Saharon Shelah- The Erdos-Rado Arrow for Singular

8/3/2019 Saharon Shelah- The Erdos-Rado Arrow for Singular

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THE ERDOS-RADO ARROW FOR SINGULAR

SAHARON SHELAH

Abstract. We prove that if cf() > 0 and 2cf() < then

(, + 1)2 in ZFC

Key words and phrases. set theory, partition calculus.First typed: August 2005

Research supported by the United States-Israel Binational Science Foundation. Publica-tion 881.

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0. introduction

For regular uncountable , the Erdos-Dushnik-Miller theorem, Theorem11.3 of [1], states that (, + 1)2. For singular cardinals, , they wereonly able to obtain the weaker result, Theorem 11.1 of [1], that (, )2.It is not hard to see that if cf() = then (, +1)2. If cf () > and is a strong limit cardinal, then it follows from the General CanonizationLemma, Lemma 28.1 in [1], that (, + 1)2. Question 11.4 of [1] iswhether this holds without the assumption that is a strong limit cardinal,e.g., whether, in ZFC,

(1) 1 (1, + 1)2.

In [5] it was proved that (, + 1)2 if 2cf() < and there is a nicefilter on , (see [3, Ch.V]: follows from suitable failures of SCH). Also proved

there are consistency results when 2cf() Here continuing [5] but not relying on it, we eliminate the extra assump-

tion, i.e, we prove (in ZFC)LABEL 0.1

Theorem 0.1. If 0 < = cf() and 2 < then (, + 1)2.

Before starting the proof, let us recall the well known definition:LABEL 0.2

Definition 0.2. Let D be an 1-complete filter on Y, and f YOrd, and

Ord {}.We define when rkD(f) = by induction on (it is well known that

rkD(f) < ):

(*) rkD(f) = iff < rkD(f) = , and for every g

Y

Ordsatisfying g 0, so the demand of1- completeness in the definition, holds for us.Recall also

LABEL 0.3Definition 0.3. Assume Y , D , f are as in definition 0.2.0.2

J[f, D] = {Z Y : Y \ Z D or rkD+(Y\Z)(f) > rkD(f)}

Lastly, we quote the next claim (the definition 0.3 and claim are from [2],0.3and explicitly [4](5.8(2),5.9)):

LABEL 0.4Claim 0.4. Assume > 0 is realized, and D is a -complete (a normal)

filter onY.Then J[f, D] is a -complete (a normal) ideal on Y disjoint to D for any

f YOrd


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THE ERDOS-RADO ARROW FOR SINGULAR 3

1. The proof

In this section we prove Theorem 0.1 of the Introduction, which, for con- 0.1venience, we now restate.

LABEL 1.1Theorem 1.1. If 0 < = cf(), 2 < then (, + 1)2.

Proof.Stage A We know that 0 < = cf() < , 2

< We will show that

(, + 1)2.So, towards a contradiction, suppose that

()1 c : []2 {red, green} but has no red set of cardinality and no

green set of order type + 1.

Choose such that:

()2 = i : i < is increasing and continuous with limit , and fori = 0 or i a successor ordinal, i is a successor cardinal. We also let0 = 0 and for i < , 1+i = [i, i+1). For < we will leti() = the unique i < such that i.

We can clearly assume, in addition, that

()3 0 > 2, for i < , i+1 ++i , and that each i is homogeneously

red for c.

The last is justified by the Erdos-Dushnik-Miller theorem for i+1, i.e., asi+1 (i+1, + 1)

2 because i+1 is regular.

Stage B: For 0 < i < , we define Seqi to be {0,...,n1 : i(0) < ...

+i .

Now we can choose Bi, an end-segment of i such that for all Seqiand all 0 < +i , if there is Bi such that rk

() = , then there arei+1 such -s. Recall that i and therefore also Bi are of order type i+1,which is a successor cardinal > +i > |Seqi| hence such Bi exists. Everythingis now in place for the main definition.


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4 SAHARON SHELAH

Stage C: (, Z, D, f) K iff

(1) D is a normal filter on ,

(2) f : Ord,(3) Z D(4) for some 0 < i < we have Seqi and Z is disjoint to i + 1 and

for every j Z (hence j > i) there is Bj such that rk() = f(j)

(so, in particular, T).

Stage D: Note that K = , since if we choose j Bj, for j < , takeZ = \ {0}, = the empty sequence, choose D to be any normal filter on and define f by f(j) = rkj (), then (, Z, D, f) K.

