} and Y = {y

Y+ : rk(B) > rk(y, B) > 0}.Without loss of generality Ymax = {y Y+ : rk(y, B) = rk(B) is a partition of

1B. For y Y+ let Dy be the ultrafilter on B

generated by {y} {1 x : x B, rk(x,B) < rk(y, B)} for each y Y. Without loss of generality Y Y andfor some Ymax Ymax we have y Y+ y < ymax. Also as B/I is isomorphicto the Boolean Algebra of finite, co-finite subsets of , y Y rk(y, B) < rk(B)

and clause (k)() of the assumption of 2.1 clearly y Y\Y {y Y : y y idrk(y,B)(B

)} is finite so without loss of generality is empty for y Y\Y

(singleton for y Y, of course), note that if > then Y is of cardinality |A|and without loss of generality |Y\Y| = .

Let g be a one-to-one function from onto X and for A A (from clause (b))let {A,k : k < } list A without repetition. Let g

: be g() = the unique < such that g() X.


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For < let i() be the unique i < 1 such that ()(1 = 1 + i). ForA A we define i(A) = Min{i < 1 : i(g()) < i for every uA}

1 we have [A] : A A

, uA well defined such that:(i) [A] <

(ii) (A) {w 1+ i : < and i(A) i < 1}

(iii) ) [A1] = [A2] A1 = A2

(iv) ) {[A1] : A A, Rang(g A) infinite} is equal to {1 + i : then

|A

| = |A

| = = |Y

| so let dy : y Y

list A

with no repetitions.Now we define our Boolean Algebra B. It is the Boolean Algebra of subsets of

X = X {x : [, )} generated by the following (recall that a may be

empty)(recall that X = {X : < }):

2(i) the sets {a B : |a| < } {a a : a B, |a| } when <

(ii) {x} for [, )


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(iii) the sets cy (for y Y) where

cy =: x X :for some < we have x X & y Dz}{x : [,

) and y Y and dy

.

Clearly

0 B is a subalgebra ofP(X), including all the singletons hence is atomic;has atoms and elements.

[Why? The least trivial is x X =


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[Why? Easy. The main point concerns (Xa)(Xa) satisfying clause(i) when it has cardinality < this holds by 1(iv) and (X a) cy hascardinality < or (X a)\cy has cardinality < which holds by 1(v).]

4 < X a id(B)[Why? First X a B by clause (i) of2 above, second if X a /id(B) then by 2 for some ordinal we have [a B & |a| < a id(B)], hence by 3 above (X a) is an atom of B/id(B) for large

enough, hence X a belong to id+1(B), contradiction.]

5 for < , B (X a) = B hence if < < then for no c such thatc B, c B X a, |c| < and c B, c B X a, |c| < dowe have B (X a\c) = B (X a\c).[Why? By clauses (f) + (e) of the assumption, the first phrase holds. Thehence follows by clause (g) of the assumption.]

Let J1 be the ideal ofB generated by


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Case 2: y id1(B).Then we can find distinct n < with i(n) = 0 for n < such that n <

n y. Then {Xn : n < } C hence c / J2. So we are done.]

We shall prove that

7 B/J2 is isomorphic to a homomorphic image ofB.

Toward proving 7 let S = {x : [, )} and define a function h as follows: its

domain is {cy : y Y Ymax} and h(cy) = y for y Y Y

max, so h is into B.

Now

() if n1 n < , m1 m < , y0, . . . , yn1 Y Ymax is with no repetitions,

then4:

in B, 1 =:


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As B is superatomic and the choice of Y Ymax clearly by () the statement

7 follows, in fact h induces an isomorphism h from B/J2 onto B. But B issuperatomic and J2 id(B) by 6(i) hence

8 B is superatomic.

Now as {{} : < } are the atoms ofB clearly and recall {X a/J1 : < }are the atoms ofB/J1 by 6(iii) and as J1 [X]


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is constantly . Now by clause (b) of the assumption without loss of generality forsome A A we have A {x : v}. So [A] < is well defined and{x X[A] a[A] : f({x}) B X[A] a[A]} does not belong to I

+[A]

; so by 3

(applied to = [A] and the properties ofc1[A], c2[A]) an easy contradiction.We can conclude that 5 hence v has cardinality < cf() hence |{x X : f(x) =x}| < . If = we are done so assume < .

Now S = {x : [, )} = X\X X satisfies:

9() (b B)(b S infinite

v

b X = 1 rk(b/J1, B/J1)) and

() if S satisfies the property of S in clause (), then |S\S| < [Why? Clause () is proved by inspecting the definition ofB. As for clause(), if |S\S| as S\S X clearly then there is A A such that

{g(i) : i A} S\S. First if =: [A] is well defined then X a B,rk((X a))/J1, B/J1) = 0 < 1 but (X a) S

a is infinite;contradiction. Second if [A] is not well defined then for some < wehave {g(i) : i A} X is infinite and we get a similar contradiction.]

