Andrzej Roslanowski and Saharon Shelah- The Yellow Cake

8/3/2019 Andrzej Roslanowski and Saharon Shelah- The Yellow Cake

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THE YELLOW CAKE

ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

Abstract. In this paper we consider the following property:(Da) For every function f : R R R there are functions g0n, g

1n : R R

(for n < ) such that

(x, y R)(f(x, y) =X

n


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2 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

Notation 0.2. (1) For two sequences , we write whenever is a properinitial segment of , and when either or = . The length of

a sequence is denoted by g().(2) The set of rationals is denoted by Q and the set of reals is called R. The

cardinality of R is called c (and it is refered to as the the continuum).The dominating number (the minimal size of a dominating family in in the ordering of eventual dominance) is denoted by d and the unboundednumber (the minimal size of an unbounded family in that order) is calledb.

(3) The quantifiers (n) and (n) are abbreviations for

(m )(n > m) and (m )(n > m),

respectively.(4) For a forcing notion P, P stands for the canonical Pname for the generic

filter in P. With this one exception, all Pnames for objects in the extensionvia P will be denoted with a dot above (e.g. A, f).

1. Fsweet forcing notion

Definition 1.1. An uncountable family F is spread if

() for each k, n < and a sequence f,n : < 1, n < n of pairwise

distinct elements of F there are an increasing sequence i : i < 1and an integer k > k such that

(i < )(n < n)(fi,n(k) < fi+1,n(k)).

Remark 1.2. (1) Note that if an uncountable family F has the propertythat its every uncountable subfamily is unbounded on every K [] thenF is spread.

(2) If is uncountable and one adds many Cohen reals c : < then {c : < } is a spread family.

(3) If there is a spread family then b = 1 (so in particular MA2(centered)fails).

Definition 1.3. Let F be a spread family. A forcing notion P is Fsweet ifthe following condition is satisfied:

()Fsweet for each sequence p : < 1 P there are A []1, k < and a

sequence f,n : n < n, A F such that (, n) = (, n) f,n =

f,n and() if i : i < is an increasing sequence of elements of A such that for

some k (k

, )

(i < )(n < n)(fi,n(k) < fi+1,n(k))

then there is p P such that p (i )(pi P).

Proposition 1.4. Assume that F is a spread family and P is an Fsweet forcing notion. Then

P F is a spread family .

Proof. First note that easily Fsweetness implies the ccc.


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THE YELLOW CAKE 3

Suppose that k+ < , f,n : < 1, n < n+ are Pnames for elements of F,

p P and

p P (, < 1)(n, n < n+)((, n) = (, n) f,n = f,n).

For < 1 choose conditions p p and functions f,n F (for n < n+) such

that p (n < n+)(f,n = f,n). Passing to a subsequence, we may assume that

(, n) = (, n) f,n = f,n .

Choose k > k+, a set A []1 and a sequence f,n : A, n+ n < nas guaranteed by ()Fsweet of 1.3 for p : < 1 (note that here, for notationalconvenience, we use the interval [n+, n) instead of n there). Shrinking the setA and possibly decreasing n (and reenumerating f,ns) we may assume that allfunctions in appearing in f,n : A, n < n

are distinct. By () of 1.1 we findk > k and an increasing sequence i : i < A such that

(i < )(n < n

)(fi,n(k) < fi+1,n(k)).But it follows from () of 1.3 that now we can find a condition q P such thatq (i )(pi P). As all conditions p are stronger than p we may demandthat q p. Now use the choice of the pi s and fi,n (for n < n

+) to finish theproof.

Theorem 1.5. Assume F is a spread family. Let P, Q : < be a finitesupport iteration of forcing notions such that for each < we have

(1) P F is spread , and(2) P Q is Fsweet .

ThenP is Fsweet (and consequently, P F is a spread family ).

Proof. We show this by induction on .Case 1: = + 1Let p : < 1 P+1. Take a condition p

P such that

p P { < 1 : p P} is uncountable

(there is one by the ccc). Next, use the assumption that Q is Fsweet and get

Pnames A [1]1 and k, n and f,n : A, n < n F such that thecondition p forces that they are as guaranteed by ()Fsweet of 1.3 for the sequencep() : < 1, p P .