Now clearly by 0.2, among the quadruples ( , Z , D , f ) K, there is0.2one with rkD(f) minimal. So, fix one such quadruple, and denote it by

(

, Z

, D

, f

). Let D

1 be the filter on dual to J[f

, D

], so by claim0.4 it is a normal filter on extending D.0.4For j Z, set Cj = { Bj : rk

() = f(j)}. Thus by the choice ofBjwe know that card(Cj) = j+1, and for every Cj the set (Rang(

){})is homogeneously green under the colouring c. Now: suppose j Z. Forevery Z \ (j +1) and Cj, let C

+ () = { C : c({, }) = green}.

Also, let Z+() = { Z \ (j + 1) : card(C+ ()) = +1}.

Stage E: For j Z and Cj , let Y() = Z \ Z+(). Since 0 > 2 and

j+1 > 0 is regular, for each j Z there are Y = Yj and C

j Cj

with card(Cj) = j+1 such that Cj Y() = Yj.

Let Z = {j Z : Yj D1}. Now the proof split to two cases.

Case 1: Z = mod D1Define Y = {j Z: for every i Z j, we have j Yi}. Notice that

Y is the intersection of Z with the diagonal intersection of sets from D1(since i Z Yi D

1), hence (by the normality of D

1) Y

= mod D1.But then, as we will see soon, by shrinking the Cj for j Y

, we can get ahomogeneous red set of cardinality , which is contrary to the assumptiontoward contradiction.

We define Cj for j Y by induction on j such that Cj is a subset of C

j

of cardinality j+1. Now, for j Y, let Cj be the set of C

j such that

for every i Y j and every Ci we have C+j (). So, in fact, Cj

has cardinality j+1 as it is the result of removing < j+1 elements from Cjwhere |Cj| = j+1 by its choice. Indeed, the number of such pairs (i, ) is

j and: for i Y j and Ci:

(a) j Yi [Why? by the definition of Y as j Y]

(b) Ci [Why? as Ci and Ci Ci by the induction hypothesis]

(c) Y() = Yi [Why? as by (b) we have Ci and the choice of C

i]

(d) j Y() [Why? by (a)+(c)](e) j / Z+() [Why? by (d) and the choice of Y() as Z \ Z+()]


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THE ERDOS-RADO ARROW FOR SINGULAR 5

(f) C+j () has cardinality < j+1 [Why? by (e) and the choice of Z+(),

as j Z Z]

So Cj is a well defined subset of Cj of cardinality j+1 for every j Y

.

But then, clearly the union of the Cj for j Y, call it C satisfies:

() it has cardinality [as j Y |Cj| = j+1 and sup(Y) = as

Y = mod D1 ]

() c[Cj]2 is constantly red [as we are assuming ()3]

() if i < j are from Y and Ci, Cj then c{, } = red [as / C+j ()]

So C has cardinality and is homogeneously red. This concludes the proofin the case Z = mod D1

Case 2:

Z = mod D1.In that case there are i Z, Ci such that Z

+() = mod D1[Why? well, Z D D1 and Z = mod D

1, hence Z

\ Z = .

Choose i Z \ Z. By the definition of Z, Yi / D1. So, if C

i then

Y() = Yi / D1 and choose C

i, so Y() / D

1 hence by the definition of

Y() we have Z \ Z+() = Y() / D1. Since Z D1, we conclude that

Z+() = mod D1].Let = , Z = Z+(), D = D + Z, it is a normal filter by the

previous sentence as D D1 and lastly we define f Ord by:

(a) if j Z then f(j) = Min{rk() : C+j () Bj}

(b) otherwise f(j) = 0

Clearly() (, Z, D, f) K, and() f


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6 SAHARON SHELAH

References

[1] Paul Erdos, Andras Hajnal, A. Mate, and Richard Rado. Combinatorial set theory:

Partition Relations for Cardinals, volume 106 of Studies in Logic and the Foundationof Math. North Holland Publ. Co, Amsterdam, 1984.

[2] Saharon Shelah. A note on cardinal exponentiation. The Journal of Symbolic Logic,45:5666, 1980. [Sh:71]

[3] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford Logic Guides. Oxford Uni-versity Press, 1994. [Sh:g]

[4] Saharon Shelah. Applications of PCF theory. Journal of Symbolic Logic, 65:16241674,2000. [Sh:589]

[5] Saharon Shelah and Lee Stanley. Filters, Cohen Sets and Consistent Extensions ofthe Erdos-Dushnik-Miller Theorem. Journal of Symbolic Logic, 65:259271, 2000.math.LO/9709228. [ShSt:419]

Institute of Mathematics The Hebrew University of Jerusalem Jerusalem

91904, Israel and Department of Mathematics Rutgers University New Brunswick,NJ 08854, USA

E-mail address: [email protected]: http://www.math.rutgers.edu/~shelah

Saharon Shelah- The Erdos-Rado Arrow for Singular

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