Saharon Shelah- Reflection Implies the SCH

8/3/2019 Saharon Shelah- Reflection Implies the SCH

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REFLECTION IMPLIES THE SCHSH794

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Mathematics DepartmentNew Brunswick, NJ USA

Abstract. We prove that, e.g., if > cf() = 0 and > 20 and every stationaryfamily of countable subsets of + reflect in some subset of + of cardinality 1 thenthe SCH for + holds (moreover, for +, any scale for + has a bad stationary setof cofinality 1). This answers a question of Foreman and Todorcevic who gets suchconclusion from the simultaneous reflection of four stationary sets.

2000 Mathematics Subject Classification. 03E04, 03E05.Key words and phrases. reflection, stationary sets, Singular Cardinal Hypotheses, pcf, set

theory.

The author would like to thank the Israel Science Foundation for partial support of this research(Grant No. 242/03)I would like to thank Alice Leonhardt for the beautiful typing.First Typed - 01/12/18Done - Dec. 2001Latest Revisions - 07/Oct/1

Typeset by AMS-TEX

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0 Introduction

In 1 we prove that the strong hypothesis (pp() = + for every singular )hence the SCH (singular cardinal hypothesis, that is + + 2) holds when:for every 1 every stationary S []

0 reflect in some A []1 .This answers a question of Foreman and Todorcevic [FoTo] where they proved

that the SCH holds for every 1 when: every four stationary S []0 , =

1, 2, 3, 4 reflect simultaneously in some A []1 . They were probably motivatedby Velickovic [Ve92a] which used another reflection principle: for every stationaryA []0 there is A []1 such that A [A]0 contains a closed unboundedsubset, rather than just a stationary set.

The proof here is self-contained modulo two basic quotations from [Sh:g], [Sh:f];we continued [Sh:e], [Sh 755] in some respects. We prove more in 1. In particular

if > cf() = 0 and pp() > + then some A [+]0 reflect in no uncountableA [+] and see more in the end.

We thank the referee and Shimoni Garti for not few helpful comments.

For the readers convenience let us recall some basic definitions.

0.1 Definition. Assume is regular uncountable (if = +, [B]


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REFLECTION IMPLIES THE SCH SH794 3

0.4 Definition. Let be a regular uncountable cardinal, and assume A is astationary subset of [B] and has uncountable cofinality then prj(A) := {sup(a) : a A}, and it is a stationarysubset of ; if = we may omit it.

0.9 Definition. Let fi be a function with domain 0 to the ordinals, for everyi I where I is a set of ordinals. We say that the sequence f = fi : i Iis free, if we can find a sequence n = ni : i I of natural numbers such that:

(i, j I) (i < j) (ni, nj n < ) fi(n) < fj(n). We say that f is -freewhen for every J [I] 0, then we can assume (without loss of generality) that ni = n() forevery i S, as we can decrease S.


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4 SAHARON SHELAH

1 Reflection in [+]0 and the strong hypothesis

1.1 The Main Claim. Assume

(A) = + and > cf() = 0 and 2 (e.g., = 2 which impliesthat below always = 1)

(B) = n : n < is an increasing sequence of regular cardinals > 1 withlimit and = tcf(n,


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6 SAHARON SHELAH

2) Let 0. We can restrict ourselves to > +, A []

+

(getting the stronghypothesis and SCH above ).

Proof. 1) As in 1.3, by 1.1 we have > cf() = 0 pp() = +, this implies

clause (a) (i.e. by [Sh:g, VIII,1], > cf() pp() = +). Hence inductively by[Sh:g, IX,1.8,pg.369], [Sh 430, 1.1] we have < cf([], ) = if cf() > and is + if > cf(). This is a consequence of the strong hypothesis.) TheSCH follows.2) The same proof. 1.4

Proof of 1.1. Let M be an algebra with universe and countably many functions,e.g. all those definable in (H(+), ,


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Proof of the subclaim 1.5. How do we choose them?Let Ai : i 1 be a sequence of infinite pairwise almost disjoint subsets of .

Let D1,i = {A : Ai\A is finite}, D2,i = {A : Aj\A is finite for all but

finitely many j < 1}, so D2,i does not depend on i. Clearly (D1,i, D2,i) : i < 1satisfies ()3.

Recall, 0.7, that by 0.7 and the fact that > > 2, there is a stationaryS S1 from I[], and so every stationary S

S has the same properties (i.e.

is a stationary subset of which belongs to I[] and is included in S1).