Hence for n = 1, 1 the set Snf =: {x : [,

) and fn({x}) X} has

cardinality < . Let Sf = S1f S

1f.

Also for every y Y letting = rk(y, B) we have cyf(cy) J1, (justrecall that the automorphism that f induced on B/J1 is the identity, and recallthat [d S & d J1 d is finite by 6], hence the symmetric difference of

{{x} : dy}\S

f and {f{x

} : dy}\S

f is finite.

As A = {dy : y Y} is a MAD family of subsets of \ as in clause (k)() of

the assumption, the set { [, ) : f({x}) = {x}} is of cardinality < ; so we

are done.

Not exactly: we have assumed 1.To eliminate this extra assumption we make some minor changes. First without

loss of generality B is a Boolean Algebra of subsets of { : < even} with thesingletons being its atoms. Second, for A A, if possible we choose u = uA as fol-lows, as we can replace uA by any infinite subset, without loss of generality [clause(c) is possible as in the justification of

0above]:

(a) if case () occurs in (b) below theng(A,k) : k u is with no repetitions

(b) either () or () where

() g(A,k) is odd for every k u

() g(A,k) is even for every k u


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(c) if case () occurs in clause (b), then there is a unique y = yA Y such that{g(A,k)} : k u converge to DyA .

Note

() if uA is not well defined then for some finite w we have

{g(A,k) : k < }

w

X.

Now we choose [A] : A A, uA well defined such that:

() [A] : A A, uA well defined is with no repetitions, each [A] is an oddordinal < and if possible it lists all of them.

Clearly without loss of generality B/id1(B) is nontrivial hence Y = so choosey Y. Now we define a function g from B into P() as follows:

g(x) = { < : is even and x or = [A] hence odd,

A A, uA and yA are well defined and x yA / idrk(yA,B)(B) or

is odd but / {[A] : A A, uA and yA are well defined} and

x y / idrk(y,B)(B)}.

Easily g is a homomorphism from B into P() as B is superatomic. Let B bethe Boolean Algebra of subsets of generated by Rang(g) {() : < }. Nowwe just replace B by B P(). 2.1

2.2 Discussion:Why do we use MAD families A []0 and not []1? If we use the latter, we

have to take more care about superatomicity as the intersections of such membersmay otherwise contradict superatomicity.

3 Sufficient conditions for the constructions assumptions

Here we shall show that the assumptions of 2.1 are reasonable. Now in 3.2 weshall reduce the clause (k) of 2.1 to Pr(, ) where Pr formalizes clause (b) there.In 3.3, 3.5 we give sufficient conditions for Pr(, ). In fact, it is clear that (highenough) it is not easy to fail it. In 3.10 we give a sufficient condition for a strongversion of clauses (e) - (f) of 2.1 (and earlier deal with the conditions appearing in


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it). So at least for some cardinals the statement not having the assumptions of2.1 with = + (for simplicity) = 2(), = 3() and such that (h) + (i)+ (j) of 2.1 holds has large consistency strength.

3.1 Definition. 1) Pr(,,) means that and for some A we have:

(a) A []0

(b) A is almost disjoint, i.e. A = B A |A B| < 0(c) |A| =

(d) (B [])(A A)[A B].

2) If we omit we mean some .3) We call A []0 saturated if for every A []0 not almost contained5 in afinite union of members ofA, almost contains a member ofA.

3.2 Fact: 1) Clause (b) of the assumption of 2.1 is equivalent to Pr(,, cf()).2) Clause (k)() + () of the assumption of 2.1 follows from Pr(, , ) & = + 20 .3) IfA []0 is almost disjoint and is saturated then Pr(|A|, , 1).4) If = 0 then Pr(, ) Pr(,,) and = Pr(,,).5) If < 1 2 and Pr(2, ) then Pr(1, ).

Proof. 1) Read the two statements.2) Let A []0 exemplify Pr(, , ). For each A A we can find BA, : (c) strong limit and 2 +

(d) n : n < satisfies the requirements from [Sh 513, 1] or at least theconclusion, i.e.

for every for some n we have: ifa Reg \ and |a| < then sup pcfn-complete(a) .

Then for every :

, we can find {A : < } such that

() each A has the form A,n : n < , it belongs to

n


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1b) Clause (c), i.e. strong limit, is needed just to start the induction.2) Similarly for 3.3(2).

We quote Goldstern Judah and Shelah [GJSh 399] which implies 3.5(1) and (2).

3.5 Claim. Assume CH + SCH + ()(cf() = 0 & 20 < +). Thenthere is a saturated MAD familyA []0 for every uncountable (of cardinality0).

Proof. This is the main result of Goldstern, Judah and Shelah [GJSh 399].

3.6 Definition. Let .

1) LetS

be the class of a = an : n < such that |an| , an an+1, [cf() =0 |an| < ] and = lim supn|an+1\an|. Let S, =: {a : a = an : n < S and an []}.

