Saharon Shelah- Existence of Endo-Rigid Boolean Algebras

8/3/2019 Saharon Shelah- Existence of Endo-Rigid Boolean Algebras

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EXISTENCE OF ENDO-RIGID BOOLEAN ALGEBRAS

In [Sh 2] we, answering a question of Monk, have explicated the notionof a Boolean algebra with no endomorphisms except the ones inducedby ultrafilters on it (see 2 here) and proved the existence of one withcharacter density 0, assuming first 1 and then only CH. The idea wasthat ifh is an endomorphism of B,not among the trivial ones, then thereare pairwise disjoint Dn B with h(dn) dn. Then we can, for someS , add an element x such that dn B with h(dn) dn Then we can,for some S , add an element x such that d x for n S, x dn = 0 forn S while forbidding a solution for {y h(dn) : n S} {y h(dn) =0 : n S}. Further analysis showed that the point is that we are omittingpositive quantifier free types. Continuing this Monk succeeded to provein ZF C, the existence of such Boolean algebras of cardinality 20 . In hisproof he

(a) replaced some uses of the countable density character by the 1-chain condition

(b) generally it is hard to omit < 20 many types but because of the

special character of the types and models involve, using 20 almostdisjoint subsets of , he succeeded in doing this

(c) for another step in the proof (ensuring indecomposability - see Def-inition 2.1) he ( and independently by Nikos) found it is in facteasier to do this when for every countable I B there is x B freeover it.

The question of the existence of such Boolean algebras in other cardinal-ities open (See [DMR] and a preliminary list of problems for the handbookof Boolean Algebras by Monk).

We shall prove (in ZF C) the existence of such B of density character

and cardinality 0 whenever > 0. We can conclude answers tosome other questions from Monks list, (combine 3.1 with 2.5). We use acombinatorial method from [Sh 3],[Sh 4], that is represented in section 1.

In [Sh 1], [Sh 6] (and [Sh 7]) the author offers the opinion that thecombinatorial proofs of [Sh 1], Ch VIII (applied thee for general first ordertheories) should be useful for proving the existence of many non-isomorphic,ad/or pairwise non-embeddable structures which have few (or no) automor-phisms or endomorphism of direct decompositions etc. As an illumination,in [Sh 6] a rigid Boolean algebra in every 0 omitting countable typesalong the way, the method is proved in ZF C, nevertheless it has features

of the diamond. It has been used (so in Gobel and Corner [CG] and Gobeland Shelah [GS1],[GS2]. See more on the method and on refinements of itin [Sh 4] and [Sh 3] and mainly [Sh 5].

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2 EXISTENCE OF ENDO-RIGID BOOLEAN ALGEBRAS

The combinatorial pronciple

Content Let > k be fixed infinite cardinal

We shall deal with the case cf 0, 0 = k, and usually k = 0. Let

M be a function symbols, each with k places, of power . Let (, k)be the vocabulary with function symbols {F i , j : i < j < k} where Fi, jis a j-place function symbol. Let M = M,k(T) the free (, k)-algebra

generated by Tdef= k>(= { : a sequence < k of ordinals < ) (We could

have as well considered T as a set of urelements, and let M be the familyH


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EXISTENCE OF ENDO-RIGID BOOLEAN ALGEBRAS 3

We will be given W = {(f, N) : < ()}, so that every branch of f converges to some (), () non decreasing (in ). We have a freeobject generated by T(B0 in our case) and by induction on we define

B and a, B increasing continuos, such that B+1 is an extension ofB, a B+1 B usually B+1 is generated by B and a is in thecompletion ofB0). Every element will depend on few ( k) members ofT,and a depends in a peculiar way: the set Y T on which it dependsis Y0 Y

1 where Y

0 is bounded below () (i.e. Y

0

> for some < ()) and Y1 is a branch of f

or something similar. See more in 1.8.

1.5 Definition of the Game. We define for W F a game Gm(W),which asts -moves.

In the n-th move:Player II: Choose fn, a tree-embedding of

n, extending


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(2) We call W a barrier if player I does not win in Gm(W) and evendoes not win in Gm(W).

(3) We call W disjoint iffor any distinct (f, N) W( = 1, 2), f1 and

f2

have no common branch.

1.7 The Existence Theorem.

(1) if 0 = k, cf > 0 then there is a strong disjoint barrier.(2) Suppose 0 = k, cd > 0. Then there is W = {(f

, N) : 0 a0 = 0], [(n k) + m > 0 a1 = 0] and except in those cases there are no zeroelements in the partition.

