Violating the Singular Cardinals Hypothesis Without Large Cardinals
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1 Violating the Singular Cardinals Hypothesis Without Large Cardinals CUNY Logic Workshop, November 18, 2011 by Peter Koepke (Bonn); joint work with Moti Gitik (Tel Aviv) Easton proved that the behavior of the exponential function 2 κ at regular cardinals κ is independent of the axioms of set theory except for some simple classical laws. The Singular Cardinals Hypothesis SCH implies that the Generalized Continuum Hypothesis GCH 2 κ =κ + holds at a singular cardinal κ if GCH holds below κ. Gitik and Mitchell have determined the consistency strength of the negation of the Singular Cardinals Hypothesis in Zermelo Fraenkel set theory with the axiom of choice AC in terms of large cardinals. Arthur Apter and I pursue a program of determining such consistency strengths in Zermelo Fraenkel set theory without AC. Moti Gitik and I showed that the following negation of the Singular Cardinals Hypothesis is relatively consistent with Zermelo Fraenkel set theory: GCH holds below the first uncountable limit cardinal ℵ ω and there is a surjection from its power set P(ℵ ω ) onto some arbitrarily high cardinal λ. This leads to the conjecture that without the axiom of choice and without assuming large cardinal strength a - surjectively modified - exponential function can take rather arbitrary values at all infinite cardinals. 1
2 Cantor s Continuum Hypothesis Theorem 1. (Georg Cantor) The power set {x x N} of N is not denumerable. Theorem 2. 2 ℵ 0 ℵ 1. Conjecture 3. (Cantor s Continuum Hypothesis, CH) 2 ℵ 0 =ℵ 1. Kurt Gödel proved the consistency of CH, assuming the consistency of the Zermelo- Fraenkel axioms ZF, by constructing the model L of constructible sets. Theorem 4. L CH. Paul Cohen proved the opposite relative consistency Theorem 5. Any (countable) model V of ZFC+2 ℵ0 >ℵ 1. ZFC can be extended to a model V[G] of 2
3 Cohen introduced the method of forcing to adjoin a characteristic function F to the ground model V satisfying 1. F:λ ω 2 for some λ ℵ 2 V 2. i<j<λ λn.f(i,n) λn.f(j,n); set A i =λn.f(i,n) A i A j ω A i A j F 0 1 i j λ 3
4 The Cohen partial order is essentially P ={(p i ) i<λ ) d [1,ω) D [λ] <ω (( i D p i :d 2) ( i D p i = ))} partially ordered by reverse inclusion: p=(p i ) i<λ ) q=(q i ) i<λ ) (p is stronger than q) iff i<λ p i q i ω 0 1 p q q i j λ If G P is a generic path through P then F = p G,i<λ i p i is as required. 4
5 AO Fuzzifying the A i Define the symmetric difference of two functions A,A : dom(a)= dom(a ) 2 by A A (ξ)=1 iff A(ξ) A (ξ). For A,A :ℵ 0 2 define an equivalence relation by A A iff A A V Let Ã={A A A} be the -equivalence class of A and let =(Ã i i<λ) be the sequence of equivalence classes of the Cohen reals. 5
6 A symmetric submodel The model N = HOD V[G] (V {AO } i<λ à i ) consists of all sets which, in V[G], are hereditarily definable from parameters in the transitive closure of V {AO }. Lemma 6. Every set X N is definable in V[G] in the following form: there are an - formula ϕ, x V, n<ω, and i 0,,i l 1 <λ such that X ={u V[G] V[G] ϕ(u,x,ao,a i0,,a il 1 )}. Lemma 7. N is a model of ZF, and there is a surjection f:p(ℵ 0 ) λ in N defined by f(z)= { i, if z à i ; 0, else. 6
