saharon shelah- more jonsson algebras

8/3/2019 Saharon Shelah- More Jonsson Algebras

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MORE JONSSON ALGEBRAS

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Department of MathematicsRutgers University

New Brunswick, NJ USA

Abstract. We prove that on many inaccessible cardinals there is a Jonsson algebra,so e.g. the first regular Jonsson cardinal is -Mahlo. We give further restrictionson successor of singulars which are Jonsson cardinals. E.g. there is a Jonsson algebraof cardinality + . Lastly, we give further information on guessing of clubs.

The author would like to thank the ISF for partially supporting this research and Alice Leonhardtfor the beautiful typing.Publ. No. 413Latest Revision - 01/Oct/5

Typeset by AMS-TEX

1


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Annotated Content

1 Jonsson algebras on higher Mahlos and idrk().

[We return to the ideal of subsets ofA of ranks < (for self-containment;see [Sh:g, IV],1.1-1.6) for < +; we deal again with guessing of clubs(1.11). Then we prove that there are Jonsson algebras on for inaccessi-ble not ( )-Mahlo (0.1, 0.25)].

2 Back to Successor of Singulars.

[We deal with = +, singular of uncountable cofinality. We give suf-

ficient conditions for + +


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MORE JONSSON ALGEBRAS 3

1 JONSSON ALGEBRAS ON HIGHER MAHLOS AND idrk()

We continue [Sh:g, III], [Sh:g, IV], see history there, and we use some theorems

from there.Our main result: if is inaccessible not -Mahlo then on there is a Jonssoncardinal. If the reader is willing to lose 0.29 he can ignore also 0.6(1), 0.7, 0.8(2), 0.9,0.11, 0.12, 0.13, 0.15, 0.16(2), 0.28, 0.29; also, 0.12 is just for pure club guessinginterest. Why < < just as = + < .

0.1 Theorem. 1) Suppose is inaccessible and is not ( )-Mahlo.Then on there is a Jonsson algebra.2) Instead of not ( )-Mahlo it suffices to assume there is a stationary setA of singulars satisfying (on idrk() see below):{ < : inaccessible , A stationary} idrk(), A / id

rk() and < .

Proof. 1) If is not -Mahlo, use [Sh:g, IV,2.14,p.212]. Otherwise this is a partic-ular case of 0.25 as there are n < and E , a club of such that E & inaccessible is not n-Mahlo. So S = { E : cf() < } is as required in0.25.2) Look at 0.25. 0.1

0.2 Definition. We say e is a strict (or strict or almost strict) +-club system if:

(a) e = ei : i < + limit,

(b) ei a club of i(c) otp(ei) = cf(i) for the strict case and otp(ei) for the strict

caseand i otp(ei) < i for the almost strict case (so in the strict

case,cf(i) < otp(ei) < and cf(i) = otp(ei) = ).

0.3 Definition. 1) For inaccessible, < +, let S idrk() iff for every1 strict

+-club system e, the following sequence Ai : i of subsets of defined belowsatisfies A is not stationary:

(i) A0 = S { < : S stationary in }

(ii) Ai+1 = { < : Ai stationary in so cf() > 0}(iii) ifi is a limit ordinal, then for the club ei of i of order type we have

2:

1equivalently some see 0.42We may consider adding a second clause: (b) if f is inaccessible, 0 < i < then cf() > i;

this influences 0.5(6); true, it has only local effect that is the two definitions agree for exceptwhen for some inaccessible i,0 < i < i + < ; in [Sh:g, IV] we use the version with clause(b)


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4 SAHARON SHELAH

Ai =

< : if j ei, and [ cf(i) = otp(j ei) < ] then Aj}

2) We define rk(A) as Min{ : A idrk()} for A .

3) id > 0, then cf() and if then is inaccessible.4) Let e be a strict +-club system. If < = cf() < cf() and Ai : i < is an increasing sequence of subsets of with union A and ( A)(cf() > ) or

( < )(cf() = A not stationary in ), then A[,e] = i


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2) Check.3) By induction on .4) We prove this by induction on . For = 0 this is trivial. For successor, by

Definition 0.4(1)(iii) this is easy by the last assumption. For limit, by clause (a) in0.3(1)(iii), if A[,e] then (j e)[ A

[j,e]], recalling < < . So for j eas A

[j,e]i : i < is increasing with union A

[j,e] by the induction hypothesis for some

i(j,) < we have i [i(j,), ) A[j,e]i . As |e| < = cf() necessarily

i() = sup{i(j,) : j e} < , so je

A[j,e]i() which means A

[,e]i() . As was

any member ofA[,e] we can conclude that A[,e] i 0 rk(A ) < n + .


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6 SAHARON SHELAH

Proof. Straight (parts (6), (7) like the proof of 0.11(6)). 0.7

Recall

0.6 Definition. 1) An ideal I on a cardinal of uncountable cofinality is calledweakly normal if it contains all bounded subsets of and: for every f : satisfying f() < 1 + and A I+, for some < we have { A : f() < } I+.2) An ideal I is -indecomposable when: for any sequence Ai : i < of subsets

of ifi


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For each < let () =i


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8 SAHARON SHELAH

(d) Ifn > 0, > 0 it is the family of A such that : < inaccessible and A /

0), < , = rk() and = cf() < ,S = { < : cf() = } then we have ()S where

()S for some < we have S /i


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6) For inaccessible and = n+, < , the ideal id()+{ < : cf() }and also id 0 then rk(S) n +

(b) if > 0, E, cf() > 0 then rk(S) < n + .

