Saharon Shelah- Every null-additive set is meager-additive

8/3/2019 Saharon Shelah- Every null-additive set is meager-additive

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Every null-additive set is meager-additive

Saharon Shelah

1. The basic definitions and the main theorem.

1. Definition. (1) We define addition on 2 as addition modulo 2 on each component,i.e., if x,y ,z 2 and x + y = z then for every n we have z(n) = x(n) + y(n) (mod 2).(2) For A, B 2 and x 2 we set x + A =df {x + y : y A}, and we define A + Bsimilarly.(3) We denote the Lebesgue measure on 2 with . We say that X 2 is null-additiveif for every A 2 which is null, i.e. (A) = 0, X + A is null too.(4) We say that X 2 is meager-additive if for every A 2 which is meager also X+ Ais meager.

2. Theorem. Every null-additive set is meager-additive.

3. Outline and discussion. Theorem 2 answers a question of Palikowsi. It will be

proved in 2. In 3 we shall present direct characterizations of the null-additive sets, andin 4 we shall do the same for the meager-additive sets.It is obvious that every countable set is both null-additive and meager-additive. Are

there uncountable null-additive sets, and even null-additive sets of cardinality 20? It willbe shown in 5 that if the continuum hypothesis holds then there is such a set. Haim Judahhas shown that there is a model of ZFC in which all the null-additive sets are countable,but there are in it uncountable meager-additive sets. This is the model obtained by addingto L more than 1 Cohen reals. In this model the Borel conjecture holds, and thereforeevery null-additive set is strongly meager and hence countable. On the other hand, in thismodel the uncountable set of all constructible reals is meager-additive.

2 The proof of Theorem 2.4. Notation. (1) we shall use variables as follow: i,j,k,l,m,n for natural numbers,f , g , h for functions from to , , , ,, for finite sequences of 0s and 1s, x,y ,z formembers of 2, A,B,X,Y for subsets of 2, and S, T for trees.(2) >2 =

n2 with U, V. For >2, U >2and x 2 we shall write + x for + x length(), and U + x for { + x : U}.(3) For , >2 we write if is an extension of .(4) A tree is a nonempty subset of >2 such that

(a) If and T then also T, and(b) If T and n > length () then there is a of length n such that and

T.(5) For a tree T Lim T = {x 2 : for every n < x n T}.(6) A tree T is said to be nowhere dense if for every T there is a >2 such that

Publ. No. 445. First version written in April 91. Partially supported by the Basic Research Fundations ofthe Israel Academy of Sciences and the Edmund Landau Center for research in Mathematical Analysis, supported

by the Minerva Foundation (Germany)

We thank Azriel Levy for rewriting the paper. The reader should thank him for including more proofs and

making the paper self contained.

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and / T. A set B 2 is said to be nowhere dense if B Lim T for some nowheredense tree T.(7) For every x, y 2 we write x y if x(n) = y(n) for all but finitely many n < . ForA 2 Afin =df {y 2 : y x for some x A}.(8) U[] =df { U : or } (read: U through ).

(9) U =df { >2 : U} (read: U above ), and for >2k =df (k + i) : i < length k.(10) For , >2 2 we write n if length () = length() and (i) = (i) forevery n i < length (). For S >2 2 we define Sn = { : n for some S}.

5. Outline of the proof. Let X 2 be null-additive. It clearly suffices to prove thatfor every A 2 which is nowhere dense X + A is meager. Given a nowhere dense treeS we shall prove in Lemma 6 a condition which is sufficient for a tree T to be such thatT+S is nowhere dense. Then we shall split X to a union X =

i=1 Xi such that for each i

Xi Lim Ti where Ti is a tree which satisfies that condition. Thus for a nowhere dense Seach set Xi +Lim S Lim(Ti +S) is nowhere dense, hence X+Lim S i=1 Lim(Ti +S)is meager.

6. Lemma. Let T be a tree such that(a) T is nowhere dense.(b) f = fT is the function from to given byf(n) = min{m : for every n2 there is a m2 such that and / T}.Thus for every sequence of length n there is a witness of length f(n) that T is nowheredense. Obviously for every n < f (n) > n, and ifn < m then f(n) f(m).

