Todd Eisworth and Saharon Shelah- Successors of Singular Cardinals and Coloring Theorems I

8/3/2019 Todd Eisworth and Saharon Shelah- Successors of Singular Cardinals and Coloring Theorems I

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SUCCESSORS OF SINGULAR CARDINALS AND

COLORING THEOREMS I

TODD EISWORTH AND SAHARON SHELAH

Abstract. We investigate the existence of strong colorings onsuccessors of singular cardinals. This work continues Section 2 of[1], but now our emphasis is on finding colorings of pairs of ordinals,rather than colorings of finite sets of ordinals.

1. Introduction

The theme of this paper is that strong coloring theorems hold at suc-cessors of singular cardinals of uncountable cofinality, except possiblyin the case where the singular cardinal is a limit of regular cardinalsthat are Jonsson in a strong sense.

Our general framework is that = +, where is singular of un-countable cofinality. We will be searching for colorings of pairs of or-dinals < that exhibit quite complicated behaviour. The followingdefinition (taken from [2]) explains what complicated means in the

previous sentence.

Definition 1.1. Let be an infinite cardinal, and suppose + . Pr1(,,,) means that there is a symmetric twoplace functionc from to such that if < and for i < , i, : < is astrictly increasing sequence of ordinals < with all i,s distinct, thenfor every < there are i < j < such that

(1.1) 1 < and 2 < = c(i,1, i,2) = .

Just as in [1], one of our main tools is a game that measures howJonsson a given cardinal is.

Recall that a cardinal is a Jonsson cardinal if for every c : []


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2 TODD EISWORTH AND SAHARON SHELAH

Definition 1.2. Assume are cardinals, is an ordinal, n ,and J is an ideal on . We define the game GmnJ[,,] as follows:

A play lasts moves.

In the th move, the first player chooses a function F : []


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COLORING THEOREMS 3

second player loses if the ordertype of the set of moves wherehe did not pass is < .

(5) If Player I has a winning strategy in GmnJ[,,] for every < where is singular and > cf() is regular, then PlayerI has a winning strategy in GmnJ[,

, ]. We can weaken therequirement that is regular and instead require that cf() >cf() and = .

In Section 2 of [1], the existence of winning strategies for Player Iin variants of the game is investigated. We will prove one such resulthere; the reader should look in [1] for others.

Claim 1.4. If 2 < < (2)+() then Player I has a winning strategy

in Gm

[,, (2

)+

].Proof. At a stage i, Player I will select a function Fi : []


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By the choice of M0 and Mi, clearly (i) is a successor ordinal or alimit ordinal of cofinality +, and

(1.6) | SkM0(Ai) i()| 2.

Since Ai Aj for i > j, we know the sequence i : i < (2)+ isnondecreasing. Furthermore, for each i we know

(1.7) i < min{ : ()} < (2)+.

This means that the sequence i : i < (2)+ is eventually constant,say with value . Let i be the least ordinal < (2)+ such that i =

for i i.

Proposition 1.5. If i i < (2)+, then SkM0(Ai+1) () is a

proper subset of SkM0(Ai) ().Proof. Note that i, , and () are all elements ofMi+1 as they aredefinable in H(+), ,


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COLORING THEOREMS 5

Let us recall that if S is a stationary subset of , then an Sclubsystem is a sequence C = C : S such that for (limit) S, C

is closed unbounded in .In this section, we will be concerned with the case where is the

successor of a singular cardinal, i.e., = + where cf() < . In thiscontext, if C is an Sclub system, then for S we define an idealJb[] on C by A J

b[] if and only if A C, and for some < and

< ,

A nacc(C) [ < or cf() < ] .

Note that it is a bit easier to understand the definition of Jb[] by

looking at the contrapositive a subset A ofC is large, i.e., not in

Jb[] , if and only if A nacc(C) is cofinal in , and the cofinalities ofmembers of any end segment of A nacc(C) are unbounded below .

Claim 2.1. Let = +, where is a singular cardinal of cofinality < . Let S be stationary, and assume that sup{cf() : S} = < . Let C be an Sclub system, and for each S, let J be theideal J

b[] . Let i : i < be a nondecreasing sequence of cardinals

such that

(2.1) =i


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2 for every club E , for each i < there are sta-tionarily many S such that the set of nacc(C)

satisfying the following h() i

B0[E, e] / I

for all < , h() ran(F

B0[E, e] \


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COLORING THEOREMS 7

the set

S := { S E : E nacc(C) / J}

is stationary.Fix S, and suppose we are given < and < . Since

E nacc(C) / J, we can find E nacc(C) such that >max{, } and cf() > . Since the ordertype ofE is > |e|,we know that B0[E, e] is unbounded in e hence a member ofI. Thisshows that the set of such is in J+ , as required.

