155 (1998)
A polarized partition relation and failure of GCH at singular strong limit
by
Saharon S h e l a h (Jerusalem and New Brunswick, N.J.)
Abstract. The main result is that for λ strong limit singular failing the continuum hypothesis (i.e. 2
λ> λ
+), a polarized partition theorem holds.
1. Introduction. In the present paper we show a polarized partition theorem for strong limit singular cardinals λ failing the continuum hypoth- esis. Let us recall the following definition.
Definition 1.1. For ordinal numbers α
1, α
2, β
1, β
2and a cardinal θ, the polarized partition symbol
α
1β
1→
α
2β
2 1,1θ
means that if d is a function from α
1× β
1into θ then for some A ⊆ α
1of order type α
2and B ⊆ β
1of order type β
2, the function d¹A×B is constant.
We address the following problem of Erd˝os and Hajnal:
(∗) if µ is strong limit singular of uncountable cofinality with θ < cf(µ),
does
µ
+µ
→
µ µ
1,1θ
?
The particular case of this question for µ = ℵ
ω1and θ = 2 was posed by Erd˝os, Hajnal and Rado (under the assumption of GCH) in [EHR, Prob- lem 11, p. 183]). Hajnal said that the assumption of GCH in [EHR] was not crucial, and he added that the intention was to ask the question “in some, preferably nice, Set Theory”.
1991 Mathematics Subject Classification: Primary 03E05, 04A20, 04A30.
Research partially supported by “Basic Research Foundation” administered by The Israel Academy of Sciences and Humanities. Publication 586.
[153]
Baumgartner and Hajnal have proved that if µ is weakly compact then the answer to (∗) is “yes” (see [BH]), also if µ is strong limit of cofinality ℵ
0. But for a weakly compact µ we do not know if for every α < µ
+:
µ
+µ
→
α µ
1,1θ
.
The first time I heard the problem (around 1990) I noted that (∗) holds when µ is a singular limit of measurable cardinals. This result is presented in Theorem 2.2. It seemed likely that we could combine this with suitable collapses, to get “small” such µ (like ℵ
ω1) but there was no success in this direction.
In September 1994, Hajnal reasked me the question putting great stress on it. Here we answer the problem (∗) using methods of [Sh:g]. But instead of the assumption of GCH (postulated in [EHR]) we assume 2
µ> µ
+. The proof seems quite flexible but we did not find out what else it is good for.
This is a good example of the major theme of [Sh:g]:
Thesis 1.2. Whereas CH and GCH are good (helpful, strategic) assump- tions having many consequences, and, say, ¬CH is not, the negation of GCH at singular cardinals (i.e. for µ strong limit singular 2
µ> µ
+, or the really strong hypothesis: cf(µ) < µ ⇒ pp(µ) > µ
+) is a good (helpful, strategic) assumption.
Foreman pointed out that the result presented in Theorem 1.2 below is preserved by µ
+-closed forcing notions. Therefore, if
V |=
λ
+λ
→
λ λ
1,1θ
then
V
Levy(λ+,2λ)|=
λ
+λ
→
λ λ
1,1θ
.
Consequently, the result is consistent with 2
λ= λ
+& λ is small. (Note that although our final model may satisfy the Singular Cardinals Hypothesis, the intermediate model still violates SCH at λ, hence needs large cardinals, see [J].) For λ not small we can use Theorem 2.2.
Before we move to the main theorem, let us recall an open problem important for our methods:
Problem 1.3. (1) Let κ = cf(µ) > ℵ
0, µ > 2
κand λ = cf(λ) ∈ (µ, pp
+(µ)). Can we find θ < µ and a ∈ [µ ∩ Reg]
θsuch that λ ∈ pcf(a), a = S
i<κ
a
i, a
ibounded in µ and σ ∈ a
i⇒ V
α<σ
|α|
θ< σ? For this it is
enough to show :
(2) If µ = cf(µ) > 2
<θbut W
α<µ
|α|
<θ≥ µ then we can find a ∈ [µ ∩ Reg]
<θsuch that λ ∈ pcf(a). (In fact, it suffices to prove it for the case θ = ℵ
1.)
