A polarized partition relation and failure of GCH at singular strong limit

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1 F U N D A M E N T A MATHEMATICAE 155 (1998) A polarzed partton relaton and falure of GCH at sngular strong lmt by Saharon S h e l a h (Jerusalem and New Brunswck, N.J.) Abstract. The man result s that for λ strong lmt sngular falng the contnuum hypothess (.e. 2 λ > λ ), a polarzed partton theorem holds. 1. Introducton. In the present paper we show a polarzed partton theorem for strong lmt sngular cardnals λ falng the contnuum hypothess. Let us recall the followng defnton. Defnton 1.1. For ordnal numbers α 1, α 2, β 1, β 2 and a cardnal, the polarzed partton symbol ( ) ( ) 1,1 α1 α2 β 1 β 2 means that f d s a functon from α 1 β 1 nto then for some A α 1 of order type α 2 and B β 1 of order type β 2, the functon d A B s constant. () We address the followng problem of Erdős and Hajnal: f µ s strong lmt sngular of uncountable cofnalty wth < cf(µ), does ( ) ( ) µ 1,1 µ? The partcular case of ths queston for µ = ℵ ω1 and = 2 was posed by Erdős, Hajnal and Rado (under the assumpton of GCH) n [EHR, Problem 11, p. 183]). Hajnal sad that the assumpton of GCH n [EHR] was not crucal, and he added that the ntenton was to ask the queston n some, preferably nce, Set Theory Mathematcs Subject Classfcaton: Prmary 03E05, 04A20, 04A30. Research partally supported by Basc Research Foundaton admnstered by The Israel Academy of Scences and Humantes. Publcaton 586. [153]

2 154 S. Shelah Baumgartner and Hajnal have proved that f µ s weakly compact then the answer to () s yes (see [BH]), also f µ s strong lmt of cofnalty ℵ 0. But for a weakly compact µ we do not know f for every α < µ : ( ) ( ) µ 1,1 α. The frst tme I heard the problem (around 1990) I noted that () holds when µ s a sngular lmt of measurable cardnals. Ths result s presented n Theorem 2.2. It seemed lkely that we could combne ths wth sutable collapses, to get small such µ (lke ℵ ω1 ) but there was no success n ths drecton. In September 1994, Hajnal reasked me the queston puttng great stress on t. Here we answer the problem () usng methods of [Sh:g]. But nstead of the assumpton of GCH (postulated n [EHR]) we assume 2 µ > µ. The proof seems qute flexble but we dd not fnd out what else t s good for. Ths s a good example of the major theme of [Sh:g]: Thess 1.2. Whereas CH and GCH are good (helpful, strategc) assumptons havng many consequences, and, say, CH s not, the negaton of GCH at sngular cardnals (.e. for µ strong lmt sngular 2 µ > µ, or the really strong hypothess: cf(µ) < µ pp(µ) > µ ) s a good (helpful, strategc) assumpton. Foreman ponted out that the result presented n Theorem 1.2 below s preserved by µ -closed forcng notons. Therefore, f ( ) ( ) λ 1,1 λ V = λ λ then ( ) V Levy(λ,2 λ) λ = λ ( ) 1,1 λ. λ Consequently, the result s consstent wth 2 λ = λ & λ s small. (Note that although our fnal model may satsfy the Sngular Cardnals Hypothess, the ntermedate model stll volates SCH at λ, hence needs large cardnals, see [J].) For λ not small we can use Theorem 2.2. Before we move to the man theorem, let us recall an open problem mportant for our methods: Problem 1.3. (1) Let κ = cf(µ) > ℵ 0, µ > 2 κ and λ = cf(λ) (µ, pp (µ)). Can we fnd < µ and a [µ Reg] such that λ pcf(a), a = <κ a, a bounded n µ and σ a α<σ α < σ? For ths t s enough to show:

3 Polarzed partton relaton 155 (2) If µ = cf(µ) > 2 < but α<µ α < µ then we can fnd a [µ Reg] < such that λ pcf(a). (In fact, t suffces to prove t for the case = ℵ 1.) As shown n [Sh:g] we have Theorem 1.4. If µ s strong lmt sngular of cofnalty κ > ℵ 0 and 2 µ > λ = cf(λ) > µ then for some strctly ncreasng sequence λ : < κ of regulars wth lmt µ, <κ λ /Jκ bd has true cofnalty λ. If κ = ℵ 0, ths stll holds for λ = µ. [More fully, by [Sh:g, II, 5], we know pp(µ) = 2 µ and by [Sh:g, VIII, 1.6(2)], we know pp (µ) = pp (µ). Note that for κ = ℵ Jκ bd 0 we should replace by a possbly larger deal, usng [Sh 430, 1.1, 6.5] but there s no need J bd κ here.] Remark 1.5. Note that the problem s a pp = cov problem (see more n [Sh 430, 1]); so f κ = ℵ 0 and λ < µ ω 1 the concluson of 1.4 holds; we allow Jκ bd to be ncreased, even there are < µ fxed ponts < λ suffces. 2. Man result Theorem 2.1. Suppose µ s strong lmt sngular satsfyng 2 µ > µ. Then: ( ) ( ) µ 1,1 µ 1 (1) for any < cf(µ). (2) If d s a functon from µ µ to and < µ then for some sets A µ and B µ we have otp(a) = µ 1, otp(b) = µ and the restrcton d A B does not depend on the frst coordnate. P r o o f. (1) Ths follows from part (2) (snce f d(α, β) = d (β) for α A, β B, where d : B, and B = µ, < cf(µ) then there s B B wth B = µ such that d B s constant and hence d A B s constant as requred). (2) Let d : µ µ. Let κ = cf(µ) and µ = µ : < κ be a contnuous strctly ncreasng sequence such that µ = <κ µ, µ 0 > κ. We can fnd a sequence C = C α : α < µ such that: (A) C α α s closed, otp(c α ) < µ, (B) β nacc(c α ) C β = C α β, (C) f C α has no last element then α = sup(c α ) (so α s a lmt ordnal) and any member of nacc(c α ) s a successor ordnal, (D) f σ = cf(σ) < µ then the set S σ := {δ < µ : cf(δ) = σ & δ = sup(c δ ) & otp(c δ ) = σ} s statonary

4 156 S. Shelah (possble by [Sh 420, 1]); we could have added (E) for every σ Reg µ and a club E of µ, for statonary many δ S σ, E separates any two successve members of C δ. Let c be a symmetrc two-place functon from µ to κ such that for each < κ and β < µ the set a β := {α < β : c(α, β) } has cardnalty µ and α < β < γ c(α, γ) max{c(α, β), c(β, γ)} and α C β & µ C β c(α, β) (as n [Sh 108], easly constructed by nducton on β). Let λ = λ : < κ be a strctly ncreasng sequence of regular cardnals wth lmt µ such that <κ λ /Jκ bd has true cofnalty µ (exsts by 1.4 wth λ = µ 2 µ ). As we can replace λ by any subsequence of length κ, wthout loss of generalty ( < κ)(λ > 2 µ ). Lastly, let χ = ℶ 8 (µ) and < χ be a well orderng of H(χ)(:= {x : the transtve closure of x s of cardnalty < χ}). Now we choose by nducton on α < µ sequences M α = M α, : < κ such that: () M α, (H(χ),, < χ), () M α, = 2 µ and µ (M α, ) M α, and 2 µ 1 M α,, () d, c, C, λ, µ, α M α,, M β,j : β < α, j < κ M α,, β a α M β, M α, and M α,j : j < M α,, j< M α,j M α,, (v) M β, : β a α belongs to M α,. There s no problem to carry out the constructon. Note that actually clause (v) follows from () (), as a α s defned from c, α,. Our demands mply that [β a α M β, M α, ] and [j < M α,j M α, ] and a α M α,, hence α <κ M α,. For α < µ let f α <κ λ be defned by f α () = sup(λ M α, ). Note that f α () < λ as λ = cf(λ ) > 2 µ = M α,. Also, f β < α then for every [c(β, α), κ) we have β M α, and hence M β M α,. Therefore, as also λ M α,, we have f β M α, and f β () M α, λ. Consequently, ( [c(β, α), κ))(f β () < f α ()) and thus f β < J bd κ f α. Snce {f α : α < µ } <κ λ has cardnalty µ and <κ λ /Jκ bd s µ -drected, there s f <κ λ such that () 1 ( α < µ )(f α < J bd κ f ).