Let g be a function from to and n = ni : i < , n = ni : i < increasing

sequences of natural numbers such that(c) fg(i)(ni) n

i < ni+1 for every i < , where f

m denotes the m-th iteration of f.

Then for every tree S which satisfies(d) S is of width (n, g), i.e., for every i < |n

i2 S| g(i),

T + S is nowhere dense.

Proof. Let ni2. We shall show the existence of an ni2 such that and

/ T + S.By (c) there is a sequence m0, . . . , mg(i) such that m0 = ni, f(mk) mk+1 for

0 k g(i) and mg(i) = ni. Let k : k < ki enumerate the set

ni2 S. ki g(i) by (d).We define k mk2 for 0 k ki by recursion as follows. 0 = . Given k mk2,for k < ki, we shall define k+1 mk+12 so that for no extension n

i2 of k+1 we

shall have + k T. We have k + k mk mk2 and by the definition of f and bythe choice of the mks k + k mk has an extension mk+12 such that / T. If wetake k+1 = + k mk+1 then k + k mk implies k k+1 , k+1

mk+12 andk+1 + k mk+1 = / T, and therefore for every n

i2 such that k

we have + k / T. Let = ki , and assume that

T + S. Then, for some k < ki g(i) + k T, contradicting our choice of k+1 = mk+1. Thus / T + S.

7. Lemma. If S, Ti, i are trees and Lim S

i Lim Ti then for some S andj S[] Tj .

Proof. Suppose that this is not the case, i.e., for every S and i < there is a

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such that S[] and / Ti. Once there is such a we can assume that andlength > length . We define now, by induction on i, i and ki so that ki = length i,k0 = 0, 0 = , i i+1, ki < ki+1, i+1 S and i+1 / Ti. Let y =

i i.

Since i S for every i y Lim S

i Lim Ti, hence for some j y Lim Tj .

However, y kj+1 = j+1 / Tj , contradicting y Lim Tj .

8. Lemma. Let S and T be trees such that Lim S (Lim T)fin. Then there are k < ,, k2, S such that S T.Proof. For n < , 1, 2

n2 and 2 T we defineT1,2 =

df { : 1} {1 : 2 T} (This is the tree T[2] with 2 replacedby 1). Clearly

(1) (Lim T)fin =

n 0.There is a tree T such that (Lim T) > 0, for every T (Lim(T[])) > 0, and((2 \ (Lim T)fin) + X) Lim T = , and then X =

T, length =length Y, where

Y, = {x X : xlength + T[] T}.Proof. Since (Lim T) > 0 then, as easily seen, ((Lim T)fin) = 1, hence(2 \ (Lim T)fin) = 0. Since X is null-additive also (X + (2 \ (Lim T)fin)) = 0. Hencethere is a tree T such that (Lim T) > 0 and (X + (2 \ (Lim T)fin)) Lim T = .Without loss of generality we can assume that T has been pruned so that for T(Lim T[]) > 0.

Let x X then 2 \ (x + (Lim T)fin) = x + (2 \ (Lim T)fin) X+ (2 \ (Lim T)fin).Hence (2 \ (x + (Lim T)fin)) Lim T (X + (2 \ (Lim T)fin) Lim T = , i.e.,Lim T x + (Lim T)fin, and therefore Lim(x + T) = x + Lim T (Lim T)fin. ByLemma 8 there are T and length 2 such that xlength + T T. Let = + , then + = and therefore xlength + T[] T[] T, hence x Y, .10. Lemma. Let X be null-additive, and let n = ni : i < , n

= ni : i < besuch that for every i < ni < n

i and n

i + i 2

ni ni+1, then we can represent X as

m 0)(k uj,ynj ) y(k) = 1}.

B is clearly a closed subset of 2 hence for T = {y n : y B n } B = Lim (T).

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The properties of T in which we are interested are(B0) T n12.(B1) For each T n

i2T[] ni+12 = 2(ni+1ni)(1 2i).

(B2) If , 0, . . . , k1 ni2, +0 , . . . ,

+k1

ni+12, + l T, l +l for l < k and

0

, . . . , k1

is with no repetitions then {

+ : +

ni+12, (

l < k)(+ + +l

T)}

2ni+1ni

1 2i

k.