As shown in [Sh:g] we have
Theorem 1.4. If µ is strong limit singular of cofinality κ > ℵ
0and 2
µ> λ = cf(λ) > µ then for some strictly increasing sequence hλ
i: i < κi of regulars with limit µ, Q
i<κ
λ
i/J
κbdhas true cofinality λ. If κ = ℵ
0, this still holds for λ = µ
++.
[More fully, by [Sh:g, II, §5], we know pp(µ) =
+2
µand by [Sh:g, VIII, 1.6(2)], we know pp
+(µ) = pp
+Jbdκ
(µ). Note that for κ = ℵ
0we should replace J
κbdby a possibly larger ideal, using [Sh 430, 1.1, 6.5] but there is no need here.]
Remark 1.5. Note that the problem is a pp = cov problem (see more in [Sh 430, §1]); so if κ = ℵ
0and λ < µ
+ω1the conclusion of 1.4 holds; we allow J
κbdto be increased, even “there are < µ
+fixed points < λ
+” suffices.
2. Main result
Theorem 2.1. Suppose µ is strong limit singular satisfying 2
µ> µ
+. Then:
(1)
µ
+µ
→
µ + 1 µ
1,1θ
for any θ < cf(µ).
(2) If d is a function from µ
+× µ to θ and θ < µ then for some sets A ⊆ µ
+and B ⊆ µ we have otp(A) = µ + 1, otp(B) = µ and the restriction d¹A × B does not depend on the first coordinate.
P r o o f. (1) This follows from part (2) (since if d(α, β) = d
0(β) for α ∈ A, β ∈ B, where d
0: B → θ, and |B| = µ, θ < cf(µ) then there is B
0⊆ B with |B
0| = µ such that d
0¹B is constant and hence d¹A × B
0is constant as required).
(2) Let d : µ
+× µ → θ. Let κ = cf(µ) and µ = hµ
i: i < κi be a continuous strictly increasing sequence such that µ = P
i<κ
µ
i, µ
0> κ + θ.
We can find a sequence C = hC
α: α < µ
+i such that:
(A) C
α⊆ α is closed, otp(C
α) < µ, (B) β ∈ nacc(C
α) ⇒ C
β= C
α∩ β,
(C) if C
αhas no last element then α = sup(C
α) (so α is a limit ordinal) and any member of nacc(C
α) is a successor ordinal,
(D) if σ = cf(σ) < µ then the set
S
σ:= {δ < µ
+: cf(δ) = σ & δ = sup(C
δ) & otp(C
δ) = σ}
is stationary
(possible by [Sh 420, §1]); we could have added
(E) for every σ ∈ Reg ∩µ
+and a club E of µ
+, for stationary many δ ∈ S
σ, E separates any two successive members of C
δ.
Let c be a symmetric two-place function from µ
+to κ such that for each i < κ and β < µ
+the set
a
βi:= {α < β : c(α, β) ≤ i}
has cardinality ≤ µ
iand α < β < γ ⇒ c(α, γ) ≤ max{c(α, β), c(β, γ)} and α ∈ C
β& µ
i≥ |C
β| ⇒ c(α, β) ≤ i
(as in [Sh 108], easily constructed by induction on β).
Let λ = hλ
i: i < κi be a strictly increasing sequence of regular cardinals with limit µ such that Q
i<κ
λ
i/J
κbdhas true cofinality µ
++(exists by 1.4 with λ = µ
++≤ 2
µ). As we can replace λ by any subsequence of length κ, without loss of generality (∀i < κ)(λ
i> 2
µ+i). Lastly, let χ = i
8(µ)
+and <
∗χbe a well ordering of H(χ)(:= {x : the transitive closure of x is of cardinality < χ}).
Now we choose by induction on α < µ
+sequences M
α= hM
α,i: i < κi such that:
(i) M
α,i≺ (H(χ), ∈, <
∗χ),
(ii) kM
α,ik = 2
µ+iand
µ+i(M
α,i) ⊆ M
α,iand 2
µ+i+ 1 ⊆ M
α,i, (iii) d, c, C, λ, µ, α ∈ M
α,i, hM
β,j: β < α, j < κi ∈ M
α,i, S
β∈aαi
M
β,i⊆ M
α,iand hM
α,j: j < ii ∈ M
α,i, S
j<i
M
α,j⊆ M
α,i, (iv) hM
β,i: β ∈ a
αii belongs to M
α,i.