3) For a S let set(a) = {w : |w| = 0 and w

n


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3) We may replace by a set A (but obviously ,A is equivalent to ,|A| and

,A to ,|A|).

3.8 Fact. 1) Assume > cf() = 0 is strong limit, = {n : n < }, n < n+1and b S,. Then we can find A S such that:

(a) if a A then (n)(m)(an bm) (so a b)

(b) if a A then |an| = n moreover otp(an) = n and an+1 is an end extensionof an

(c) if a A then an : n < is increasing

(d) a1 = a2 then set(a1) set(a2) = (e) if c S is compatible with b then it is compatible with a for some a A.

2) If ( < n)(|| < n = cf(n)) and


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(a) b S,

(b) < set(b) set(b) =

(c) c a()

(d) () = Min{ : a incompatible with b for every < }.

Arriving to choose () by clause (d), we note that < () c =

n

a()n

n

bn has cardinality < , hence we can find b, a() for < 2 with set(b,) :

< 2 pairwise disjoint. So for all but + || of the < 2, b = b, is asneeded.2) After reading [Sh 460] this is easy and anyhow in subsequent work we give fulleranswers.

3) As in [GJSh 399]. 3.9

3.10 Claim. 1) Assume

,, is strong limit, 0 = cf() < and 22 , = 2 and, (from

3.7) holds so = 0 .

Then some B = B : < satisfies clauses (c) - (g) of 2.1; in fact B is asubalgebra ofP() with 2 levels and id cf() = 0 or = 0+ there is no infinite MAD familyA []0 of cardinality < the continuum.

Proof. 1) Let =

n


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(v) i f a, b S, and

n


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2 if i0, i1 < and b0, b1 []

and h is a one to one mapping from b0 onto b1such that Dom(h) h() = , then for some t0 A+i0 , t

1 A+i1 wehave: t0 b0, t

1 b1 and h maps t0 into a co-infinite subset of t1

[why? for some 0 < the set b0 n


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automorphism moving atoms.2) Assume: is uncountable strong limit of cofinality 0, ppJbd () = 2

(see [Sh:g,

Ch.IX,5] why this is reasonable) and = (2)+ 22

, < (2)+, = 2 and

< < Ded+(), e.g. = 2 for = Min{ : 2 > }. Then there is asuperatomic Boolean Algebra of cardinality and atoms, with no automorphismmoving atoms.

3) In part (2) we can replace = (2)+ by = 22

, if some very weak pcfhypothesis (whose negation is not known to be consistent and also of 4), e.g.

() if a is a countable set of regular cardinal then pcf(a) is countable (or just n()).

Proof. 1) We, of course, use Lemma 2.1 with

+

here standing for there, so wehave to show that the assumptions there holds.Clause (a) of 2.1 holds trivially.Clause (b) of 2.1 follows from , (every (A []

)(B A)(B A)) rather

than just (A []+

). There is a sequence B : < satisfying clauses (c) - (g)of 2.1 by 3.10. There is a Boolean Algebra B satisfying clauses (h), (i), (j) of 2.1because < , so there is a tree T with nodes and branches, let Y be aset of branches ofT and let B be the Boolean Algebra of subsets ofT generatedby {a : a T, a is linearly ordered by


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2) The question is whether inductively we can get for many s the parallel of 3.10.3) We can under weak assumptions add , ()0 and demand thatthe Boolean algebra has atoms. For this we need to check condition (k)(). We

probably can omit the demand (

)0

in the generalization of 3.11 indicatedabove, for this we just need to weaken A is MAD in 2.1.

3.14 Claim. 1) Let > 0. A sufficient condition for the existence of a saturatedMAD familyA []0 is

, letting i.e. = Min{|A| : A []0 is an infinite MAD family, then forevery satisfying 20 < 0 we have (a), and 0 < = cf() (b), where

(a), there is a setb Reg \20 of cardinality such that b/[b]|{i < : bi c}|

(b), is regular, S { < : cf() = cf()} is stationary, A = A : S, A , otp(A) = , 2 = 2 A1 A2 finite.

2) Similarly concerning,

Proof. As in [Sh 668].

4 On independence

In the bound 4(), the last exponentiation was really sa() where

4.1 Definition. 1) sa+() = sup{|B|+ : B is a superatomic Boolean Algebra with atoms}.

2) sa() = sup{|B| : B is a superatomic Boolean Algebra with atoms}.3) sa+(, ) = sup{|B|+ : B is a superatomic Boolean subalgebra ofP() extend-ing {a : a finite or cofinite} such that a B |a| < |\a| < }.4) sa(, ) = sup{|B| : B is as in (3)}.5) sa() = Min{ : cf() and if < then sa+(, ) }.

That is, by the proof of Theorem 1.1


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28 SAHARON SHELAH

4.2 Claim. If B is a superatomic Boolean Algebra with no automorphism mov-ing atoms, = cf() > 0 then |B| < sa

+(3(< )), moreover |B|

saharon shelah- superatomic boolean algebras: maximal rigidity

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