(B3) For every ni2 we have : ni T implies T.

These properties can be established by an obvious counting argument.By (B0), (B1) and (B3) we have

(Lim T) =

i=1

{x 2 : x ni T}

= ({x 2 : x n1 T})

i=1

({x 2 : x ni+1 T})

({x 2 : x ni

T})

= 1

i=1

|T ni+12|/2ni+1

|T ni+12|/2ni=

i=1

(1 2i) > 0

For the T which we constructed let T and Y, be as in Lemma 9. For length 2let Y,, = {y Y, : y length = }. Clearly

(2) X =

T, length =length =length Y,,

Since there are only countably many Y,,s they can be taken to be the Xms we arelooking for, provided we show that every such Y,, is a subset of Lim(S) for some treeS of width n, ga for some real 0 < a < 1. We shall see that this is indeed the case if wetake S = {y m : y Y,,, m < } and a = (T

[]). a > 0 by what we assumed aboutT. As, obviously, Y,, Lim(S) all we have to do is to show that S is of width n, ga.

We can choose a set W S nj+12 such that the function mapping W to njis one to one and onto S n

j2

We fix now , , and denote Y,, by Y and the length of , , by n. Let z 2

be such that z n = + and z(i) = 0 for i n. Then for every y such that y n = wehave y + z = yn. Therefore, by the definition of Y we have

(3) Y = {y 2 : yn = , (yn)+T[] T} = {y 2 : yn = , y+z+T[] T}

for every y Y there is a unique W such that nj = y nj ( may be y nj+1).Clearly |W| = |Sn

j2| and we denote |W| with s, so it suffices to prove s ga(j). Ifn

j n

then the only member ofS nj2 is nj hence s = 1, so s ga(j). We shall now deal with

the case where nj > n. Let 0, . . . s1 be the members ofW. For m < s m = y nj+1 forsome y Y, hence, by (3), m + z + T

[] T and therefore (z + T[]) nj+12 m + T.Since this holds for every W we have

(4) z + T[] nj+12

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Let us find out the size of

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(c)(d). Let S be a tree almost of width f. Then for some k we have |T n2| f(n) forall n k. By (1) of Lemma 8 Lim (S)fin =

1,2k2, 2S Lim(S1,2). Each S1,2 is of

width f since for n k we have |S1,2 n2| = 1 and for n > k we have |S1,2

n2| |Sn2| f(n). Therefore, ifX

m2 by defining S m2 by recursion on m. S 02 = {}.For ni < m ni+1 let

S m2 = { m2 : ni Sni2 and S

njj for some j < i j = 0}.

S can be easily seen to be a tree, and clearly Lim (S)fin n 0 and > 0 there is an N suchthat for every n N there is a t n2 T such that |t| 2n(a ) and for each t(Lim(T[]) > 2n(1 ).

Using Lemma A one can prove

Lemma B. For every tree T with (Lim(T)) > 0, every > 0 and every sequencei : 0 < i < of positive reals there is a subtree T of T and an increasing sequenceni : i < of natural numbers such that n0 = 0, (Lim(T

)) > (Lim (T)) and

(6) for i > 0 and every ni2 T (Lim (T[])) > 2ni(1 i)

By basic mesure theory (Afin) = 0 so there is a tree T such that (Lim(T)) > 0 andLim(T) Afin = hence Lim (T)fin A = . Given < (Lim(T)) and i : i < as inLemma B we obtain a subtree T of T as in that lemma with (Lim (T)) > 0. The unionof sufficiently many finite translates of T, i.e., trees T1,2 as in (1) of Lemma 8 is atree T satisfying (6) with (Lim(T)) 12 . Lim (T

)fin = Lim (T)fin Lim(T)fin andhence Lim (T) A Lim(T)fin A = . We take now i = 14(i+1)3 and take T to be Tand we get (Lim (T)) 12 and

(7) for i > 0 and every ni2 T (Lim (T[])) > 2ni(1 1

4(i + 1)3)

Let f be the corset given by f(n) = i +1 for ni n < ni+1. By (d) there are trees Smof width f such that X