Let A be the set of all < 1 such that there is a condition stronger thanboth p and p which forces that p() is in A. Clearly |A| = 1. For each A choose a condition q P stronger than both p and p which forces

that p() A and decides the values of k

, n

and f,n : n < n

. Next we maychoose A [A]1 , k, n and f,n : A

, n < n F such that (for each

A and n < n) q k = k & n = n & f,n = f,n . Moreover we may

demand that the f,ns are pairwise distinct (for A, n < n).Apply the inductive hypothesis to the sequence q : A (and P ) to get

A [A]1 , k+, n+ > n and f,n : A, n n < n+. For simplicity wemay assume that there are no repetitions in the sequence f,n : A, n < n(we may shrink A and decrease n reenumerating f,ns suitably). We claim thatthis sequence and max{k, k+} satisfy the demand in () if 1.3. So suppose that


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i : i < is an increasing sequence of elements ofA such that for some k > k, k+

we have

(i < )(n < n+

)(fi,n(k) < fi+1,n(k)).Clearly, by our choices, we find a condition p+ P stronger than p such thatp+ (i )(qi P ). Next, in V

P , we look at the sequence pi() : qi P , i < . We may find a P name p

+() such that (p+ forces that)

p+() Q (i )(qi P & pi() Q ).

Look at the condition p+p+().

Case 2: is a limit ordinal.If p : < 1 P then, under the assumption of the current case, for some

A [1]1 and < , the sets {supp(p) \ : A} are pairwise disjoint. Apply

the inductive hypothesis to P and the sequence p : A.

Conclusion 1.6. Suppose that > 1 is a regular cardinal such that


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THE YELLOW CAKE 5

(1) A approximation is a sequence g = g : < 2, > such that:

(a) g : Q (for < 2, >),

(b) if < then ( < c)(k )(gk() = 0 & k ),(c) if < , > and neither nor g() is an initial segment of ,

then g0() = g1() = 0.

(2) If 0 < 1 and gk = g,k : < 2,

> (for k = 0, 1) are k

approximations such that g,0 g,1 (for all < 2 and

>) then we

say that g1 extends g0 (in short: g0 g1).(3) We say that a approximation g agrees with the function f if

(, < )

f(, ) =

>

g0() g1() and the series converges absolutely

.

Proposition 2.2. If g are approximations (for < ) such that the sequence

g : < is increasing and =

be a approximation which agrees with f .

We define a forcing notion Pg,f as follows:

a condition is a tuple p = Zp, jp, rp, : < 2, jp> such that

() jp < and Zp is a finite subset of , rp, Q (for < 2, jp>),

() the set { jp> : rp0, = 0 or r

p1, = 0} is finite, and if

jp> and

neither nor g() is an initial segment of then rp, = 0,

() if Zp then

|f(, )

{g0() rp1, : jp

>}| < 2jp

,|f(, )

{rp0, g

1() :

jp>}| < 2jp

, and

|f(, )

{rp0, rp1, :

jp>}| < 2jp

(note that by demand () all the sums above are finite),() if , Zp {} are distinct then (, ) < jp;

the order is defined by p q if and only if

(a) jp jq, Zp Zq and rp, = rq, for

jp>, < 2,


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(b) if Zp then

{|r

p

0, g1()| : j

q

> \j

p

>} < 41 2j

pjq

2jp1 ,{|g0() r

p1, | :

jq> \jp>} < 4

1 2jpjq

2jp1, and

{|rp0, r

p1, | :

jq> \jp>} < 4

1 2jpjq

2jp1

.

Proposition 2.4. Suppose that < c, f : ( + 1) ( + 1) R and g is aapproximation that agrees with f . Then:

(1) Pg,f is a (non-trivial) Fsweet forcing notion of size || + 0.

(2) InVPg,

f , there is a(+1)approximation g such that g g and g agreeswith f.

Proof. (1) First note that (Pg,f , ) is a partial order and easily Pg,f = (re-

member that Zp may be empty). Before we continue let us show the following claimthat will be used later too.

Claim 2.4.1. For each j < , < and the sets

Ijdef= {p Pg,f : j

p j},

Idef= {p Pg,f : Z

p}, and

Ijdef= {p Pg,f : j < j

p & ( < 2)(k (j,jp))(rp,j = 0 & j )}

are dense subsets ofPg,f .