8) Assume S and S+ = { : is inaccessible and S ( S is stationary)}.Then rk(S) rk

(S) + .

9) If rk(S) = + 1 then for some club C of , { < : rk(S C) } is a

stationary nonreflecting subset of .

Proof. Let e be a strict +-club system as in 0.3(4).0) Should be clear.1) Clearly also id() is an ideal which includes all bounded subsets of. We provethe equality by induction on and then by induction on .

So if < , A ; let for any B, B[i] be defined as in Definition 0.3 (for e), we candiscard the case = 0; and without loss of generality = sup(A) & A(+1) = ;now (ignoring the case is inaccessible for simplicity)

A id()

< : > and A /

and

and


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< :

0 & = then A id

h()rk ()

() if is inaccessible, = n , n > 1 (so = 0) then A

id(n1)+h()rk () and h() <

() if is inaccessible, = n + > n, n 1 then A

idn+h()rk () and h() < .

Now we can case by case prove that A id(), using the induction hypothesis on

and on (or part (1)) and the definition of id

().3), 4) Check.5) For the second statement note that by parts (1) + (2) we have rk(S) rk(S) < so =: rk(S) is as required.6) We prove this by induction on and for a fix by induction on .

Case 1: < .


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By part (1) we know that id() = idrk() and the latter +{ < : cf() }is weakly normal by 0.5(7) and is -indecomposable for any regular (||+, )by 0.5(6). Alternatively, the proofs are similar to those of case (3).

Case 2: = n, 1 n < .By Definition 0.9 clause (c) obviously id() contains the family of bounded

subsets of and is even normal hence -complete hence -indecomposable for any < .

Case 3: = n + , 1 n < , 1 < .First we prove the indecomposability part, so let = cf() [||+, ) and

assume Ai : i is an increasing continuous sequence of subsets of and assumeA / id() and we should prove that for some i < we have Ai / id().

Let us define for i :

Bi =: { < : inaccessible and A / and < we apply the inductionhypothesis with = , = n + and Ai : i = Ai : i andget: for every B for some i(, ) < we have Ai(,) / id

n+(), but < hence i() =: sup{i(, ) : < } < , and clearly Bi(), as the Aj sare increasing. As < and B stationary (by assumptions) we have: B is a

stationary subset of and B

i i, and A /


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7) By induction on the rank.8) By induction on .9) Easy. 0.11

0.12 Claim. Suppose is inaccessible, S a stationary set of inaccessibles > ,S1 { < : a limit cardinal > of cofinality > 0 and = } is stationary, > = cf() and for S the ideal I is a weakly normal -indecomposableideal on S1 and J is a weakly normal -indecomposable ideal on S, (and ofcourse both are proper ideals which contains the bounded subsets of their domain;of course we demand S = sup(S1 ) so S > ). Further letC1 = C1 : S1 be a strict S1-club system satisfying:

() for every club E of S : { S1 : E \C

1 unbounded in } I

+

J+.

Then: (1) We can find an S1-club system C2 = C2 : S1 such that for every

club E of the set of S satisfying the following is not in J:

< : S1 E and {cf() : nacc(C

2) and E}

is unbounded in

I+ .

(2) Suppose in addition{cf() : S1} < . Then we can demand that for some < , S1 |C2| < . Also ifC

1 is almost strict then we can demand that C2

is almost strict.(3) Suppose {cf() : S1} < and for arbitrarily large regular < we have{ S : I not -indecomposable} J.

Then we can strengthen the conclusion to: C2 is a nice strict S1-club systemsuch that for every club E of the set of S satisfying the following is not in J:

< : S1 E and C2\E is bounded in

= mod I.

(4) In part (1) (and (2), (3)) instead of I weakly normal -indecomposable itsuffices to assume: if belongs to S and h1 : S1 is pressing down andh2 : S1 then for some j1 < , < we have


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{ S1 : h1() < j and h2() < } I+ .

5) We can replace { : < , cf() } : < by S : < such that

(i) 0 and S stationary subset of ;and

(iv) Min(S) > .

6) Assume A is stationary such that A[0,e] = A (any e will do).Then in part (1) we can add nacc(C2) A and waive S cf() > 0.

0.13 Remark. 1) This is similar to [Sh:g, IV,1.7,p.188]. We can replace S is aset of inaccessibles > by S is a set of cardinals of cofinality = and get ageneralization of [Sh:g, IV,1.7,p.188].2) Note that () of 0.12 holds if S1 is a set of singulars and otp(C

1) < for every

S1.Concerning () see [Sh 276, 3.7,p.370] or [Sh:g, III,2.12,p.134], it is a very weakcondition, a strong version of not being weakly compact.3) This claim is not presently used here (but its relative 0.14 will be used) but stillhas interest.