Now we return to the proof of Claim 2.1.

Proof of Claim 2.1. Let = cf() be a regular cardinal < that isgreater than and . For each limit < , if there is an i such that Player I wins the version of GmI [e, i,

+] where we allow

Player II to pass, then we let h() be the maximal such i note thati exists by (5) of Claim 1.3 and let Str be a strategy that witnessesthis.

Note that since < + and J = Jb[] for S, we have that for

S and i < that

{ nacc(C) : Str is defined and i h()} = nacc(C) mod J.

We will make + attempts to build F witnessing the conclusion.In stage < +, we assume that our prior work has furnished us

with a decreasing sequence E : < of clubs in , and, for each < where Str is defined, an initial segment F

, A

: < of a

play of GmI [e, h(),

+] in which Player I uses Str. (Note that our

convention is that if Player II chooses to pass at a stage, we let A beundefined.)

For each such , let F : [e]


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is a () S for which

A1 := { nacc(C()) : B0[E, e] / I} / J().

By assumption (2), we have

A2 := { A1 : Str is defined } / J().

For A2, look at the play F, A

: <

+. Since Player I wins,there is a <

+ such that Player II passed at stage for all .Since > and J() is

based, for some < +,

A3 = { A1 : Str is defined and } / J().

Now E was defined so that for some i, for all S,

(2.3) Bi [E, C, I, e, F] J,

but (again by assumption (2))

A4 = { A1 : Str is defined, , and i h()} / J().

For A4, we know that at stage of our play of GmI [e, h(),

+]

the set B0[E, e] was not a legal move. Since our sequence of clubsis decreasing, we know that B0[E, e] is a subset of B0[E, e] for all

< ], so we have

B0[E, e]


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COLORING THEOREMS 9

3. Building the Coloring

We now come to the main point of this paper; we dedicate this sectionand the next to proving the following theorem.

Theorem 1. Assume = +, where is a singular cardinal of un-countable cofinality, say 0 < = cf() < . Assume i : i < isnondecreasing with supremum , and there is a < such that

for eachi, for every large enough regular < , Player I has a winningstrategy in the game Gm[, i,

]. Then Pr1(,,, ) holds.

Let Si : i < be a sequence of pairwise disjoint stationary subsetsof { < : cf() = }. For i < , let Ci be an Siclub system suchthat

/ idp(Ci, Ji), where Ji = Jb[]Ci : Si

for Si, otp(Ci) = cf() = = cf()

Such ladder systems can be found by Claim 2.6 (and Remark 2.6A(6)) of [2] for the second statement to hold, we need that hasuncountable cofinality.

Claim 3.1. There is a club system e such that |e| cf() + c f (),and e swallows each Ci, i.e., if Si (e {}), then Ci e.

Proof. Let S = i


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For each i < , there are hi and Fi = Fi : < , limit as in the

conclusion of Claim 2.1 applied to Ci and e; note that we satisfy the

assumptions of Claim 2.1 by way of Claim 2.2.Let i : i < be a strictly increasing sequence of regular cardinals

> and cofinal in such that

(3.4) = tcfi


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COLORING THEOREMS 11

Finally, we define our coloring by

(3.6) c(, ) = F

h1(i(,))

(,) (w(, )).If the attempt to define c(, ) breaks down at some point for some

specific < , then we set c(, ) = 0.We now prove that this coloring works, so suppose t : < are

pairwise disjoint subsets of such that |t| = 1 < and j < , and

without loss of generality < min t and 1 . We need to find 0and 1 such that

(3.7) t0 and t1 < and c(, ) = j.

Let j1 be the least j such that j < j , and let S, C, and F denote

Sj1, Cj1, and Fj1 respectively.

Given < , we define the envelope of t (denoted env(t)) by theformula

(3.8) env(t) =t

{(, ) : k(, )}.

The envelope of t is the set of all ordinals obtained by walking downto from some t using the ladder system e. This makes sense aswe have arranged that < min t. Note also that | env(t)| |t| = 1.

Next we define functions gmin and gmax in

i such that gmax Jbd gmin .

Now let

= (2

)+

, and let M : < be a sequence of elemen-tary submodels ofH(), ,


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By the choice of C and F, for some S E we have the set(3.12)

A = { nacc(C) : ( < ) ran(F

B0[E, e] \ |e(,)|.Claim 3.4.