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We define

T = { >2 : + T for every S of the same length as }

We do not show that T

is a tree but obviously if T

then T

, thusLim(T) is defined. If (Lim (T)) > 0 then, by a well-known property of the mea-sure, (Lim (T)fin) = 1, hence in order to prove (Lim(S) + A) = 0 it suffices toprove (Lim (S) + A) Lim(T)fin = . Assume y (Lim(S) + A) Lim(T)fin. Sincey Lim(T)fin there is a y 2 such that y(n) = y(n) for all sufficiently big ns andy Lim(T). Since y Lim(S) + A there is an x Lim(S) such that y + x A, hencey + x / Lim(T)fin, hence y + x / Lim(T). Therefore, for some n y n + x n / T, hence,by the definition of T, y n / T contradicting y Lim(T).

We still have to prove that (Lim(T)) > 0. We shall prove, by induction on i, that

(8) ni n ni+1 |(T \ T) n

2| 2n

j 0 we have, by the induction hypothesis |T \ T ni2| 2ni

j


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Let ni : i and T be as in Lemma 14. As in the proof of Lemma 10 we get(Lim T) > 0. Let T and Y, be as in Lemma 9 and let Y,, , S and z be as in theproof of Lemma 10. All we have to do is to show that S is almost of width f. Let us fix, and . We shall now see that

(9)If T[] ni2 then

|{+ : + T ni+12}|/2(ni+1ni) (1 2(i+1))|Sni2|1

Let + T[] ni+12, then, by the definition of S (see (3)), if + S ni+12 then+ + + + z T. Thus{+ : + T ni+12} {+ : + ni+12, (+ S)+ + + + z T}.Let us take in (B2) = , k = |S ni2|, {l : l < k} = S ni2, {+l : l < k} S

ni+12,and for l < k +l ni = l, l = l + z,

+l =

+l + z, hence l =

+l ni for l < k. Since

for l < k +l + z = +l S

ni+12 we have

{+

: +

ni+1

2, (+

S)(+

+ +

+ z T} {+ : + ni+12, (l < k)(+l + + T)},

therefore by (B2)|{+ : + ni+12, (+ S)(+ + + + z T}| 2ni+1ni(1 2(i+1))|S

ni2|1,which establishes (9).

(9) tells us how T grows from the level ni to the level ni+1 and therefore|T ni2| 2ni

j 0 and we can assume c0 < 1. Then

< log c0 log(|T ni2| 2ni)

j c 2j + 1 |S nj2|, hence S is almost of width f.

Lemma 14 follows immediately from the following Lemma.

15. Lemma. For every n and 0 < p < 1 there is an N > n such that for everyn N and t n2 there is a t n

2 which satisfies the following (i)(iii).(i) For each t n t.(ii) For each t |t[]| 2n

n p.(iii) If 0 < k 2n , 0, . . . , k1 n2, +0 , . . . ,

+k1

n2, +j = +l for j < l < k,

+l t, l = +l n for l < k then |{

+ : + n

2, (l < k)+++l t}| 2n

npk1.

Proof. We shall prove the lemma by the probabilistic method. Let n > n and letA = {+ n

2 : + n t}. We construct a subset A of A as follows. We take a coinwhich yields heads with probability p. For each + A we toss this coin and we put +

in A iff the coin shows heads. We shall see that if we take t = A then, for sufficientlylarge n, the probability that (ii) holds has a positive lower bound which does not dependon n while the probability that (iii) holds is arbitrarily close to 1. Hence there is an Nand a t as claimed by the lemma. We prove first two lemmas.

Lemma 16. For k , , 0, . . . , k1, +0 , . . . , +k1 as im Lemma 15 there are reals c1, c2 > 0

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which depend only on p, n and k such that

Pr

|{+ : + n

2,l 0.

(10)Pr

|j < m :l


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Therefore the event |{j < m :

l m(pk + d) is incompatible with (11),so we continue the inequality (10) by

(12)

Pr

ik2, |Bi| dm2k2+2|{j Bi :

l |Bi|(pk +d

2)

ik2, |Bi| dm

2k2+2

Pr

|{j Bi :l |Bi|(pk +d

2)

For a fixed j < m the events +j + +l A

for different ls are independent hence

Pr

l


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where k , , 0, . . . , k1, +0 , . . . , +k1 are as in (iii) of Lemma 15.