Proof. 1) Let e be a strict -club system.It suffices to show that for some regular < and club E2 of the sequence

C2,E2, = C2,E

2, = g

1(C

1, E

2, e) : < S1 satisfies the conclusion (on g1

see [Sh 365], Definition 2.1(2) and uses in 2 there). So we shall assume that thisfails. This means that for every club E2 of and regular cardinal < some

club E = E(E2, ) exemplifies the failure of C2,E2,. This means that for some

Y = Y(E2, ) J for every S\Y we have

< : S1 E and {cf() : nacc(C2,E2, ) and E} is

unbounded in

I.

We now define by induction on a club E of :

for = 0: E =:


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14 SAHARON SHELAH

for limit: E =: and :

= cf() < E(E, )

.

Let E+ =

i < : i a cardinal , i E, moreover i = otp(E i)

.

By () (in the assumption)

B =: { S : A I+ } J

+

and let

A =S

A

where for S

A =: { S1 : E+ \C1 unbounded in }.

Note that if B or A then = sup( E+) E+; note also that A S1and B S. Now as S1 cf() = , for each A there are () < and() = cf[()] < such that:

()0 () = cf() < & () <

= sup

cf() : nacc(C

2,E , ) E+1

.

[Why? We can find an increasing sequence i, i : i < cf(), i increasing with iwith limit , i C

1, i E, i < cf(i) i < Min

C1\(i + 1)

(possible by

the definition of the set A and of the club E+). For each i < cf() we can find

i < , i sup(C3

)], and C3

: S1 is as required. Soassume that for every club E of for some club E = E(E) this fails. We chooseby induction on < , a club E of , as follows:

E0 =

E+1 = E(E)

E =


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Then for some S-club system C = C : S we have

for every club E of { S : = sup(E nacc(C) A)} J

+.

Proof. As usual let e = e : < be a strict -club system but such that forevery limit \A we have e A = . For any set C and club E of wedefine g2n(C,E, e, A) by induction on n < as follows: for n = 0, g

2n(C,E, e, A) =

{sup( E) : C} and

g2n+1(C,E, e, A) =g2n(C,E, e, A) {sup( E) : for some

nacc(g2n(C,E, e, A)) we have / A, and

sup( E) > sup( g2n(C,E, e, A)) and

sup( E) sup( e) and e}

and

g2(C,E, e, A) =n


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18 SAHARON SHELAH

\E J. Now for each E S, choose an increasing sequence ,i : i 0 or at least () < rk(S),(note: if = 0 this holds trivially; similarly for clause ())

() if > 0, then for some () we have () + rk(S+) () < rk(S)

(recall = ()), and id()rk () S is -complete (of course, = ()).

()() C is an S-club system,

() /

idp

(C, I), see definition below, where I =I

:

S, I

=:

{A

C

: for some < and < , ( A)( < cf() < acc(C)},moreover

() for every club E of we have () < rk({ S : for every < we have =sup(E nacc(C) { < : cf() > })).

Then idj(C) is a proper ideal (see 0.18 below).2) Like part (1) using id , rk instead of id

rk, rk respectively.

0.17 Remark. The ideals idj(C), idj

(C) are defined below; they are from [Sh:g,

IV,Definition 1.8(2),(3),p.190]4 but idj() = idj0

() and the definition of rkj()

is repeated in the proof below, and the ideal idp(C, I) in [Sh:g, III,3.1,p.139] is.

0.18 Definition. For regular > 0, C = C : S, C = sup(C), S = sup(S), I = I : S, I an ideal on C let idp(C, I) be the family {A :for some club E of for no Dom(C) acc(E) do we have A E C / I}.

0.19 Definition. 1) For an inaccessible Jonsson cardinal, C = C : S, C , S = sup(S) and = cf() < let idj(C) be the family of A such that:

for every > and x H() there is a sequence M exemplifying A idj() forx (and C, ) where:

2) M exemplify A idj() for x H() (and > and ) if:

0 M = M : < , < ,

4but the same x in line 4 should be every x


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1 < , + 1 M (H(), , 0

by clause ()(), if = 0 trivially.Without loss of generality < rk(S ) < and even rk(S ) < n + (rk(S) n) < n + (in part (2) the first inequality is ).

Toward contradiction assume idj(C) let x = , C, S and let M : <

exemplify idj(C) for x which means that 0,1,2 of Definition 0.19(2) holdand let be as in 2.

Let: E = { < : M and = sup(M ) for every < and > forthe from 2 of 0.19(2)} and let

S = { S : for every < , { E nacc(C) : cf() > } is unbounded in }.

So E is a club of with every member a limit cardinal, S S is stationary (as

/ idp(C, I)) and even S

/ id()

rk () (see clause ()() in the assumption) andusing 2 of Definition 0.19(2) we shall look only at S.

For each i < and < let i =: Min(M\i). As M : < exemplifies

idj(C), we have

3 for each S for some < , = cf( ) > hence

is inaccessible.

Proving this will take some steps. First for some < we have:

4 nacc(C)\ & cf() ( < )[Min(M\) is an inaccessible> ].