Let N (H((2)+), ,


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8 SAHARON SHELAH

As N we can find an increasing continuous sequence : < 1 N ofordinals with limit . So by ()2

()3 for every 1 < 1 for some 2 (1, 1) the set { S0

: f(n()) [1 , 2)} is stationary.

Hence for some unbounded subset u of 1 we have

()4 for every u the set { S0 : f(n()) [, +1)} is a stationarysubset of .

If 21 then u = 1 recalling we demand : < 1 N.We define h : 1 by h() = iff for some u we have = otp(u

f(n())) and/or = 0 & f(n()) sup(u).Clearly h N is as required. So S = Sh as required exists. But maybe 21 > ,

then after ()3 we continue as follows. Let C = C : S3

1 be such that Cis a club of of order type 1 which guess clubs, i.e. for every club C of 3 forstationarily many S31 we have C C, exists by [Sh:g, III]. Without loss of

generality C N.Now let acc(C) has cofinality 3. Again belongs to N hence

some increasing continuous sequence : < 3 N has limit . Now foreach S31 we could choose above = hence for some n < we have

( < )( < )[

< (stat S0)( f(n) <

]. So for some

n < the set S := { < 3 : cf() = 1 and n = n} is a stationary subset ofS

31

.

It follows that ( < 3)( < 3)[ < (stat S0)( f(n) < +1)].

Let be the minimal as required above, so C = { < 3: if < then < and is a limit ordinal} is a club of 3. Hence for some () S

31

we haveC() C. Let u := { : C()} so clearly : u belongs to N. 1.5

1.7 Remark. Why cant we, in the proof of 1.5, after ()3, put the instead assuming21 use as N (H(2)+), ,


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A = {A : S}

A= {A : < 1}.

SoA []0 .

As the case = 2 was the original question and its proof is simpler we first proveit.

1.8 Subclaim. A does not reflect in any A []1 .

Proof. So assume A []1 , let ai : i < 1 be an increasing continuous sequence

of countable subsets of A with union A, and let R = {i < 1 : ai A

}, andassume toward contradiction that R is a stationary subset of 1. As every a Ais M-closed, necessarily A is M-closed and so without loss of generality each aiis M-closed.For each i R as ai A by the definition ofA we can find i < 1 and i Sisuch that ai A

ii

hence by the definition of Aii we have otp(ai) i. Butas A = {ai : i < 1} with ai countable increasing with i and |A| = 1, clearlyfor some club E of 1 the sequence otp(ai) : i E is strictly increasing, hencei E otp(i E) otp(ai) so without loss of generality i E i otp(ai)and without loss of generality i < j E i < j otp(aj).

Now j E R j otp(aj) j so i : i E R is strictly increasing

but S : < 1 are pairwise disjoint and i Si so i : i E R is withoutrepetitions; but i = sup(ai) and for i < j from R E we have ai aj whichimplies that i = sup(ai) sup(aj) = j so necessarily i : i R E is strictlyincreasing.

As sup(ai) = i for i R E, clearly sup(A) = {i : i E R} and leti = Min(A\i) for i < 1, it is well defined as j : j R E is strictlyincreasing. Thinning E without loss of generality

1 i < j E R i < j & i aj .

Note that, by the choice of M

,

2 i E R i < j E R i aj n

(fi(n) aj) n

(fi(n) + 1

aj).

As i : i E R is (strictly) increasing continuous and R E is a stationarysubset of1 clearly by clause (D) of the assumption of 1.1 we can find a stationary


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R1 E R and n() such that i R1 j R1 i < j n() n < fi(n) : T}.]Let = (2)+ and N B = (H(), , f(n)}.

No problem to carry the induction.

[Clearly if n is well defined then n+1(n) is well defined (by clause (c) or (d) or(e) according to the case; hence n+1 T

n+1 is well defined by why clause (c)holds, i.e. assume n [nk1, nk), why 2n, 2n+1 Nk,n?

Case 1: If n = 0, then 2n = Nk,n trivially.

Case 2: 2n is O.K. hence Nk,n and show 2n+1 Nk,n .

[Why? Because Nk,n B, if k = 0 as 2n+2(2) is defined from 2n and T both ofwhich belongs to Nk,n. If k > 0 we have to check that also k1 Nk,n whichholds by ()0.

Case 3: 2n+1 is O.K. so Nk,n and we have to show 2n+2 Nk,n+1.As 2n+2(n) < n Nk,n+1 this should be clear.]Let = {n : n < }. Clearly lim(T) hence u =: |M| []

0 andM M

, hence it is enough to prove that u A.Now

1 sup(u) [Why? As n belongs to the Skolem hull of N N

hence Mn N N

and N = as E.]