(2) The scheme is simple ifm 0.(3) The endomorphism of the scheme is the unique endomorphism T :

B such that: (i) T z = 0 when x < a0 or x I, < k, orx I0 , < m.

(ii) T z = x when x a1 or x I, k < n or x I1 , < m.

(iii) T(b) = b

when < k.(iv) T(b) = b b when k < n.

(v) T(C) = c c when < m.

2.4 Claim. Ifh is an endomorphism of a Boolean AlgebraB, andB/ExKer(h)is infinite then there are pairwise disjoint dn B(n > ) such thath(dn) dn. By easy manipulation we can assume that h(dn) dn+1 = 0, and if Bsatisfies the c.c.c then {dn : n < w} is a maximal antichain. So, an endo-morphism of a scheme is a trivial endomorphism generated by ideals.

2.5 Lemma.

(1) Every endo-rigid Boolean Algebra B is a Hopfian and dual Hopfian.Even B + B is Hopfian (and dual Hopfian) but not rigid.

Proof Easy to check using 2.2, 2.3.

The Construction

3.1 Main Theorem. Suppose > 0. Then there is a B.A. (BooleanAlgebra) B such that:

(1) B satisfies the c.c.c.(2) B has power 0 and T1(B) = where T1 is the density character.(3) B is endo-rigid and indecomposable.

Proof We concentrate on the case cf() 1 (on the case cf = 0 see[Sh 5, 2, 3]) we shall use Theorem 1.7, and let W = {(f, N) : < },the function , M and T = > be as there.

We will think of the game as follows: player I tries to produce a nontrivial endomorphisms h. Player II supplies (via range (f1)) elements isB0 and challenges player I for defining h on them. So player I plays models229 5 6.10.2003


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Ni in the language Li with a distinguished function symbol h which is anendomorphism of Boolean Algebras. In the end, as W is a strong barrier, wewill get a model N W, in the language i if it is based on {x : J} [i.e.x = nyn, each yn isin the subalgebra generated by {x : J}] and let d(x) be the minimalsuch J. We shall now define by induction on < , the truth valuesof J,, and members , b

n, c

m, d

m,

m of B

c0 such that, letting

B = B0, i i < , i JBc0 :(1) is a branch of Rang (f

), = for < (2) if J, then for some xi < ():

a = m(m d

m) where d

m : m < is a maximal antichain

of non zero elements (of Bc0) md(dm)

>, m xp : m p,p TBc0 , and

m d

m > 0.

(3) if J, then bn, dn N

0 , c

n,

m N

(hence each is based on{x :

>, N}), and bn bm = 0 for n = m.

(4) for < , J, B omits p = {x bn = c

n : < }.

Remark. Many times we shall write < < or w < instead J, w J.

Before we carry the construction note:

3.2 Crucial Fact. : For anyx B there are k, < , and0 < < ksuch that (0) = (1) = (2) = = (k) = , x is based on {x : > or d(m , for some k, m < }.

Stage B Let us carry the construction. For < , w let

I,w = { : w>or m, and there are a branch

of Rang (f) and m N

(m < ) as in 1),2) above, such that ifwe add n


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The desired Boolean algebra B is B. We shall investigate it and even-tually prove it is endo-rigid (in 3.11) and indecomposable (in 3.12) (3.1(1)),3.1(2) are trivial).

Note also3.3 Fact.

(1) For >, x is free over {xn : >, = } hence also over

the subalgebra of Bc0 of those elements based on {x : >, =

}.(2) For every branch of F such that = for < ,(); and

finite there is k such that { : k T} is disjoint toW> {N T : w, + 20 } {d(n ) : n < w}

From 3.2 we can conclude:

3.4 Fact. If < (), < , I T finite then every element of B, basedon I > is in B.

3.5 Notation.

(1) Let B be the set of a Bc0 supported byw>

(2) For x Bc0, < let pr(x) = {a B : x a}.

(3) For < let () = M in{ : () > }.(4) For , let B = {x :

>()} {a : < }.(5) For < let B[] = {x :

W>} {a : () }Bc0

.

3.6 Fact.

(1) B is a complete Boolean subalgebra of Bc0.(2) pr(x) si well defined for x B

c0

(3) if 0 < 1 < , x Bc0 then pr0(pr1(x) = pr0(x).