Proof. By our requirements on k , , 0, . . . , k1, +0 , . . . , +k1 there are at most 2

n possi-

ble ks and s and (2n

)2n

sequences +0 , . . . , +k1, while 0, . . . , k1 are determined by

+0 , . . . , +k

1 and n. Therefore we get, by Lemma 16,

Pr

k,,0,...,k1,+0 ,...,

+k1

|{+ : + n

2, (l < k) + + +l A}| 2n

npk1

k,,0,...,k1,+0 ,...,

+k1

Pr

|{+ : + n

2, (l < k) + + +l A}| 2n

npk1

2n 2n (2n

)(2n) c1e

c22n

Proof of Lemma 15 (continued). For each + n

2 such that + + A if the

coin shows heads and different tosses are independent |A[]

| is a binomial random variablewith expectation 2nnp. By the central limit theorem of probability theory (see, e. g.,

Feller [1, Ch. 7]) the limit, as n , of Pr

|A[]| 2nnp

is

012

ex2

2 dx = 12

,

hence there is an N such that for every n N Pr

|A[]| 2nnp

13 . For different

t the random variables |A[]| are idependent, hence

(14) Pr

t(|A[]| 2n

np)

1

3|t|

1

32n

The right-hand side of (13) clearly vanishes as n , let us take N to be such that forn N the right-hand side of (13) is < 32n . Therefore we have, by (13) and (14),

(15)

Pr

k,,0,...,k1,+0 ,...,

+k1

(|{+ : + n

2, (l < k) + + +l A}| < 2n

npk1)

t

(|A[]| 2nnp)

> 0

By (15) there is a t as required by the lemma.4 Characterization of the meager-additive sets

18. Theorem. For every X 2 the following conditions are equivalent:a. X is meager additive.b. For every sequence n0 < n1 < n2 < . . . of natural numbers there is a sequencei0 < i1 < . . . of natural numbers and a y 2 such that for every x X and for everysufficiently big k < there is an l [ik, ik+1) such that x [nl, nl+1) = y [nl, nl+1).

Proof. Throughout this proof, if x 2 >2, k, l and k < l then x [k, l) willdenote the sequence lk2 such that (i) = x(k + i) for all i < l k.

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(b)(a). In order to prove (a) it clearly suffices to show that X + Lim T is meager forevery nowhere dense tree T.

For a nowhere dense tree T let ni : i < be an ascending sequence of naturalnumbers such that n0 = 0 and for every i there is a sequence i ni+1ni2 suchthat for every ni2 i / T. Let ij : j < and y be as in (b), then, by (b),X =

k Xk where Xk = {x X : (m k)(l [im, im+1))x[nl, nl+1) = y (nl, nl+1).

It clearly suffices to prove that Xk + Lim T is nowhere dense.

Let nim2 for some m k; we shall show that has an extension which is not inXk + Lim T. Let = im im+1 . . . im+11 and let = y [nim, nim+1) + . Weshow that no extension z of is in Xk +Lim T. Suppose z Xk + Lim T thenz = x + w, x Xk w Lim T. Therefore = 1 + 2 and = 1 + 2 such that 11xand 2 2 w, hence 2 2 T. Let nim2 be such that (j) = 0 for every j < nim,and let = , 1 = 1,

2 = 2. Clearly

= 1 + 2. Since x Xk there is,

by (b), an l [im, im+1) such that x [nl, nl+1) = y [nl, nl+1). Since 1 1 x we have1 [nim, nim+1) = x [nim , nim+1) and hence 1 [nl, nl+1) = x [nl, nl+1) = y [nl, nl+1)

Therefore, by the definition of and

y [nl, nl+1) + 2 [nl, nl+1) =

1 [nl, nl+1) +

2 [nl, nl+1) =

[nl, nl+1)= y [nl, nl+1) + l

hence 2 [nl, nl+1) = l. By the definition of l 2 2 / T, contradicting 2 2 T.