(1) If t, and = k(, ) 1, then nacc(e(,)).

(2) If t and < , then

(, ) = (, ) for < k(, ), and

k(,)(, ) =

Proof. For the first clause, note that is an element of e(,) andhence by our choice of e, C e(,). Thus

e(,), and sincecf() > |e(,)|, we know that

cannot be an accumulation point of

e(,).The first part of the second statement follows because of the defini-

tion of . As far as the second part of the second statement goes, itis best visualized as follows:

We walk down the ladder system e from to , we eventually hit aladder that contains this happens at stage k(, ) 1. Since C isa subset of this ladder, the next step in our walk from to must bedown to because < < .


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We can visualize the preceding claim in the following manner: ischosen so that for all sufficiently large < , all the walks from some

element of t to are funnelled through acts as a bottleneck.This will be key when want to prove that our coloring works.

Since A, we can choose a finite increasing sequence 0 fb(i

), and therefore (a, b) = a.

Subclaim 4. w(a, b) = {b0, . . . bn}.

Proof. Our previous subclaims imply that an ordinal eb is relevantif and only if the ladder ea meets the interval (sup(eb ), ). Sinceh0(i

) = n +1, we know that w(a, b) consists of the last n +1 relevant

ordinals in eb

.For i n, clearly bi eb and sup(bi eb)

bn. We have made sure

that ea (bi ,

bi ) = (for example,

ai is an element in this intersection)

and so each bi is relevant.Since a < bn, it is clear that there are no relevant ordinals larger

than bn.Given i < n, if eb (

bi ,

bi+1), then

bi sup( eb) bi+1.


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Since ai+1 < bi <

bi+1 <

ai+1, it follows that

[sup( eb), ) [ai+1,

ai+1),

and so is not relevant. Thus {b0, . . . , bn} are the last n + 1 relevant

elements of eb, as was required.

To finish the proof of Claim 3.5, we note that as h1(i) = j, we have

(3.25) c(a, b) = Fj1b

({b0, . . . , bn}) = j

.

4. Finding the required ordinals

The whole of this section will be occupied with showing that forall sufficiently large i < , it is possible to find objects satisfying therequirements of Table 1.

We begin with some notation intended to simplify the presentation.

a(z0, z1) abbreviates the formula (a0 , . . . ,

an, z0, z1)

b(yn, z0, z1) abbreviates the formula (b0,

nb , yn, z0, z1)

For i < , (i, z1) abbreviates the formula

(4.1) ( env(tz1))[gmin

z1 [i, ) f [i, ) gmax

z1 [i, )]

We have arranged things so that the sentence

(4.2) (za0 < )(za1 < )(

ybn < )

(zb0 < )(zb1 < )[

a(za0 , za1)

b(ybn, zb0, z

b1)]

holds.There are far too many alternations of quantifiers in the above for-

mula for most people to deal with comfortably; the best way to viewthem is as a single quantifier that asserts the existence of a tree of

5tuples with the property that every node of the tree has succes-sors, and every branch through the tree gives us five objects satisfyinga b.

Let (i, za0 , . . . , zb1) abbreviate the formula

a(za0 , za1)

b(ybn, zb0, z

b1) (i, z

a1) (i, z

b1)

gmaxza1 [i + 1, ) < gminzb1

[i + 1, )

.


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By pruning the tree so that every branch through it is a strictly in-creasing 5tuple, we get

(4.3) (za0 < )(za1 < )(

ybn < )

(zb0 < )(zb1 < )(

i < )[(i, za0 , . . . , zb1)].

We now make a rather ad hoc definition of another quantifier in anattempt to make the arguments that follow a little bit clearer. Giveni < , let the quantifier ,izb0 < mean that not only are there un-boundedly many zb0s below satisfying whatever property, but alsothat for each < i, we can find unboundedly many suitable z

b0s for

which fzb0(i) is greater than .

Claim 4.1. If we choose

a

<

a

<

b

n such that(4.4) (zb0 < )(

zb1 < )(i < )[(i, a, a, bn, z

b0, z

b1)],

then

(4.5) (i < )(,izb0 < )(zb1 < )[(i,

a, a, bn, zb0, z

b1)].

Proof. Suppose that we have a < a < bn such that (4.4) holds but(4.5) fails. Then there is an unbounded I such that for each i I,

(4.6) (,izb0 < )(zb1 < )[(i,

a, a, bn, zb0, z

b1)].