5 Polarzed partton relaton 157 Let, for α < µ, g α κ be defned by g α () = d(α, f ()). Snce κ < µ < µ = cf(µ ), there s a functon g κ such that () 2 the set A = {α < µ : g α = g } s unbounded n µ. Now choose, by nducton on ζ < µ, models N ζ such that: (a) N ζ (H(χ),, < χ), (b) the sequence N ζ : ζ < µ s ncreasng contnuous, (c) N ζ = µ and κ> (N ζ ) N ζ f ζ s not a lmt ordnal, (d) N ξ : ξ ζ N ζ1, (e) µ 1 N ζ, α<ζ, <κ M α, N ζ and M α, : α < µ, < κ, f α : α < µ, g, A and d belong to the frst model N 0. Let E := {ζ < µ : N ζ µ = ζ}. Clearly, E s a club of µ, and thus we can fnd an ncreasng sequence δ : < κ such that () 3 δ S µ proof). acc(e) ( µ ) (see clause (D) at the begnnng of the For each < κ choose a successor ordnal α nacc(c δ ) \ {δ j 1 : j < }. Take any α A \ <κ δ. We choose by nducton on < κ an ordnal j and sets A, B such that: (α) j < κ and µ j > λ (so j > ) and j strctly ncreasng n, (β) f δ [j, κ) < f α 1 [j, κ) < f α [j, κ) < f [j, κ), (γ) for each 0 < 1 we have c(δ 0, α 1 ) < j 1, c(α 0, α 1 ) < j 1, c(α 1, α ) < j 1 and c(δ 1, α ) < j 1, (δ) A A (α, δ ), (ε) otp(a ) = µ, (ζ) A M δ,j, (η) B λ j, () otp(b ) = λ j, (ι) B ε M α,j for ε <, (κ) for every α ε A ε {α } and ζ and β B ζ {f (j ζ )} we have d(α, β) = g (j ζ ). If we succeed then A = ε<κ A ε {α } and B = ζ<κ B ζ are as requred. Durng the nducton at stage concernng (ι), f ε 1 = then for some j < κ, B ε M α,j has cardnalty λ jε, hence we can replace B ε by a subset of the same cardnalty whch belongs to the model M α,j f j s large enough such that µ j > λ ; f ε 1 < then by the demand for ε 1, we have j<κ B ε M α,j. So assume that the sequence (j ε, A ε, B ε ) : ε < has already been defned. We can fnd j (0) < κ satsfyng requrements (α), (β), (γ) and (ι) and such that ε< λ j ε < µ j (0). Then for each ε < we have δ ε a α j (0) and

6 158 S. Shelah hence M δε,j ε M α,j (0) (for ε < ). But A ε M δε,j ε (by clause (ζ)) and B ε M α,j (0) (for ε < ), so {A ε, B ε : ε < } M α,j (0). Snce κ> (M α,j (0)) M α,j (0) (see ()), the sequence (A ε, B ε ) : ε < belongs to M α,j (0). We know that for γ 1 < γ 2 n nacc(c δ ) we have c(γ 1, γ 2 ) (remember clause (B) and the choce of c). As j (0) > and so µ j (0) µ, the sequence M := M α,j (0) : α nacc(c δ ) s -ncreasng and M α M α,j (0) for α nacc(c δ ) and M α,j (0) appears n t. Also, as δ acc(e), there s an ncreasng sequence γ ξ : ξ < µ of members of nacc(c δ ) such that γ 0 = α and (γ ξ, γ ξ1 ) E, say β ξ (γ ξ, γ ξ1 ) E. Each element of nacc(c δ ) s a successor ordnal, so every γ ξ s a successor ordnal. Each model M γξ,j (0) s closed under sequences of length µ, and hence γ ζ : ζ < ξ M γξ,j (0) (by choosng the rght C and δ s we could have managed to have α = mn(c δ ), {γ ξ : ξ < µ } = nacc(c δ), wthout usng ths amount of closure). For each ξ < µ, we know that (H(χ),, < χ) = ( x A )[x > γ ξ & ( ε < )( y B ε )(d(x, y) = g (j ε ))] because x = α satsfes t. As all the parameters,.e. A, γ ξ, d, g and B ε : ε <, belong to N βξ (remember clauses (e) and (c); note that B ε M α,j (0), α < β ξ), there s an ordnal β ξ (γ ξ, β ξ ) (γ ξ, γ ξ1 ) satsfyng the demands on x. Now, necessarly for some j (1, ξ) (j (0), κ) we have β ξ M γ ξ1,j (1,ξ). Hence for some j < κ the set A := {β ξ : ξ < µ & j (1, ξ) = j } has cardnalty µ. Clearly A A (as each βξ A ). Now, the sequence M γξ,j : ξ < µ M δ,j s -ncreasng, and hence A M δ,j. Snce µ j > µ = A we have A M δ,j. Note that at the moment we know that the set A satsfes the demands (δ) (ζ). By the choce of j (0), as j > j (0), clearly M δ,j M α,j, and hence A M α,j. Smlarly, A ε : ε M α,j, α M α,j and sup(m α,j λ j ) = f α (j ) < f (j ). Consequently, ε A ε {α } M α,j (by the nducton hypothess or the above) and t belongs to M α,j. Snce ε A ε {α } A, clearly ( (H(χ),, < χ) = x ) A ε {α } (d(x, f (j )) = g (j )). ε Note that A ε {α }, g (j ), d, λ j M α,j and f (j ) λ j \sup(m α,j λ j ). ε