Proof of the claim. Let j < , < , and p Pg,f .

If j j

p

then p I

j

, so suppose that j

p

< j. Let m : m < m

enumerate Z

p

.Choose pairwise distinct j,m : < 2, m < m (j,) such that j,m and

gmj,m(m) = 0 (remember 2.1(1b)). Fix j > j such that j is not an initial

segment of any m (for m < m). Let jq = j + max{j,m : < 2, m < m} + j,

Zq = Zp and define rq0,, rq1, as follows.

(1) If jp> then rq, = r

p,.

(2) If jq> \j

p> \ {mj,m : m < m} \ {j

}, < 2 then rq1, = 0.

(3) If = j then rq0, , r

q1, Q \ {0} are such that |r

q0, r

q1, | < 2

jp and

|f(, )

{rp0, rp1, :

jp>} rq0, rq1, | < 2

2jq .

(4) If = mj0,m, m < m then rq1, Q is such that |g

0(m) r

q1,| < 2

jp

and|f(m, )

{g0 (m) r

p1, :

jp>} g0(m) rq1, | < 2

2jq ;

if = mj1,m, m < m then rq0, Q is such that |r

q0, g

1(m)| < 2

jp

and

|f(, m)

{rp0, g1 (m) :

jp>} rq0, g1(m)| < 2

2jq .

One easily checks that q = Zq, jq, rq, : < 2, jq> is a condition in Pg,f

stronger than p (and q Ij ).


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THE YELLOW CAKE 7

Now suppose that / Zp. Take j0 > jp such that ( Zp {})((, ) < j0).

Let m : m < m enumerate Zp {} and let j,m : < 2, m < m (j0, ) be

pairwise distinct and such that j,m & gmj,m(m) = 0. Let j

> jp

be suchthat j is not an initial segment of any m . Put Z

q = Zp{}, jq = jp+max{j,m : < 2, m < m} + j, and define rq, like before, with one modification. If m =

and = j0,m then rq1, Q is such that |f(, ) g

0() r

q1,| < 2

2jq ; if m =

and = j1,m then rq0, Q is such that |f(, ) r

q0, g

1()| < 2

2jq .

Similarly one builds a condition q Ij stronger than p (just choose j suitably).

Now we are going to show that Pg,f is Fsweet. So suppose that p : <

Pg,f . Choose A [1]

1 such that

Zp : A forms a system with kernel Z, for each , A, |Zp | = |Zp |, jp = jp and

rp, : < 2, jp> = r

p, : < 2,

jp>

(remember 2.3()), if , A and : Zp Zp is the order preserving bijection then Z

is the identity on Z and ( Zp)(jp = ()j

p ).

Let k = jp , n = |Zp \ Z| for some (equivalently: all) A. For A letf,n : n < n enumerate { : Zp \ Z}. Clearly there are no repetitions inf,n : n < n, A. We claim that this sequence is as required in () of 1.3. Sosuppose that i : i < A is an increasing sequence such that for some k > k

we have

(i < )(n < n)(fi,n(k) < fi+1,n(k)).

Passing to a subsequence we may additionally demand that for each m < k, for everyn < n, the sequence fi,n(m) : i < is either constant or strictly increasing.For n < n let kn j

p be such that the sequence fi,nkn : i < is constantbut the sequence fi,n(kn) : i < is strictly increasing. Take j > k such thatif m fi,nkn, n < n

then m < j. Fix an enumeration m : m < m ofZp0 (so m = |Z| + n) and choose j, j,m > j + 2 with the properties as in thefirst part of the proof of 2.4.1 (with p0 in the place of p there). Put Z

q = Zp0

and define jq, rq, exactly as there (so, in particular, for each j> \

jp0>

we have rq, = 0). We claim that q (i )(pi Pg,

f). So suppose

that q q, i0 < . Choose i > i0 such that for each n < n and k > kn, if

m = fi,nk then m > jq

. Moreover, we demand that if kn < k < jq

, n < n

then rq

0,fi,nk

= rq

1,fi,nk

= 0 (remember 2.3()). Then we have the effect that

( jq> \j

pi>)( < 2)( Zpi \ Z)(rq

, g1() = g

0() r

q

1, = 0)).