7 ,A Approximating N Lemma 8. (Approximation Lemma) Let X N and X Ord. Then there are i 0,, i l 1 <λ such that X V[A i0,,a il 1 ]. Proof. Let X ={u Ord V[G] ϕ(u,x,ao,a i0,,a il 1 )}. Define X = {u Ord there is p=(p i ) P such that p i0 A i0,,p il 1 A il 1, and p ϕ(ǔ,xˇ,τ,a i 0,,A i l 1 )}, where τ,a i0,,a il 1 are canonical names for AO i0,,a il 1 resp. Then X V[A i0,,a il 1 ]. X X is obvious. 7
8 X X uses an automorphism argument to show: whenever p=(p i ) and p =(p i ) are conditions with p i0 A i0,,p il 1 A il 1 and p i0 A i0,,p il 1 A il 1 then we cannot have p ϕ(ǔ,xˇ,τ,a i 0,,A i l 1 ) and p ϕ(ǔ,xˇ,τ,a i 0,,A i l 1 ). Wlog, p,p have the shape: p p i 0 i l 1 i 0 i l 1 Define an automorphism π:{r P r p} {r P r p }: r r p p i 0 i l 1 i 0 i l 1 Since the names uˇ, xˇ, τ, A i0, π(p)=p ϕ., A il 1 are invariant under π, we cannot have p ϕ and 8
9 Lemma 9. If λ (2 ℵ 0 ) V then there is no surjection P(ℵ 0 ) λ + in N. Proof. By the Approximation Lemma the ground model V has λ names for elements in P N (ℵ 0 ). A surjection P(ℵ 0 ) λ + in N would yield a surjection λ (λ + ) N in V[G]. But cardinals are preserved between V, N and V[G]. Theorem 10. There is a model of ZF + AC with a surjection P(ℵ 0 ) ℵ ω and with no surjection P(ℵ 0 ) ℵ ω+1. Hence θ:=θ(ℵ 0 ):=sup{ξ there is a surjection P(ℵ 0 ) ξ}=ℵ ω+1. 9
10 Hausdorff s Generalized Continuum Hypothesis Felix Hausdorff conjectured an extension of CH Conjecture 11. (Generalized Continuum Hypothesis, GCH) α 2 ℵ α =ℵ α+1 Since GCH holds in Gödel s model L, Theorem 12. GCH is independent of ZFC. 10
11 William B. Easton proved Theorem 13. Let E: Ord Ord be a sufficiently absolute function such that E(α)>α α<β E(α) E(β) Lim(E(α)) cof(e(α))>ℵ α Then there is a model V[G] such that α(ℵ α is regular 2 ℵ α =ℵ E(α) ) 11
12 The Singular Cardinals Hypothesis is / implies the statement (SCH) if κ is a singular strong limit cardinal then 2 κ =κ + Moti Gitik and Bill Mitchell showed Theorem 14. The following two theories are equiconsistent: ZFC+ SCH ZFC + there are many measurable cardinals 12
13 SCH without the Axiom of Choice Theorem 15. The following theories are equiconsistent: ZF ZF + GCH holds below ℵ ω some fixed big ordinal α + there is a surjection from P(ℵ ω ) onto ℵ α, for This is a surjective failure of SCH, without requiring large cardinals. Injective failures possess high consistency strengths. 13
14 The forcing Fix a ground model V of ZFC+GCH and let λ=ℵ α be some cardinal in V. The forcing P 0 =(P 0,, ) adjoins one Cohen subset of ℵ n+1 for every n<ω. P 0 ={p (δ n ) n<ω ( n<ω:δ n [ℵ n,ℵ n+1 ) p: n<ω [ℵ n,δ n ) 2)}. The forcing (P, P, ) is defined by 14
15 P ={(p,(a i,p i ) i<λ ) (δ n ) n<ω D [λ] <ω ( n<ω:δ n [ℵ n,ℵ n+1 ), p : n<ω [ℵ n,δ n ) 2 2, i Dp i : n<ω [ℵ n,δ n ) 2 p i, i Da i [ℵ ω \ℵ 0 ] <ω n<ω card(a i [ℵ n,ℵ n+1 )) 1, i Da i =p i = )} ℵ n+1 ℵ n ℵ 2 ξ a i ℵ 1 ℵ 0 ξ p i p j p 15
16 P is partially ordered by p =(p,(a i,p i ) i<λ ) P (p,(a i,p i ) i<λ )=p iff a) p p, i<λ(a i a i p i p i ), b) i<λ n<ω ξ a i [ℵ n,ℵ n+1 ) ζ dom(p i \p i ) [ℵ n,ℵ n+1 ) p i (ζ)=p (ξ)(ζ), and c) j supp(p) (a j \a j ) i supp(p),ij a i =. equal endsegments ξ a i ξ p i p i p j p p 16
17 Lemma 16. P satisfies the ℵ ω+2 -chain condition. Let G be V -generic for P. 17
18 Define G = {p P (p,(a i,p i ) i<λ ) G} A = G : [ℵ n,ℵ n+1 ) 2 2 n<ω A (ξ) = {(ζ,a (ξ,ζ)) ζ [ℵ n,ℵ n+1 )}:[ℵ n,ℵ n+1 ) 2, for ℵ n ξ<ℵ n+1 A i = {p i (p,(a j,p j ) j<λ ) G}:[ℵ 0,ℵ ω ) 2 equal endsegments ξ a i ξ A i A j A 18
19 Fuzzifying the A i For functions A,A :(ℵ ω \ℵ 0 ) 2 define an equivalence relation by A A iff n<ω((a A ) ℵ n+1 V[G ] (A A ) [ℵ n+1,ℵ ω ) V). Let Ã={A A A} be the -equivalence class of A. 19