There is no problem to carry out the construction. Note that actually clause (iv) follows from (i)–(iii), as a
αiis defined from c, α, i. Our demands imply that
[β ∈ a
αi⇒ M
β,i≺ M
α,i] and [j < i ⇒ M
α,j≺ M
α,i] and a
αi⊆ M
α,i, hence α ⊆ S
i<κ
M
α,i. For α < µ
+let f
α∈ Q
i<κ
λ
ibe defined by f
α(i) = sup(λ
i∩ M
α,i). Note that f
α(i) < λ
ias λ
i= cf(λ
i) > 2
µ+i= kM
α,ik. Also, if β < α then for every i ∈ [c(β, α), κ) we have β ∈ M
α,iand hence M
β∈ M
α,i. Therefore, as also λ ∈ M
α,i, we have f
β∈ M
α,iand f
β(i) ∈ M
α,i∩ λ
i. Consequently,
(∀i ∈ [c(β, α), κ))(f
β(i) < f
α(i)) and thus f
β<
Jbd κf
α. Since {f
α: α < µ
+} ⊆ Q
i<κ
λ
ihas cardinality µ
+and Q
i<κ
λ
i/J
κbdis µ
++-directed, there is f
∗∈ Q
i<κ
λ
isuch that (∗)
1(∀α < µ
+)(f
α<
Jbdκ
f
∗).
Let, for α < µ
+, g
α∈
κθ be defined by g
α(i) = d(α, f
∗(i)). Since |
κθ| < µ <
µ
+= cf(µ
+), there is a function g
∗∈
κθ such that
(∗)
2the set A
∗= {α < µ
+: g
α= g
∗} is unbounded in µ
+. Now choose, by induction on ζ < µ
+, models N
ζsuch that:
(a) N
ζ≺ (H(χ), ∈, <
∗χ),
(b) the sequence hN
ζ: ζ < µ
+i is increasing continuous, (c) kN
ζk = µ and
κ>(N
ζ) ⊆ N
ζif ζ is not a limit ordinal, (d) hN
ξ: ξ ≤ ζi ∈ N
ζ+1,
(e) µ + 1 ⊆ N
ζ, S
α<ζ, i<κ
M
α,i⊆ N
ζand hM
α,i: α < µ
+, i < κi, hf
α: α < µ
+i, g
∗, A
∗and d belong to the first model N
0.
Let E := {ζ < µ
+: N
ζ∩ µ
+= ζ}. Clearly, E is a club of µ
+, and thus we can find an increasing sequence hδ
i: i < κi such that
(∗)
3δ
i∈ S
µ+i
∩ acc(E) (⊆ µ
+) (see clause (D) at the beginning of the proof).
For each i < κ choose a successor ordinal α
∗i∈ nacc(C
δi) \ S
{δ
j+ 1 : j < i}.
Take any α
∗∈ A
∗\ S
i<κ
δ
i.
We choose by induction on i < κ an ordinal j
iand sets A
i, B
isuch that:
(α) j
i< κ and µ
ji> λ
i(so j
i> i) and j
istrictly increasing in i, (β) f
δi¹[j
i, κ) < f
α∗i+1
¹[j
i, κ) < f
α∗¹[j
i, κ) < f
∗¹[j
i, κ),
(γ) for each i
0< i
1we have c(δ
i0, α
∗i1) < j
i1, c(α
∗i0, α
∗i1) < j
i1, c(α
∗i1, α
∗)
< j
i1and c(δ
i1, α
∗) < j
i1, (δ) A
i⊆ A
∗∩ (α
∗i, δ
i), (ε) otp(A
i) = µ
+i, (ζ) A
i∈ M
δi,ji, (η) B
i⊆ λ
ji, (θ) otp(B
i) = λ
ji,
(ι) B
ε∈ M
α∗i,ji
for ε < i, (κ) for every α ∈ S
ε≤i
A
ε∪ {α
∗} and ζ ≤ i and β ∈ B
ζ∪ {f
∗(j
ζ)} we have d(α, β) = g
∗(j
ζ).