(a)(b). Let X be meager-additive. Let ni : i < be an ascending sequence of naturalnumbers. Let B = {x 2 : j(k [nj, nj+1)) x(k) = 0} and T = {x n : x B, n }.Clearly B = Lim T is nowhere dense, so X + Lim T is meager, hence there are nowhere

dense trees Sn, n such that for every n Sn Sn+1 and X + Lim T

n Sn. Wedefine now the sequences il : l < , which is an ascending sequence of natural numbers,and l : l < by recursion as follows. i0 = 0. Given il let l and il+1 be such thatl

nil+1nil2 and for every nil2 l / Sl; there are such l and il+1 since Sl isnowhere dense. Let y 2 be given by y [nil , nil+1) = l for every l < . We shall nowprove that il : l < and y are as required by (b).

Let x X, so Lim(x + T) = x + Lim T X + Lim T

n Sn. Therefore,by Lemma 7 (where we take x + T for S) there is an T and n such thatx + T[] Sn. Let k be such that k n and ik length . By x + T

[] Sn we havex nik+1 + (T

[] nik+1 2) Sn Sk. Thus for every T[] nik+12 x nik+1 + Sk,

hence, by the definition ofk and y, x[nik , nik+1)+[nik , nik+1) = k = y[nik , nik+1) and

therefore x[nik , nik+1)y[nik , nik+1) = [nik , nik+1), i.e., x[nik , nik+1)y[nik , nik+1) /{ [nik , nik+1) : T

[]}. Since ik > length this can happen, by the definition of T,only if for some ik j < ik+1 x [nj , nj+1) y [nj , nj+1) is identically zero, and this iswhat we had to prove.

5. An uncountable null-additive set.

19. Theorem. If the continuum hypothesis holds then there is an uncountable null-additive set.

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Proof. Let f : < 1 be a sequence containing all corsets and let T : < 1 bea sequence containing all perfect trees. Let E be the set of all limit ordinals < 1 suchthat for every , < and n < there is a < such that

T T, T n2 = T

n2 and for all m |T m2| max(|T

m2|, f(m))

Clearly E is closed. For every , < 1 there is a perfect tree T such that T T,T n2 = T n2 and for all m < |T m2| max(|T n2|, f(m)). This tree T is Tfor some < 1. By a simple closure argument this implies that E is unbounded.

We need now the following lemma which will be proved later.

20. Lemma. There is an increasing and continuous sequence : < 1 of ordinalsin E such that for every < 1, k < and < there is an ordinal which is good for( ,,k), where by is good for ( , , k) we mean that

(16)

(i) < +1

(ii) T T, T k2 = T k2(iii) for all such that > and for every < , there is a <

such that T T T and T is almost of width f

For < 1 let be the which is good for (, 0, 0). We choose Lim T \ { : < },and let X = { : < 1}. X is clearly uncountable. We shall prove that X is null-additiveby proving that X satisfies condition (c) of Theorem 13. For a given corset f f = f forsome < 1. Let < 1 be such that > . Let Z = { < +1 : T is almost of width f}.We shall see thatX { : } Z Lim T . Since Z and are countable condition (c) of Theorem 13holds.

Let > , it suffices to prove that Lim T for some Z. < and since is good for = k = 0 hence there is a < such that T T , and T is of width f.Thus Z and Lim T Lim T .

Proof of Lemma 20. We define : < 1 as follows. 0 is the least member of E.For a limit ordinal =


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(f) kn+1 > kn and every Tn+1 kn2 has at least two extensions in Tn+1

kn+12.There are indeed such sequences n : n < and kn : n < . (a) determines k0 and 0;if < 1 then there is an 0 as in (a) by the induction hypothesis. in E and let ustake n, n, kn, n+1 for ,,n , in the definition of E, then E says that there is ann+1 which satisfies (b)(e). Since Tn+1 is perfect there is a kn+1 as in (f).

Let T =

n Tn . By (c),(d),(f) T is a perfect tree, hence it is T for some < 1.Since T, and therefore also , depend on and k we denote with (, k). As is easilyseen T(,k) T, T(,k)

k2 = T k2, and for every < T(,k) Tl+1 T,where l is such that = l. This means that (iii) of (16) holds for = . We shall have toshow that (iii) holds for < and to deal with the case where is a limit ordinal.

If is a limit ordinal let n : n < be an increasing sequence such that 0 > andn

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