[Why? In the definition of idj, i.e. clause (b) of2 of Definition 0.19(2) we do notspeak on for S, we speak on

, for nacc(C) E. As S

we have

E so > hence cannot satisfy (a) + (b) of2, but as E it satisfies(a) hence for some < , we have 4.]Next note

5 = & E nacc(C)

= .


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[Why? So we have = M hence C M so ( M)[Min(C\) M ], and now for every E nacc(C) we can find M satisfying > sup(C ) so = Min(C\) M as required in 5.]

6 singular & E nacc(C) & cf() > cf(

)

= .

[Why? Fix such . There is a club e of of order type cf( ) which belongs

to M ; also cf( ) M so cf(

) < . Also for every

e0 = { e S : / acc(C)} there is such that sup(C ) < < , hence = sup{ :

e0} < (as cf() > cf( ) by assumption). As acc(E)

there is 1 M , 1 > . So is the minimal ordinal satisfying

1 < & ( e S)[ nacc(C)] & ( e S)[ nacc(C)

sup( C) < 1]

hence M hence = as required.]

Of course, [ singular cf( ) < ] as cf(

) M

= M ; so together

3 actually holds.

Letting S =: { S : = cf(

) > }, we have S

= 0 by clause () of () in our

assumption and if = 0 by 0.5(0) (for the idrk case) or 0.11(0) (for the id

case)we can choose () such that rk(S

()) > ().

So to get the contradiction it suffices to prove rk

S()

(). Stipulate

() = .

Let () =: rk()

S+ ()

for .

Let () =

() n

() +

() where

() <

() (see the assumption in the

beginning of the proof). For < , as {, S} M() and () M() clearly

() M() hence

() M() hence

() < .

We now prove by induction on i E {} that

rki

S() i E

i ni() +

i().

This suffices as for i = (as i() ()) it gives: rk

S()

= rk(S() E) =

rk(S() E)

() rk(S

+) (), contradicting the choice of ()

(and ()).

Proof of . The case cf(i) 0 i nacc(E) i nacc(acc(E)) is trivial; so weassume


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1 i acc( acc(E)) & cf(i) > 0 hence rki

S() i E

= rki

S() i

.

For a given i, clearly for every club e of i() which belongs to M() we have

i = sup(e i) (as M think e is an unbounded subset ofi() and i = sup(i M)

as i E) and for a given i, by the definition of rk there is a club e ofi() satisfying

Min(e) > i() such that one of the following occurs:

(a) i() = 0 and e rk(S+ ) = 0 & S+ e =

(b) i() > 0 and e rk(S

+ ) < ni() + i().

As S+, i() M() without loss of generality e M() hence i acc(e). Neces-

sarily

2 if i acc(e) acc(E), then () e.

[Why? Otherwise sup(() e) is a member of e (as e is closed, () acc(e)

so () > Min(e)), is as acc(e)) and is < () and it belongs to M()

(as e, () M()), contradicting the choice of ().]

Hence one of the following occurs:

(A) i() = 0 and e is disjoint to S

+

(B) i()

> 0 and rk()S+

() <

() ni

()+ i

()for every

acc(e) acc(E).

First assume (A). Now for any acc(E) S() we have () is inaccessible (as

S() and the definition of S()) and

() S is stationary in

() (otherwise

there is a club e M() of () disjoint to S, but necessarily e

but our

present assumption is S() S, contradiction); together () S

+ hence

() / e (e from above, after 1), so necessarily = i() / acc(e). So

acc(e) acc(E) i is a club of i disjoint to S() hence rki S() i

= 0 which

suffices for .If (B) above occurs, then for acc(e) acc(E) we have () n

() +

() hence clearly j < S() j < & =sup(M()

()) j < & = sup(M(),j

())

() = Min(M(),j

\). Now for j < let Wj = {w : w belongs to M(),j and w S} and forw Wj we let w

+ = { < : inaccessible and w is a stationary subset of },

let i(),j,w =

i(),j = Min(M(),j \i). Also for j < , w Wj and i >

j(),j,w

let i(),j,w = rk

i(),j,w

(w+ i(),j,w), so as w+ S+ necessarily i(),j,w =

i(),j,w n

i(),j,w +

i(),j,w with n

i(),j,w < and

i(),j,w <

i(),j. By the

definition of M,j and i(),j,w clearly

i(),j,w decrease with j and by 8 we have

j() < i E & cf(i) >

j()

i(),j,w =

i(). Now we prove by induction on

i E {} that

+ if j < , j() < i E, w Wj then

rki(S() w i E) i n

i(),j,w +

i(),j,w.

This clearly suffices (for w = S we shall get for each M(),j which is more thanenough).

Proof of +. The case cf(i) 0 i nacc(E) i nacc(acc(E)) is trivial; so we

assume

3 i acc(acc(E)) & cf(i) > 0 hence rki

S() w i E

= rki

S() w i

.