If we succeed then A = S
ε<κ
A
ε∪{α
∗} and B = S
ζ<κ
B
ζare as required.
During the induction at stage i concerning (ι), if ε + 1 = i then for some j < κ, B
ε∩M
α∗i,j
has cardinality λ
jε, hence we can replace B
εby a subset of the same cardinality which belongs to the model M
α∗i,j
if j is large enough such that µ
j> λ
i; if ε + 1 < i then by the demand for ε + 1, we have W
j<κ
B
ε∈ M
α∗i,j
. So assume that the sequence h(j
ε, A
ε, B
ε) : ε < ii has already been defined.
We can find j
i(0) < κ satisfying requirements (α), (β), (γ) and (ι) and such that V
ε<i
λ
jε< µ
ji(0). Then for each ε < i we have δ
ε∈ a
αj∗ii(0)
and
hence M
δε,jε≺ M
α∗i,ji(0)
(for ε < i). But A
ε∈ M
δε,jε(by clause (ζ)) and B
ε∈ M
α∗i,ji(0)
(for ε < i), so {A
ε, B
ε: ε < i} ⊆ M
α∗i,ji(0)
. Since
κ>
(M
α∗i,ji(0)
) ⊆ M
α∗i,ji(0)
(see (ii)), the sequence h(A
ε, B
ε) : ε < ii belongs to M
α∗i,ji(0)
. We know that for γ
1< γ
2in nacc(C
δi) we have c(γ
1, γ
2) ≤ i (remember clause (B) and the choice of c). As j
i(0) > i and so µ
ji(0)≥ µ
+i, the sequence
M
∗:= hM
α,ji(0): α ∈ nacc(C
δi)i
is ≺-increasing and M
∗¹α ∈ M
α,ji(0)for α ∈ nacc(C
δi) and M
α∗i,ji(0)
ap- pears in it. Also, as δ
i∈ acc(E), there is an increasing sequence hγ
ξ: ξ < µ
+ii of members of nacc(C
δi) such that γ
0= α
∗iand (γ
ξ, γ
ξ+1) ∩ E 6= ∅, say β
ξ∈ (γ
ξ, γ
ξ+1) ∩ E. Each element of nacc(C
δ) is a successor ordinal, so every γ
ξis a successor ordinal. Each model M
γξ,ji(0)is closed under se- quences of length ≤ µ
+i, and hence hγ
ζ: ζ < ξi ∈ M
γξ,ji(0)(by choos- ing the right C and δ
i’s we could have managed to have α
∗i= min(C
δi), {γ
ξ: ξ < µ
+i} = nacc(C
δ), without using this amount of closure).
For each ξ < µ
+i, we know that
(H(χ), ∈, <
∗χ) |= “(∃x ∈ A
∗)[x > γ
ξ& (∀ε < i)(∀y ∈ B
ε)(d(x, y) = g
∗(j
ε))]”
because x = α
∗satisfies it. As all the parameters, i.e. A
∗, γ
ξ, d, g
∗and hB
ε: ε < ii, belong to N
βξ(remember clauses (e) and (c); note that B
ε∈ M
α∗i,ji(0)
, α
∗i< β
ξ), there is an ordinal β
ξ∗∈ (γ
ξ, β
ξ) ⊆ (γ
ξ, γ
ξ+1) satisfying the demands on x. Now, necessarily for some j
i(1, ξ) ∈ (j
i(0), κ) we have β
ξ∗∈ M
γξ+1,ji(1,ξ). Hence for some j
i< κ the set
A
i:= {β
ξ∗: ξ < µ
+i& j
i(1, ξ) = j
i}
has cardinality µ
+i. Clearly A
i⊆ A
∗(as each β
ξ∗∈ A
∗). Now, the sequence hM
γξ,ji: ξ < µ
+ii
_hM
δi,jii is ≺-increasing, and hence A
i⊆ M
δi,ji. Since µ
+ji> µ
+i= |A
i| we have A
i∈ M
δi,ji. Note that at the moment we know that the set A
isatisfies the demands (δ)–(ζ). By the choice of j
i(0), as j
i> j
i(0), clearly M
δi,ji≺ M
α∗,ji, and hence A
i∈ M
α∗,ji. Similarly, hA
ε: ε ≤ ii ∈ M
α∗,ji, α
∗∈ M
α∗,jiand
sup(M
α∗,ji∩ λ
ji) = f
α∗(j
i) < f
∗(j
i).