7 Hence the set B := Polarzed partton relaton 159 { ( y < λ j : x ) } A ε {α } (d(x, y) = g (j )) ε has to be unbounded n λ j. It s easy to check that j, A, B satsfy clauses (α) (κ). Thus we have carred out the nducton step, fnshng the proof of the theorem. 2.1 Theorem 2.2. Suppose µ s a sngular lmt of measurable cardnals. Then ( ) ( ) µ µ (1) f = 2 or at least < cf(µ). ( ) ( ) µ (2) Moreover, f α < µ α and < cf(µ) then (3) If < µ, α < µ and d s a functon from µ µ to then for some A µ, otp(a) = α, and B = <cf(µ) B µ, d A B s constant for each < cf(µ). P r o o f. Clearly (3) (2) (1), so we shall prove part (3). Let d : µ µ. Let κ := cf(µ). Choose sequences λ : < κ and µ : < κ such that µ : < κ s ncreasng contnuous, µ = <κ µ, µ 0 > κ, each λ s measurable and µ < λ < µ 1 (for < κ). Let D be a λ -complete unform ultraflter on λ. For α < µ defne g α κ by g α () = γ ff {β < λ : d(α, β) = γ} D (as < λ t exsts). The number of such functons s κ < µ (as µ s necessarly strong lmt), so for some g κ the set A := {α < µ : g α = g } s unbounded n µ. For each < κ we defne an equvalence relaton e on µ : αe β ff ( γ < λ )[d(α, γ) = d(β, γ)]. So the number of e -equvalence classes s λ < µ. Hence we can fnd an ncreasng contnuous sequence α ζ : ζ < µ of ordnals < µ such that: () for each < κ and e -equvalence class X, ether X A α 0, or for every ζ < µ, (α ζ, α ζ1 ) X A has cardnalty µ. Let α = <κ a, a = µ, a : < κ parwse dsjont. Now, by nducton on < κ, we choose A, B such that: (a) A {(α ζ, α ζ1 ) : ζ a } A and each A (α ζ, α ζ1 ) s a sngleton, (b) B D, (c) f α A, β B j, j then d(α, β) = g (j). Now, at stage, (A ε, B ε ) : ε < are already chosen. Let us choose A ε. For each ζ a choose β ζ (α ζ, α ζ1 ) A such that f > 0 then for some.

8 160 S. Shelah β A 0, β ζ e β, and let A = {β ζ : ζ a }. Now clause (a) s mmedate, and the relevant part of clause (c),.e. j <, s O.K. Next, as j A j A, the set B := {γ < λ : d(β, γ) = g ()} j β A j s the ntersecton of µ < λ sets from D and hence B D. Clearly clause (b) and the remanng part of clause (c) (.e. j = ) holds. So we can carry out the nducton and hence fnsh the proof. 2.2 References [EHR] P. Erdős, A. Hajnal and R. Rado, Partton relatons for cardnal numbers, Acta Math. Acad. Sc. Hungar. 16 (1965), [BH] J. Baumgartner and A. Hajnal, Polarzed partton relatons, preprnt, [J] T. Jech, Set Theory, Academc Press, New York, [Sh:g] S. Shelah, Cardnal Arthmetc, Oxford Logc Gudes 29, Oxford Unv. Press, [Sh 430], Further cardnal arthmetc, Israel J. Math. 95 (1996), [Sh 420], Advances n cardnal arthmetc, n: Fnte and Infnte Combnatorcs n Sets and Logc, N. W. Sauer et al. (eds.), Kluwer Acad. Publ., 1993, [Sh 108], On successors of sngular cardnals, n: Logc Colloquum 78 (Mons, 1978), Stud. Logc Found. Math. 97, North-Holland, Amsterdam, 1979, Insttute of Mathematcs Department of Mathematcs The Hebrew Unversty Rutgers Unversty Jerusalem, Israel New Brunswck, New Jersey E-mal: shelah@math.huj.ac.l U.S.A. Receved 4 November 1995; n revsed form 18 October 1996