For a given w Wj and i E\j(),j,w clearly for every club e of

i(),j,w which

belongs to M(),j we have i = sup(i e); (this because M thinks e is anunbounded subset of i

() and i E implies i = sup(i M) as a limit ordinal);


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24 SAHARON SHELAH

so i acc(e) even i acc(acc(e)), etc. By the definition of rki(),j,w

, for our i,

there is a club e of i(),j,w with Min(e) >

i(),j,w and h (for case (c)) such that

one of the following cases occurs:

(a) i(),j,w = 0 & n

i(),j,w = 0 that is

i(),j,w = 0 and w

+ e = so

e rk (w+ ) = 0 &

(b) i(),j,w > 0 and e rk

(w

+ ) < ni(),j,w + i(),j,w

(c) i(),j,w = 0 & ni(),j,w > 0, h a pressing down function on w

+ i such

that for each j < i we have j < e & h() = j rk(w+ ) 0 and rk

(),j,w

w+ (),j,w

<

(),j,w ni(),j,w +

i(),j,w

for every acc(e) acc(E)


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(C) i(),j,w = 0, n

i(),j,w > 0, h M(),j a pressing down funtion on e such

that: < e & ( inaccessible) rk({ < : w+ e and h() =

}) < ni(),j,w (read Definition 0.9(1) clause (c) and use diagonal inter-

section; remember that for singular , rk() = rk() < ).

First assume (A). Now for any acc(E) S() w necessarily (),j,w is inac-

cessible (as S() and the definition of S()) and

(),j,w w is stationary in

(),j,w (otherwise there is a club e

M(),j of (),j,w disjoint to w, but neces-

sarily e and w, contradiction); together (),j,w w+ hence

(),j,w / e

(e from above), so as e M(),j necessarily = i(),j,w / acc(e). So

acc(e) acc(E) i is a club ofi disjoint to S() w hence rki

S() w i

= 0

which suffices for +.

Secondly, assume clause (B) occurs; then for every acc(e) acc(E) we have(),j,w n

(),j,w +

(),j,w <

(),j,w n

i(),j,w +

i(),j,w. Since

i(),j,w

Min(e) we have (n(),j,w, (),j,w)

(ii) S { < : < cf() < }

(iii) rk(S) = = n + where < , n <


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26 SAHARON SHELAH

(b)(i) J an 1-complete ideal on containing the singletons

(ii) if A J+, (i.e. A , A / J) and f A then fJA < (if e.g. J = Jbd , regular, then A = suffices as J A

= J)

(iii) if A J+ and f A() then fJA < .

Then id


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(possible as fn : An and hypothesis (b)(ii)).

Now we shall prove for S[0] {} that recalling S[0] = { : S or S isstationary in }:

rk(S E0 ) n() + fn()JAn()

< n() + .

Why does this suffice? For = , first note: if n() < n then rk(S) n() + fn()JAn()

(n 1) + fn()JAn()< (n 1) + n

rk(S) = rk(S) [why? first inequality by , second inequality by n() < n

(see above), third inequality by assumption (b)(ii), as for i An(), fn()(i), that

is fn()(i) is i < by our assumption toward contradition; the fourth inequality

is an ordinal addition and the fifth we have assumed] and this is a contradiction.

So we can assume n() = n

, but then by , we know rk(S) n() +fn()JAn().

But for i An() = An , by the definition of the A

ns we know that ni = n() =

n() = n, and so we know ni + i = rk(Si) < rk(S) = = n +

so we know fn()(i) = rk(Si ) n() = i < so by assumption (b)(iii),

fn()JAn()< , so by , rk(S) < n

+ , contradiction.

So it actually suffices to prove . We prove it by induction on .

If cf() = 0, or / acc(E0) or more generally S is not a stationary subset ,then rk(S ) = 0, and rk(Si ) = 0 hence f

n()

= 0 so the inequality holds trivially.

So assume otherwise; for each i < , for some club ei of we have:

()2 (1) ei (m(1),i < m,i) (m(1),i = m,i & (1),i < ,i).

Without loss of generality ei E0. As S is a stationary in (as we are assumingotherwise) by hypothesis (a)(ii) of the claim, cf() Min{cf() : S} > ,

so e =:

iAn()

ei is a club of .

As ,i < (see its choice) and cf() > (by hypothesis (a)(ii)) clearly =supi n() +fn()JAn()

, hence


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28 SAHARON SHELAH

B = (1) e : rk(1)(S E0 (1)) (1) n() + fn()JA

n()is a stationary subset of ; note that(1) B (1) e fn()JAn()

< min(e) fn()JAn()< (1).

But by the induction hypothesis

(1) B rk(1)(S E0 (1))

(1) n((1)) + f(1)n((1))JA(1)

n(()(1))

< (1) n((1)) + (1).

Let (1) B; putting this together with the definition of (1) B we get

()3 (1) n() + fn()JAn() (1) n((1)) + f

(1)n((1))JA(1)

n((1))

.

Now by ()2 necessarily n((1)) n() so by ()3 we have n((1)) = n() (remem-

ber f(1)n((1))JA(1)

n((1))

< (1) by the induction hypothesis). So

()4 fn()JAn() f

(1)n((1))JA(1)

n((1))

.