So we may proceed as in the proof of 2.4.1 and build a condition q+ stronger thanboth q and pi .

(2) Let G Pg,f be generic over V. For > define

g, () = rp, where p G I

g()+1,

g, () = g() for < .


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It follows immediately from 2.4.1 (and the definition of the order on Pg,f ) that the

above conditions define a + 1approximation g = g, : < 2, > which

agrees with f and extends g.

Theorem 2.5. Assume that is an uncountable cardinal such that (by

finite approximations; so, in VQ0 , c = and the family F = { : < } isspread; we use it in the clauses below),

(2) < , f is a Pname for a function from ( + 1) ( + 1) to R, g isa Pname for a approximation (for the family F added by Q0) whichagrees with f( ),

(3) P1+ Q1+ = Pg,f

(for F),

(4) if f is a P name for a function from ( + 1) ( + 1) to R, < and g

is a Pname for a approximation which agrees with f( ) then forsome < , > we have

g = g, f = f, = .

Clearly P is a ccc forcing notion (with a dense subset) of size . It follows from2.4(2), 2.2 that P (

Da) (and clearly P c = ). Moreover, by 2.4(1), 1.5 weknow that, in VQ0 , for each [1, ] the forcing notion P[1, ) is Fsweet, so

P F is a spread family of size (by 1.4).

3. When (Da) fails.

In this section we will strengthen the result of [Sh 675, 3.6] mentioned in 0.1(2)giving its combinatorial heart.

Definition 3.1. (1) For a function h such that dom(h) X Yand rng(h) Z and a positive integer n we define

(h, n) = min{|A0| + |A1| : A0 P(X) & A1 P(Y) &(w [X]n)(A A0)(w A) &(w [Y]n)(A A1)(w A) &

(A0 A0)(A1 A1)(h[A0 A1] = Z)}.

If X = Y and h is as above, and n is a positive integer then we define

(h, n) = min{|A| : A P(X) & (w [X]n)(A A)(w A) &(A A)(h[A A] = Z) }.

(2) For c = cn : n < R and d = dn : n < R let h(c, d) =n


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THE YELLOW CAKE 9

Definition 3.2. For a function h : R R R let (Dah ) mean:

(Dah ) For each f : R R R there are functions g0n, g

1n : R R (for n < )

such that

(x, y R)

f(x, y) = h(g0n(x) : n < , g1n(y) : n < )

.

(So (Da) is (Dah

), where h is as defined in 3.1(2).)

Proposition 3.3. Assume that a function h : R R R is such that on ofthe following condition holds:

(A) (h, 1) < 2(h,1) = c, or(B) (h, 1) < c for some regular cardinal , or(C) (h, 2) < c for some regular cardinal .

Then (Dah ) fails.

Proof. First let us consider the case of the assumption (A). Let A0, A1 P(R)

exemplify the minimum in the definition of (h, 1), A = {A : < (h, 1)} (we

allow repetitions). Choose a sequence r : < (h, 1) of pairwise distinct reals

and fix enumerations s : < c of R and : < c of (h, 1)(h, 1). Letf : R R R be such that

( < c)( < (h, 1))

f(s, r) / h[A0 A

1()

]

.

We claim that the function f witnesses the failure of (Dah ). So suppose thatg0n, g

1n : R R. For < (h, 1) let b = g

1n : n <

R and let () < (h, 1)be such b A1(). Take < c such that = and let a = g

0n() : n < . Fix

< (h, 1) such that a A0 and note that h(a, b) h[A0 A

1(0)

], so

f(s, r) = h(a, b) = h(g0n(s) : n < , g

1n(r) : n < ).

Suppose now that we are in the situation (B). Let c0, c1 : + + (h, 1) be

such that for any sets X0, X1 [+]+ we have

(0, 1 < (h, 1))(0, 1 X0 X1)(c0(0, 1) = 0 & c1(0, 1) = 1)

(see e.g. [Sh:g, ch III]). Let A0, A1 P(R) exemplify (h, 1), A = {A :

Andrzej Roslanowski and Saharon Shelah- The Yellow Cake

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