In (4.4), we can move the quantifier i < past the quantifiers toits left, i.e.,

(4.7) (i < )(zb0 < )(zb1 < )[(i,

a, a, bn, zb0, z

b1)],

so without loss of generality, for all i I,

(4.8) (zb0 < )(zb1 < )[(i,

a, a, bn, zb0, z

b1)].

Since (4.6) holds for all i I, it must be the case that for each i I,there is a value g(i) < i such that for all sufficiently large < , if

(4.9) (zb1 < )[(i, a, a, bn, , z

b1)],

then

(4.10) f(i) g(i).Since {f : < } witnesses that the true cofinality of

i


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and

(4.13) (zb1

< )(j < )[(j,a, a, bn

, b, zb1)].

(Note that we have quietly used the fact that |I| < = cf() to get ab that is large enough so that (4.9) implies (4.10) for all i I forthis particular b.) This last equation implies

(j < )(zb1 < )[(j,a, a, bn,

b, zb1)],

so it is possible to choose i I large enough so that

g(i) < fb(i)

and(zb1 < )[(i,

a, a, bn, b, zb1)].

This is a contradiction, as (4.9) holds for our choice of i and = b

,yet (4.10) fails.

Notice that an immediate corollary of the preceding claim is

(4.14) (za0 < )(za1 < )(

ybn < )(i < )

(,izb0 < )(zb1 < )[(i,

a, a, bn, zb0, z

b1)].

Claim 4.2. If a < is chosen so that

(4.15) (za1 < )(ybn < )(

i < )

(,izb0 < )(zb1 < )[(i,

a, za1 , ybn, z

b0, z

b1)],

then(i < )(v < i)(

za1 < )[ ]

where := gmaxza1 (i) < v,

and

:= (ybn < )(zb0 < )

v < fzb0(i) and (

zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

.

Proof. In (4.15), we can move the quantifier (i < ) past the otherquantifiers to its left, so

(4.16) (i < )(za1 < )(ybn < )(,izb0 < )(

zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

holds. The claim will be established if we show that for each i < forwhich

(4.17) (za1 < )(ybn < )

(,izb0 < )(zb1 < )[(i,

a, za1 , ybn, z

b0, z

b1)]


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holds, it is possible to find v < i such that

(4.18) (za1 < )

gmaxza1 (i) < v and

(ybn < )(zb0 < )

v < fzb0(i) and (

zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

.

Despite the lengths of the formulas involved, this is not that hard toaccomplish. Since i < = cf(), we can find v < i such that

(za1 < )

gmaxza1 (i) < v and

(ybn < )(,izb0 < )(

zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

,

and now the result follows from of the definition of ,izb1 < .

Thus there are unboundedly many za0 < for which there is a func-tion g

i


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Since i < = cf(), the quantifier (w < i) can move to the leftpast the quantifiers (za1 < )(

ybn < ). This tells us that for our a

and g,

(4.22) (i < )(w < i)(za1 < )(

ybn < )(zb0 < )

gmaxza1 (i) g(i) < fzb0(i) < w and

(zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

.

When we put all this together, we end up with the statement

(4.23) (za0 < )(i < )(v < i)(

w < i)(za1 < )

(ybn < )(zb0 < )gmaxza1 (i) v < fzb0(i) < w

and (zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

.

Since both and i are less than = cf(), we can move some quan-tifiers around and achieve

(4.24) (i < )(w < i)(za0 < )(v < i)(

za1 < )

(ybn < )(zb0 < )

gmaxza1 (i) v < fzb0(i) < w

and (zb1 < )[(i, a, za1 , y

bn, z

b0, z

b1)]

.

Thus there is a function h i


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So now we choose a such that h(i) < fa(i) and for some < i,

(za1 < )(ybn < )(zb0 < )

gmaxza1 (i) < fzb0(i) < h(i) and

(zb1 < )[(i, za0 , . . . , z

b1)]

.

Now we choose a, bn, b, and b such that

a < a < bn < b

gmaxa (i) < fb(i

) < h(i) < fa(i)

(i, a, a, bn, b, b)

It is straightforward to check that these objects satisfy all the require-ments listed in Table 1, so by Claim 3.5, we are done.

5. Conclusions

In this final section, we will deduce some conclusions in a few concretecases.

Theorem 2. If is a singular cardinal of uncountable cofinality thatis not a limit of regular Jonsson cardinals, then Pr1(

+, +, +, cf())holds.