2 sup(u) n, for every n < [by clause (d) of]

3 sup(u) = [Why? By 1 +2]


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4 S and > = otp(u)

[Why? By the choice of ]

5 if n n0, n > 0 and n B1 then u n f(n)

[Why? By the choice of2n+2(2n+1), i.e., let k be such that n [nk1, nk),so 2n+1 Nk,n by clause (c) and by clause (e) of we have 2n+2(2n+1) n Nk,n hence by above, as lim(T) we have (2n+2) = 2n+2 =sup(u n) and as M Nk,n we have 2n+2 Nk,n so sup(u n) =2n+2 < sup(Nk,n n) but the latter is equal to sup(Nk n) by ()7which is equal to gk(n) which is < f(n) by ()8, as required.]

6 if n n1 and n B2 then u n f(n)[Why? By the choice of 2n+2(2n + 1).]

So we are done. 1.9

This (i.e., 1.8 + 1.9) is enough for proving 1.1 in the case = 2. In general weshould replace 1.8 by the following claim.

1.10 Claim. The familyA does not reflect in any uncountable A [] (an) & n S(an) & (an) =n sup(an) sup(c) = (c) for each n < . So if (an) = n = (c), n < necessarily (an) = (c) contradiction so n = (c); hence (c) > (an) and, ofcourse, (c) so n < n < (c) . As n for n < were any membersof , clearly has no last element, and let = sup(). Similarly cf() = 0 is

impossible, so clearly cf() > 0 and let = cf() so |A| < and is aregular uncountable cardinal.As a A sup(a) = and A

[A]0 is stationary clearly A = sup(A) =sup(). Let i : i < be increasing continuous with limit

and if i S thenwe let i = .

For i < let i = Min(A\i), so i i < , i A and i < j < i j .

But i < i < (j)(i < j < i < j) so for some club E0 of we


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16 SAHARON SHELAH

have i < j E0 i < j j ; as we can replace i : i < by i : i E0without loss of generality i < i+1 hence i : i < is strictly increasing.Let A := {i : i < } and let H : []

0 be H(b) = sup{i : i b} and let

J := {R : the family {b A

: H(b) R} = {b A

: sup({i < : i b}) R} is not a stationary subset of [A]0}.Clearly

1 J is an 1-complete ideal on extending the non-stationary ideal and / Jby the definition of the ideal

2 if B J+ (i.e., B P()\J) then {a A : H(a) B} is a stationary

subset of []0 .

By clause (D) of the assumption of 1.1, for some stationary R1 J+ and ni <

for i R1 we have

3 if i < j are from R1 and n ni, nj (but n < ) then fi(n) < fj (n).

Recall that

4 i < j R1 i < j .

Now if i R1, let j(i) = Min(R1\(i + 1)), so fi Jbd fi and R2 R1 from J+ we have i R2 ni, mi n(),so

5 for i < j in R2 we have

() fi [n(), ) fi [n(), )

() fi [n(), ) < fj [n(), ).

Let f n f(n) a n f(n)}. As

R2 J+ clearly A is a stationary subset of [A]0 .Let R3 = {i R2 : i = sup(i R2)} so R3 R2, R2\R3 is a non-stationary subsetof (hence belongs to J) and a A sup(a) {i : i R3}.Let


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A = a [A]

0 :(a) min(R2) a and a is M-closed

(b) if i R2 & j = Min(R2\(i + 1)) then [a i a j ] and

n [n(), ) & a n fi(n) a n\fj (n) =

(c) if i < & n [n(), ) then ()(i a)

(j)(i < j < & j a) ()(fi(n) a f(n)) and

(d) if A n f(n) then a n f

(n)

but ( a)(+ 1 a)

hence sup (a n) > f(n)

.

Clearly A is a club of [A]0 (recall that A is M-closed). But ifa AA, thenfor some limit ordinal i R3 we have a sup(a) = i and n [n(), ) sup(a f(n)) = sup(a {fj (n) : j R2}).Let

B1 = {n : n() n < and A n f(n) = sup(A n)}.

B2 = {n : n() n < and f(n) < sup(A n)}.

Clearly B1, B2 are disjoint with union [n(), ) recalling A + 1 A by0.

By the definition ofA, for every a A A, we have

6 n B2 Cha(n) f(n) > f(a)(n)

7 n B1 Cha(n) = {f(n) : R2 (a)} f(a)(n).

But this contradicts the observation below.

1.11 Observation. If B , then for some < 1 we have:if a A is M-closed and {n < : sup(a n) fsup(a)(n)} = B mod J

bd ,

then otp(a) < .