(4) If < , w T is finite then the function pr,w(x) = {y B

{x : w} : x y} is well defined.

3.7 Fact.(1) For x B , < , the element pr(x) belongs to B[](2) for x B , , , w

>( + 1), the element prw(x) belongs toB(>, w).

Proof We prove this for x B, by induction on (for all )Note that pr(


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Case ii: limit.Trivial as B =

{ : k >} and each

dn(n < ) is based on>0. By case i, we can assume < () hence

w.l.o.g. < 0, and by the induction hypothesis and 3.6(3) it suffices toprove pr0(e1 a) B[]. W.l.o.g. e1 d

m = 0 for m < k and now clearly

pr0(e1 a) = e1 as pr0(e1 dm

m) = e1 d

m for m k, (because

dm, e1 are based on J, >0 J and m is based on > J and is > 0)2) Same proof.

3.8 Lemma. Suppose I, w satisfy:()I,wI

>, w a, I is closed under initial segments, and for every < if m, |I() I| 0 and if w() w,then m, d

m B(I), w()). Let w() w = { : < }, and we define

by induction on a natural number k < , such that the sets { > :

appears is m for some m > k} are pairwise disjoint to I. Now we canextend the identity on B(I, w) to a projection h0 from B(I(), w) ontoB(I, w) such that of < , m > k, then h0(

m d

m ) = 0. Now we

can define by induction on (w() w) {0, } a projection h fromB(I(), w (w() )) onto B(I, w) extending h for < and (w() w {0}). For = 0 we have it defined, for = we get the

conclusion, and in limit stages take the union. In successive stages there isno problem by the choice of h0, and the ks).

3.9 Claim. If B is an uncountable subalgebra of B then there is anantichain{dn : n < } B

and for no x B, xd2n = 0, xd2n+1 = dn+1for everyn provided that

(*) no one countable I > is a support for every a B.

Proof We now define by induction on < 1, d, I, such that:

(i) I >

is countable.(ii)


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By (iii) for each there are 0 a : IBc0 , 1,

2 a :

I+1 IBc0 such that 1

2 = 0,

0

1 d,

0

2 1 d.

By Fodors lemma w.o.l.g. 0 = 0 (i.e. does not depend on ). For

each there is n() < such that0 a : I

n()Bc0 , 1,

2 a : (I+1 I)

n()Bc0Again by renaming w.o.l.g n() = n() for every . Let for n < , dn =

dn


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Now by the definition ofX = pr(x), [ B[] x = 0 x = 0]. As (1 B[], x 1) hence by (i) (for = g(y)d2n):

(iii) g(y) d2n x = 0

Also by the definition of x

= pr(x):1, 2 B[]1x = 2x x

= 2x (as 12 B[], x

1 (1 2)) hence by (ii)(iv) g(y) d2n+1 x

= d2n+1 x.

But dn x, so from (iii) and (iv) (g(y) d2n = 0, (g(y) x)

d2n+1 = d2n+1, and g(y) Ba hence g(y) x Bb. So Bb is

1-compact this contradicits the minimality of , so we finish PartI.

Part II: If B1

is 1-compact B1

B2

, B2

= B1

{z} then B2

is1-compact.

The proof is straightforward. [If dn B2 are pairwise disjoint, let dn =

(d1nz)(d2nz) for some d

1n, d

2n B

1. Now w.l.o.g. D1nd1m = 0 for n = m-

otherwise replace then by d1n which support

every a B. w.l.o.g.I is closed under initial segments and k = |I >

is minimal. Now part I can be applied with B[], {a : w}Bc0 , for anyfinite w I of power < k instead B[] (using 3.7(2) instead 3.7(1)). So byapplying Part I (to B[], {a : w}Bc0) we can add to its conclusion:

d) for every finite w I, |w| < |I >| and x B and x B forwhich {y B : y x} is infinite, there is x1 B

, x1 x such that for noy B[] {a : w}Bc

0

, y x = x1.Now I> is infinite [otherwise let B = {a; I>}Bco , easily

it is infinite and 1-compact by Part II and then we apply Part I: for

I> = {0, . . . , k1} and for u {0, . . . , k 1}, let xudef= {x :

u}{1xx : < k, u} so xu B, 1 = {xu : u {0, . . . , k1}}, hencefor some u, {y B : y xu} is infinite; , xu contradict the conclusion ofPart I.