Now by ()2 (as we have n() = n((1)))

i An() : i / A(1)n((1))

n f(1)n((1))

(An() A(1)n((1)))J(A

n()A

(1)

n((1)))

f(1)n((1))

A(1)n((1))JA(1)

n((1))

contradicting ()4. 0.20


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0.23 Claim. IfJ is an ideal on, < , a limit ordinal, J is -complete, < ,then I = id || this implies that {i : rk(Si ) } = mod J. So let us carry the induction; if = 0 this is trivial and for limituse < < and the induction hypothesis (and J being -complete). So assume = + 1, < , cf() = , rk(S ) 2 = 2 + 2, hence S

=: { < :rk(S

) 2 + 1} is a stationary subset of .So S & cf() = A

2+1by the induction hypothesis so if

{ S : cf() = } is a stationary subset of we are done. Otherwise, still[ S { < : A2} is a stationary subset of

] hence S = { < :cf() < and A2} is a stationary subset of , and we can finish as before.

0.23

0.24 Remark. 1) It is more natural to demand only J is -complete and > ; andallow to be a successor, but this is not needed and will make the statement morecumbersome because of the problematic cofinalities in [, ].2) We can prove more in 0.23:

if < , rk(S) > then {i < : rk(Si) } = mod J.

0.25 Theorem. Assume is inaccessible and there is S stationary such thatrk({ < : is inaccessible and S is stationary in }) < rk(S).

Then on there is a Jonsson algebra.

Proof. Assume toward contradiction that there is no Jonsson algebra on . LetS+ =: { < : inaccessible and S is stationary in }.Note that without loss of generality

S is a set of singulars and rk(S) is a limit ordinal.

[Why? Let S = { S : a singular ordinal }, S = { S : is a regular cardinal},so rk(S) = rk(S

S) = Max{rk(S), rk(S)} by 0.5(0). Now if rk(S) rk(S

+) and clearly S stationary S+, so neces-sarily rk(S

) is finite hence has a stationary set which does not reflect and weare done; see [Sh:g]. If rk(S) is a successor ordinal we are done similarly.]

By the definition of rk,

=: rk(S) < + rk(S+

), but we have assumedrk(S

+) < rk(S) so rk(S) < + rk(S), which implies rk(S) < . So forsome n < we have n rk(S) < n + .

Let rk(S+) = = m+ with < . We shall now prove 0.25 by induction

on . By [Sh:g, Ch.III], without loss of generality > 0. By 0.5(9) we can find aclub E of such that:

(A) E rk(S ) < n + ( rk(S) n

)

(B) E rk(S+ ) < m + .

Note that m

+

> 0 for E (or just > 0) as

> 0. Let A =: { E : inaccessible, < and rk(S ) m + }.Clearly A implies S is a stationary subset of . By the induction hypothesisand the choice of A and clause (B) every member of A has a Jonsson algebra on itand by the definition of A (and 0.5(9)) we have [ < & A is stationary in A]; note that as A is a set of inaccessibles, any ordinal in which it reflectsis inaccessible. If A is not a stationary subset of , then without loss of generalityA = , and we get rk(S) m

+ = < rk(S), a contradiction. Sowithout loss of generality (using the induction hypothesis on ):

A is stationary, A[0] A, i.e. ( < )(A is stationary in A),each A is an inaccessible with a Jonsson algebra on it.

So by [Sh:g, IV,2.12,p.209] without loss of generality for arbitrarily large < (even inaccessible):

= cf() > 0, < and for every f we have fJbd < .

So choose such < satisfying > rk(S) n. We shall show that

() id and we can use 0.20and the statement

above to get (). If < use 0.23. So () and () holds.


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Note that S+ satisfies the assumptions on A in 0.14, i.e. clause (b) there and letting

= , the ideal id


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32 SAHARON SHELAH

Let be large enough, M an elementary submodel of (H(), , MARTIN WARNS: Label 1.16 on next line is also used somewhere else (Perhaps

should have used scite instead of stag?

0.27 Question. 1) Can the first which is -Mahlo be a Jonsson cardinal?2) Let be the first -Mahlo cardinal; is []2 consistent?3) Is it enough to assume that for some set S of inaccessibles rk(S) <

+ to deducethat there is a Jonsson algebra on (or even have Pr1(,, 0))?

0.28 Remark. 1) Instead of Jbd we could have used []


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As in the proof of 0.25 without loss of generality S cf() < and weprove this by induction on .If rk(S) < , then also rk

(S

+) < so rk(S) = rk(S), rk(S

+) = rk(S+)

and so 0.25 apply so we are done, so we can assume rk(S) . Let

=rk(S) be n

+ , < and let (0 + ||+, ) be regular. Now

rk(S[+1]) as , so without loss of generality we have ( S)(cf() >

). By 0.11(6), the ideal id


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34 SAHARON SHELAH

2 Back to Successor of Singulars

Earlier we have that if = +, > cf() and is small in the alephs sequence,then on there is a Jonsson algebra. Here we show that we can replace small inthe aleph sequence by other notions of smallness, like small in the beth sequence.This shows that on + there is a Jonsson algebra. Of course, we feel that being aJonsson cardinal is a large cardinal property, and for successor of singulars it isvery large, both in consistency strength and in relation to actual large cardinals.We have some results materializing this intuition. If = + is Jonsson > cf(),then is a limit of cardinals close to being measurable (expressed by games). If in

addition cf() > 0, 2(cf())+

< , then is close to being cf()-compact, i.e. thereis a uniform cf()-complete ideal I on that is close to being an ultrafilter (thequotient is small).