Proof. Read the definition ofA (and A,A ) and subclaim 1.5 particularly ()3.1.11,1.10,1.1


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Remark. Clearly 1.11 shows that we have much freeness in the choice ofA s.

We can get somewhat more, as in [Sh:e]

1.12 Claim. In Claim 1.1 we can add to the conclusion

() A satisfies the diamond, i.e. A.

Proof. In 1.5 we can add

()5 {2n + 1 : n < } = mod D, for < 2, < 1.

This is easy: replace D, by D

, = {A : {n : 2n A} D,}. We can fixa countable vocabulary and for < 1 choose a function F from P() onto{N : N is a -model with universe } such that F(A) = F(B) if A = B modfinite.

Case 1: > 20 .Lastly, for a A let a, a be such that a A

aa

and let Aa = {n : sup(a 2n+1) < fa(2n)}, and let Na be the -model with universe a such that the one-to-one order preserving function from onto a is an isomorphism from F(N) ontoN. Note that in the proof of A []0 is stationary, i.e. of 1.9, given a -

model M with universe without loss of generality 0 > 20

and so can demandthat the isomorphism type of M is the same for all lim(T) and, of course,M M. Hence the isomorphic type of M u is the same for all lim(T)where u is the universe of M. Now in the choice for B1 we can add the demandF({n : 2n + 1 B1}) is isomorphic to M u for every lim(T).Now check.

Case 2: 20 .Similarly letting {2n + 1 : n < } be the disjoint union of Bn : n < , each

Bn infinite. We use Aa Bn to code model with universe for some < 1, by

a function Fn

. We then let Na

be the model with universe a sucht hat the orderpreserving function from a onto a countable ordinal is an isomorphism from Naonto {Fn(Aa B

n) : n < } when the union is a -model with universe .

Now we cannot demand them all M, lim(T) has the same isomorphismtype but only the same order type. The rest should be clear. 1.12

We can also generalize


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1.13 Claim. We can weaken the assumption of 1.1 as follows

(a) = cf() > instead = + (still necessarily )

(b) replace Jbd by an ideal J on containing the finite subsets, n = cf(n) >

1, = limJn : n < but not necessarily n < n < n+1 and addP()/J is infinite (hence uncountable).

Proof. In 1.5 in ()3 we choose A : < 1, a sequence of subsets of suchthat A/J : < 1 are pairwise distinct. This implies some changes and waivingn < n+1 requires some changes in 1.9, in particular for each n using B : Sn0 with B = { lim(T

) : a n } and the partition theorem [Sh:f,XI,3.7,pg.549].

1.13


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REFERENCES.

[FoTo] Matthew Foreman and Stevo Todorcevic. A new Lowenheim-Skolem theorem. Transactions of the American Mathematical Society,357:16931715, 2005.

[RuSh 117] Matatyahu Rubin and Saharon Shelah. Combinatorial problems ontrees: partitions, -systems and large free subtrees. Annals of Pureand Applied Logic, 33:4381, 1987.

[Sh:E3] S. Shelah. On some problems in general topology. In Set Theory, BoiseID, 19921994, volume 192 of Contemporary Mathematics, pages 91101. 0708.1981.

[Sh:E12] Saharon Shelah. Analytical Guide and Corrections to [Sh:g].

math.LO/9906022.[Sh:e] Saharon Shelah. Nonstructure theory, accepted. Oxford University

Press.

[Sh 420] Saharon Shelah. Advances in Cardinal Arithmetic. In Finite and Infi-nite Combinatorics in Sets and Logic, pages 355383. Kluwer AcademicPublishers, 1993. N.W. Sauer et al (eds.). 0708.1979.

[Sh 410] Saharon Shelah. More on Cardinal Arithmetic. Archive for Mathemat-ical Logic, 32:399428, 1993. math.LO/0406550.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[Sh 430] Saharon Shelah. Further cardinal arithmetic. Israel Journal of Mathe-matics, 95:61114, 1996. math.LO/9610226.

[Sh:f] Saharon Shelah. Proper and improper forcing. Perspectives in Mathe-matical Logic. Springer, 1998.

[Sh 755] Saharon Shelah. Weak Diamond. Mathematica Japonica, 55:531538,2002. math.LO/0107207.

[Sh 775] Saharon Shelah. Middle Diamond. Archive for Mathematical Logic,

44:527560, 2005. math.LO/0212249.

[Ve92a] Boban Velickovic. Forcing axioms and stationary sets. Advances inMathematics, 94:256284, 1992.

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Saharon Shelah- Reflection Implies the SCH