20 The symmetric submodel Set T =P(<κ) V[A ], setting κ=ℵ ω V ; AO =(Ã i i<λ). The final model is N = HOD V[G] (V {T,AO } T i<λ Ã i ) consisting of all sets which, in V[G] are hereditarily definable from parameters in the transitive closure of V {T,AO }. 20
21 Lemma 17. Every set X N is definable in V[G] in the following form: there are an - formula ϕ, x V, n<ω, and i 0,,i l 1 <λ such that V X ={u V[G] V[G] ϕ(u,x,t,ao,a (ℵ n+1 ) 2,A i0,,a il 1 )}. Lemma 18. N is a model of ZF, and there is a surjection f:p(κ) λ in N defined by f(z)= { i, if z à i ; 0, else; 21
22 Approximating N Lemma 19. Let X N and X Ord. Then there are n<ω and i 0,,i l 1 <λ such that V X V[A (ℵ n+1 ) 2,A i0,,a il 1 ]. Proof. Let V X ={u Ord V[G] ϕ(u,x,t,ao,a (ℵ n+1 ) 2,A i0,,a il 1 )}. By taking n sufficiently large, we may assume that For j<l set j<k<l m [n,ω) δ [ℵ m,ℵ m+1 ):A ij [δ,ℵ m+1 ) A ik [δ,ℵ m+1 ). a ij ={ξ m n δ [ℵ m,ℵ m+1 ):A ij [δ,ℵ m+1 )=A (ξ) [δ,ℵ m+1 )} where A (ξ)={(ζ,a (ξ,ζ)) (ξ,ζ) dom(a )}. 22
23 Define X = {u Ord there is p=(p,(a i,p i ) i<λ ) P such that V p (ℵ n+1 ) 2 V A (ℵ n+1 ) 2, a i0 a i0,,a il 1 a il 1, p i0 A i0,,p il 1 A il 1, and p ϕ(ǔ,xˇ,σ,τ,a (ℵˇn+1) 2,A i 0,,A i l 1 )}, where σ,τ,a,a i0,,a il 1 are canonical names for T,AO,A,A i0,,a il 1 resp. Then X V V[A (ℵ n+1 ) 2,A i0,,a il 1 ] and X X. The converse direction, X X, uses an automorphism argument to show: whenever p = (p i ) and p =(p i ) are conditions as in the definition of X then we cannot have p ϕ(ǔ,xˇ,σ,τ,a i 0,,A i l 1 ) and p ϕ(ǔ,xˇ,σ,τ,a i 0,,A i l 1 ). 23
24 possibly different same since A il 1 possibly different p [ℵ n,ℵ n+1 ) 2 A = p i0 p i0 p i p i p ℵ 1 2 = p 2 ℵ 1 a i0 a il 1 i supp(p) identities outside supp(p) corresponding grey parts are equal black extensions determined by linking ordinals = q ℵ 1 2 = q 2 ℵ 1 a i0 a il 1 i supp(p) 24
25 Wrapping up Lemma 20. Let n < ω and i 0, V[A V (ℵ n+1 ) 2,A i0,,a il 1 ]., i l 1 < λ. Then cardinals are absolute between V and Lemma 21. Cardinals are absolute between N and V, and in particular κ=ℵ ω V =ℵ ω N. Proof. If not, then there is a function f N which collapses a cardinal in V. By Lemma V 19, f is an element of some model V [A (ℵ n+1 ) 2,A i0,,a il 1 ] as above. But this contradicts Lemma
26 Lemma 22. GCH holds in N below ℵ ω. V Proof. If X ℵ n and X N then X is an element of some model V[A (ℵ n+1 ) 2,A i0,, A il 1 ] as above. Since A i0,,a il 1 do not adjoin new subsets of ℵ n we have that V X V[A (ℵ n+1 ) 2 ]. Hence P(ℵ V V n ) N V[A (ℵ n+1 ) 2 V ]. GCH holds in V[A (ℵ n+1 ) 2 ]. Hence there is a bijection P(ℵ V V V n ) N ℵ n+1 in V[A (ℵ n+1 ) 2 ] and hence in N. 26
27 Discussion and Remarks To work with singular cardinals κ of uncountable cofinality, various finiteness properties in the construction have to be replaced by the property of being of cardinality < cof(κ). This yields choiceless violations of Silver s theorem. Theorem 23. Let V be any ground model of ZFC + GCH and let λ be some cardinal in V. Then there is a model N V of the theory ZF+ GCH holds below ℵ ω1 + there is a surjection from P(ℵ ω1 ) onto λ. Moreover, the axiom of dependent choices DC holds in N. 27
28 Conjecture 24. Let E: Ord Ord be a sufficiently absolute function such that E(α) α+2 α<β E(α) E(β) Then there is a model V[G] in which for all α θ(ℵ α ):=sup{ξ there is a surjection P(ℵ α ) ξ}=ℵ E(α). 28
29 THANK YOU 29
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