2.1 Definition. We define the game Gmn(,,) for cardinals, an ordinaland n . A play last moves; in the -th move the first player chooses a functionF from []


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2.3 Claim. 1) If []


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36 SAHARON SHELAH

Remark. On 2.4, 2.6, 2.7 see more in [EiSh 535], particularly on colouring theorems(instead of, e.g., no Jonsson algebras).

Proof. 1) Let A : < be a list of distinct subsets of , and defineF(, ) =: Min{ : A / A}.2) Easy, too, but let us elaborate.

First Case. There is a set a of regular cardinals < , with no last element, < min(a) and sup(a) [, ) such that 1 a max pcf(a 1) < 1 andmax pcf(a) = . Clearly it suffices to prove []2supa, and there is an ideal J extending Jbda such thattcf(a/J) = ; without loss of generality max pcf(a) = and pcf(a) has nolast element. If J max pcf{j : j < i}. As

J


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element but pcf(a)\{}\ does not, so we can choose i recalling that pcf({j : j ]

/ idp(C

1, J).


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38 SAHARON SHELAH

[We can build together C1, C2 like this as in the proof of 0.12 or use [Sh:g, III,2.6]as each J is cf()-based.]

Let =

i


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a legal move and B ; otherwise the second player makes it void; i.e. pass (see2.2(b)).

Having our + moves we shall get a contradiction. Let Ebe ]

.

For < + define

i() = Min{i : i cf [Min(B\())]}.

Since B is closed under F and F codes enough set theory, the proof of [Sh:g,III,1.9], (similar things are in 1 here) shows that

() if A1, cf() > i() then B and ()[ nacc(C2 ) E B ].

Now as cf() (whereas there are cf() cardinals i) for some i() < cf() wehave

+ = sup(U) where U =: { < + : i() i()}.

Choose A1 with cf() > i() (why is this possible? if cf() = 0 as () =

sup(A1) and C1 is nice; if not as A1 J

+() see [Sh:g, III,1.1]). By () we have

U B and by the choice ofE and (), clearly E nacc(C2 ) has cardinality

cf(); so for every U the second player (in the play of Gm(C2 , ,

+)) makea non-void move. As |U| = +, this contradicts St is a winning strategy for thefirst player in Gm(C

2 , ,

+).(2) Similar proof (for = see [Sh:g, II,355].) 2.6

An example of an application is

2.7 Conclusion. 1) On + there is a Jonsson algebra.2) Ifn+1() < n+2() then the first player wins in Gmn+2 (, +, (2)+).

3) If is singular not strong limit, <


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40 SAHARON SHELAH

then [] cf(), cf() < < + < , pp+ () > . Then []


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Dom(I) cannot be partitioned to sets not in I. The case we think of in 2.9 is = +, singular of uncountable cofinality.

2.9 Claim. Suppose

(a) = cf() > (2+

)+ and =

(b) C is an S-club system, S stationary and I = I : S, I an idealon C containingJ

bdC

and idp(C, I) is (see 0.17, a proper ideal and) weakly

+-saturated and

(c) ()2,

Iif A Dom(I), A / I then for some Y A, |Y| , Y / I

hence |P(Y)/I| 2.

Then:

(i) P()/idp(C, I) has cardinality 2

(ii) for every A P()\idp(C, I), there is B A, B P()\idp(C, I) andan embedding ofP()/

idp(C, I) + (\B)

into some P(Y)/I with

S, Y C, Y / I,

(iii) moreover, in (ii) we can find h : B such that for everyB B for someA we have B h1(A)mod idp(C, I). (In fact for some g : Y and idealJ on for everyB B we have: B idp(C, I) g

1(h(B)) J.)

2.10 Remark. 1) The use of and though = is to help considering the casethey are not equal.2) The point of 2.9 is that e.g. if = +, > cf(), S , then we can findC = C : S and I = I : C such that / idp(C, I) and I is(cf())-based and S, < , < { C : acc(C) or < orcf() < } I. Now if idp(C, I) is not weakly -saturated then []


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42 SAHARON SHELAH

For each A P which is in idp(C, I), choose a club EA of witnessing it (and ifA P\idp(C, I) let EA = ).

As (2+

)+ < clearly |P| < hence E =: AP

EA is a club of .

So S = { S : EC / I} is a stationary subset of. For proving (i) supposei = (2)

+and eventually we shall get a contradiction. We now choose by induction

on < + ordinals i1(), i2() < i and S and sets Y Ai2()\Ai1()EC

such that Y / I , |P(Y)/I | 2, |Y| , Ai2()\Ai1() / idp(C, I) and 0,

+ if cf() = 0 and S0, : < be a sequence of pairwise disjointstationary subsets of S0. For every club E of

+, let


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E = { < : for every < , = sup(E S0,)}, it too is a club of +.