Proof. The proof of this theorem occurs in two stageswe first showthat Pr1(

+, +, , cf()) holds, and then we show that this result canbe upgraded to obtain Pr1(+, +, +, cf().

Let be as hypothesized, and let us define = + and = cf().

Claim 5.1. Pr1(,,,) holds.

Proof. Let i : i < be a strictly increasing continuous sequencecofinal in . Let S { [, ) : cf() = } be stationary. Standardclubguessing results tell us that there is an Sclub system C such

that idp(C, J) is a proper ideal, where J is the ideal Jb[]C

for S,and furthermore, satisfying |C| = . (Note that this last requires that = cf() is uncountable.)

At this point, we have satisfied all of the assumptions of Claim 2.2except possibly for clause (8). It suffices to show that for each i < ,for all sufficiently large regular < , Player I has a winning strategyin the game Gm[, i, 1]. Since is not a limit of regular Jonssoncardinals, it follows that for all sufficiently large regular < , Player Ihas a winning strategy in Gm[,, 1]. This implies, by Lemma 1.3 (1),that for all sufficiently large regular , Player I has a winning strategyin Gm[, i, 1], and so clause (8) of Claim 2.2 is satisfied.


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To finish the proof of Theorem 2, it remains to show that we can in-crease the number of colors from to = + we need Pr1(,,,)

instead of Pr1(,,,).

Lemma 5.2. There is a coloring c1 : []2 such that whenever we

are given

< ,

t : < a sequence of pairwise disjoint elements of [],

t for < , and

< ,

we can find < such that t min(t) and

(5.1) ( t)[c1(, ) = ].Proof. Let c : []2 be a coloring that witnesses Pr1(,,,). Foreach < , let g be a onetoone function from into . We define

(5.2) c1(, ) = g1 (c(, )).

Suppose now that we are given objects , t : < , : < ,and as in the statement of the lemma. Clearly we may assume thatmin(t) > .

For i < , we define Xi := { [, ) : g() = i}. Since is aregular cardinal, it is clear that there is i < for which |Xi| = .Since c exemplifies Pr1(,,,), for some < in Xi we havet min(t) and

(5.3) ( t)[c(, ) = i].

By definition, this means

(5.4) ( t)[c1(, ) = g1(c(, )) = g1(i) = ],

hence and are as required.

To continue the proof of Theorem 2, we define a coloring c2 : []2

by

(5.5) c2(, ) = c1(, (, )),

where (, ) is as in the proof of Theorem 1.It remains to check that c2 witnesses Pr1(,,,). Toward this

end, suppose we are given < , t : < a sequence of pairwisedisjoint members of [], and < . We need to find a and b lessthan such that

(5.6) a ta b tb = c2(

a, b) = .


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Lemma 5.3. There is a stationary set of 1 < such that for some0 < 1 and [1, ), if0 < 1, then the function is constant

on t t.

Proof. Let E be an arbitrary closed unbounded subset of, and let Wbe the set of ordinals < satisfying the properties of 1. In the proofof Theorem 1, without loss of generality we can have E M0. Thismeans that the ordinal found in the course of that proof will be inE, so we finish by observing that W.

An application of Fodors Lemma gives us a single ordinal 0 and astationary W W such that for all W, there is a [, ) suchthat for all [0, ), (t t) is constant.

Using properties of the coloring c1, we can find and such that

0 <

W \ (sup(t) + 1), and

t = c1(, ) = .

Now given a t and b t , we find

(5.7) c2(a, b) = c1(

a, ) = ,

and therefore c2 exemplifies Pr(,,,).

Theorem 2 strengthens results in [1] as clearly Pr1(+, +, +, cf())

implies that + has a Jonsson algebra (i.e., + is not a Jonsson cardi-

nal). The question of whether the successor of a singular cardinal canbe a Jonsson cardinal is a wellknown open question.

We note that many of the results from Section 2 of [1] dealing withthe existence of winning strategies for Player I in Gm[,,] can becombined with Theorem 1 to give new results. For example, we havethe following result from [1].

Proposition 5.4. If 2 but ( < )[2 < ], then Player I has awinning strategy in the game Gm( , , +).

Proof. See Claim 2.3(1) and Claim 2.4(1) of [1].

Armed with this, the following claim is straightforward.

Claim 5.5. Let be a singular cardinal of uncountable cofinality.Further assume that is a cardinal such that 2


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If = 2

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Todd Eisworth and Saharon Shelah- Successors of Singular Cardinals and Coloring Theorems I