As B is 1-compact, for any x B such that {y x} is infinite, x

can be splitted in B to two elements satisfying the same i.e. x = x1 x2, x1 x2 = 0, {y B : y x} is infinite for the = 1, 2. Let I> ={ : < }, so we can find pairwise disjoint en B

, such that en =

d2n d2n+1, d2n d2n+1 = 0 and that for no y B[B ] {an : < n}, y (d2n d2n+1) = d2n+1 for every n. So for no n y B[] {an : < n}Bc0 .

As y B clearly y B[+1], but y is based on > {a : < } soby 3.7(2) y B[] {a : < }Bc0 , contradiction to y (d2n d2n+1)0 =d2n+1.229 10 6.10.2003


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Part IV: Let B, satisfy (*) of Part I. by 3.9 for some countable I >,every b B is based on I. By Part III is not a succesor ordinal, sonecessarily cf() = 0, let F i(B

) = {x B : {y B : y x} is finite}.

Next we shall show:(**) for some finite w { : () = } and x B F i(B) for every

y < x from B, for some Z


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omits p). So we shall try to find such which satisfies the requirements inStage B for belonging to J. We assume N = (|N|, h), |N| B, h =h N, h maps N

B onto itself, and N0 contains some elements we

need and somewhat more (see latter). As W is a barrier this is posible.We then will choose , an -branch of f

, distinct from for < [if + 20 this follows, the rest exclude < 20 branches of f but thereare 20 such branches], a maximal antichain dn : n < of B, dn N

0 ,

and n N in x : n < TBc

), and let bn = h(dn), cn =

h(dn n), p = {x bn = cn : n < }, and a = n} (for < ) and

c ( < ), e0, e1, e2, e3 are based on J = { T : k }, where k < also satisfies such that (k) > , k N .

We claim:(*) there is m < such that bm (e1 e2) nkdn = 0

For suppose (*) fail, then as a[ , d] (nkdn) B, w.l.o.g. (e 1 e2) nkdn = 0 (otherwise let

e0 = e0 (e1 a[ , d] nkdn) (e 2 nkdn a[ , d])e1 = e1 nkdn,e2 = e2 nkdn).

So for every M < W, bM (e1 e2) = 0.So ifx realizesp then so does e0, but e0 B contradicting an induction

hypothesis. So (*) holds.Now as dn : n < is a maximal antichain in B, for some < , d

(bm(e1e2nkdn)) = 0. Necessarily > k. So for some {1, 2}, d

bme = 0. As x realizesp, x(dbme) = dcne which is based onJ. But we know that x(db

me) is db

me1a[, d] = db

me1

(if = 1) or d bm e2 (1 a([ , d]) = d b

m e2 1 (if = 2).

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As d bm e z = 0 is based on J, > k, (k) > , is free over J, (see

Fact(2)). Necessarily x (d bm ez) is not based on J, contradiction.

Case II: < < + 20

We shall prove that if , are appropiate (for = 1, 2) and 1 = 2

then p cannot be realized in both B, a[, d]Bc0 . (So as < < +20 ,

there are less then 20 non appropiate ?, 1).As there is a perfect set of appropiate s it will suffice to prove that

for each -branch of Rang(f) for some appropiate 1B, aBc0 omits

p = p[ , d] which will be done in Case III.Note that I = {e B : for some x e for every n x b

n e = c

n e}

is an ideal.The details are easy.

Case III: = This case is splitted into several subcases. Let be any -branch of

f, = whenever < < + 20 . Let I = {d(h(x)); x B}. We

shall assume that |I| 0 I N0 , so in this case p is omitted by

B+1 or B iff ?? omitted by B (by 3.7(1)). As acomplishing this aim iseasier we shall ?? this case (work as in III 4 and use quite arbitrary p).

Subcase III 1.: For some T, and a B ExKer(h) for every

T for some x : < TBc0 , a a = 0 = h( a).

As we are interested not in f, N) itself, but in h, by using Gm(W),w.l.o.g. Range (f). By 3.10 (for rang (h), which by assumption,is infinite) ?? easy manipulations (see 2.4 and [Sh 2]) there is maximalantichain ?? : n < of B such that for no x B, x h(22n = h(d2n)and h(d2n+1) = O W.l.o.g.{dn : n < } N

0

It suffices to prove the conclusion for any -branch ofRange(f),

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Saharon Shelah- Existence of Endo-Rigid Boolean Algebras