Now for every E S and < for some nE(, ) < we have =

sup(S0, E nacc(CnE(,) )) hence (as cf(

) > 0, see its choice) for some

nE() < , uE =: { < : nE(, ) = nE()} has cardinality . Without loss ofgenerality, nE(, ), nE() are minimal. So for some n

for every club E of +, forstationarily many E S, we have E and nE() = n. Now if cf() = 0,for some () < for every club E of + for stationarily many E S wehave nE() = n

and |uE ()| = . If cf() > 0 let () = . Now there is aclub E of + such that: if E0 E is a club then for stationarily many S E,nE() = nE0() = n

, uE() = uE0

() and it has cardinality (just remember

() < in all cases so after tries of E0 we succeed). As < =


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48 SAHARON SHELAH

is a stationary subset of +, so

E+

= { < +

: acc(E) is divisible by 2

and S

Ehas order type for every < }

is a club of+.Let us choose SE+, and let e = {

i : i < } (

i increasing continuous).

We shall show that for some n, is in En and is as required in () for En, thus

deriving a contradiction. Let for <

A = {i < : (i ,

i+1) S

= }.

As = otp( S E) clearly A is an unbounded subset of ; hence we canchoose by induction on < , a member i() A such that i() > & i() >


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cardinal (for inaccessible we can get by the proof a less neat result, see more [Sh572]).

Proof. We know that for some strict S-club system C0 = C0 : S we have / idp(C

0) (exists, e.g. as in 3.1). Let C0 = { : < } (increasing continuously

in ). We claim that for some sequence of functions h = h : S with h : we have:

()h for every club E of for stationarily many S acc(E),for some < the following subset of is stationary

A,E = < : E and the ordinal Min{

: > , h() = }

belongs to E

.

This suffices: for each < let C, be the closure in C0 of {

E : and h() = } / En+1.

Choose hn+1

: such that hn+1

() = hn+1

() hn

() = hn

() and

[ = & < & < & } = Min{ e,n, : > }]

hn+1 () = hn+1 ()

.


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50 SAHARON SHELAH

Note that we can do this as = +.

Lastly let en+1 = Min(en ) for each n. Now sup(e

n ) : n <

is non-increasing (as en decreases with n) hence for some n() < : n > n()

sup(en ) = sup(en() ); and for n() + 1 we get a contradiction. 3.5

3.6 Remark. If we omit = + in 3.5, we can prove similarly a weaker statement(from it we can then derive 3.5):

() if = +, = cf() > 0, S { < : cf() = } is stationary, C0 is astrict S-club system, C0 = {, : < } (with , strictly increasing with), and / idp(C

0) then we can find e = e : S such that:

(a) e is a club of with order type

(b) for every club E of for stationarily many S we have acc(E)and for stationarily many < we have: e and ()[ < + 1 < Min(e\(+ 1)) & ,+1 E]

3.7 Remark. In 3.5 we can for each S have h : such that for everyclub E of , for stationarily many S, for every < , for stationarily many h1 ({}) we have ,+1 E.Use Ulams proof.


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3.8 Claim. Suppose = +, S stationary, C = C : S anS-club system, / idp(C), > =: sup{cf()+ : nacc(C), S}.Then there is e, a strict -club system such that:

() for every club E of , for stationarily many S, = sup{ nacc(C) : E, moreover e E}.

Proof. Let e be a strict -club system.Clearly for some < for every club E of , for stationarily many S, =

sup{ : E, nacc(C) and cf() = }. For any club E of and < we let eE = e

E, : < be: {sup( E) : ei and otp( ei) < } if

acc(E) & cf() = and eE, = e otherwise. It is enough to show that for

some club E of and < the sequence e

E is as required. If this fails, we chooseby induction on < a club E of such that 1 < 2 E2 acc(E1).For + 1, for each < , < , let E, be a club of such that e

E

is not as

required. Let E, a club of disjoint to { S : = sup{ nacc(C): cf() =

and eE , E\ min(sup(C )} and lastly E+1 =


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52 SAHARON SHELAH

()b for every club E E1 of for some S we have E C = E1 C

and for arbitrarily large i < (),

= supcf() : C [(),i , (),i+ ) E.4) In part (1), ifS I[] then without loss of generality |{C

() : S and

nacc(C() )}| < for every < .

Proof. 1) Let =


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Let from ()1 be ().

()2 There is a club E of + such that for every club E of the set {

S()[E

] : a

() E

E} is stationary recalling

S[E] = { S :a E

is unbounded in

and otp(a E) is divisible by ()}

[Why? If not, we choose by induction on < +() a club E of

+ as

follows:

(a) E0 =

(b) if is limit, E =


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54 SAHARON SHELAH

Choose a regular cardinal < , > + + |()|. We choose by induction on < a club E of as follows:

for = 0, E0 =

for limit, E =


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shall prove that for some club E of , b0[E] : S satisfies: for every club E of

for stationarily many S, E b0[E] is an unbounded subset of of order type ; this clearly suffices.

First note

() for some < for every club E of for some S acc(E) we have:

= sup{ < :for some [, + 1, ,+1) we have

C2 E and f++2() < }.

[Why? If not, then for every < there is a club E of for which the

above fails, let E = cf() = > 0, = = cf() < ,

saharon shelah- more jonsson algebras

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