Consequently, S
ε≤i
A
ε∪ {α
∗} ⊆ M
α∗,ji(by the induction hypothesis or the above) and it belongs to M
α∗,ji. Since S
ε≤i
A
ε∪ {α
∗} ⊆ A
∗, clearly (H(χ), ∈, <
∗χ) |= “
∀x ∈ [
ε≤i
A
ε∪ {α
∗}
(d(x, f
∗(j
i)) = g
∗(j
i))”.
Note that [
ε≤i
A
ε∪ {α
∗}, g
∗(j
i), d, λ
ji∈ M
α∗,jiand f
∗(j
i) ∈ λ
ji\sup(M
α∗,ji∩ λ
ji).
Hence the set B
i:=
n
y < λ
ji:
∀x ∈ [
ε≤i
A
ε∪ {α
∗}
(d(x, y) = g
∗(j
i)) o
has to be unbounded in λ
ji. It is easy to check that j
i, A
i, B
isatisfy clauses (α)–(κ).
Thus we have carried out the induction step, finishing the proof of the theorem.
2.1Theorem 2.2. Suppose µ is a singular limit of measurable cardinals.
Then (1)
µ
+µ
→
µ µ
θ
if θ = 2 or at least θ < cf(µ).
(2) Moreover , if α
∗< µ
+and θ < cf(µ) then
µ
+µ
→
α
∗µ
θ
.
(3) If θ < µ, α
∗< µ
+and d is a function from µ
+× µ to θ then for some A ⊆ µ
+, otp(A) = α
∗, and B = S
i<cf(µ)
B
i⊆ µ, d¹A × B
iis constant for each i < cf(µ).
P r o o f. Clearly (3)⇒(2)⇒(1), so we shall prove part (3).
Let d : µ
+× µ → θ. Let κ := cf(µ). Choose sequences hλ
i: i < κi and hµ
i: i < κi such that hµ
i: i < κi is increasing continuous, µ = P
i<κ
µ
i, µ
0> κ + θ, each λ
iis measurable and µ
i< λ
i< µ
i+1(for i < κ). Let D
ibe a λ
i-complete uniform ultrafilter on λ
i. For α < µ
+define g
α∈
κθ by g
α(i) = γ iff {β < λ
i: d(α, β) = γ} ∈ D (as θ < λ
iit exists). The number of such functions is θ
κ< µ (as µ is necessarily strong limit), so for some g
∗∈
κθ the set A := {α < µ
+: g
α= g
∗} is unbounded in µ
+. For each i < κ we define an equivalence relation e
ion µ
+:
αe
iβ iff (∀γ < λ
i)[d(α, γ) = d(β, γ)].
So the number of e
i-equivalence classes is ≤
λiθ < µ. Hence we can find an increasing continuous sequence hα
ζ: ζ < µ
+i of ordinals < µ
+such that:
(∗) for each i < κ and e
i-equivalence class X, either X ∩ A ⊆ α
0, or for every ζ < µ
+, (α
ζ, α
ζ+1) ∩ X ∩ A has cardinality µ.
Let α
∗= S
i<κ
a
i, |a
i| = µ
i, ha
i: i < κi pairwise disjoint. Now, by induction on i < κ, we choose A
i, B
isuch that:
(a) A
i⊆ S
{(α
ζ, α
ζ+1) : ζ ∈ a
i} ∩ A and each A
i∩ (α
ζ, α
ζ+1) is a singleton,
(b) B
i∈ D
i,
(c) if α ∈ A
i, β ∈ B
j, j ≤ i then d(α, β) = g
∗(j).
Now, at stage i, h(A
ε, B
ε) : ε < ii are already chosen. Let us choose A
ε. For
each ζ ∈ a
ichoose β
ζ∈ (α
ζ, α
ζ+1) ∩ A such that if i > 0 then for some
β
0∈ A
0, β
ζe
iβ
0, and let A
i= {β
ζ: ζ ∈ a
i}. Now clause (a) is immediate, and the relevant part of clause (c), i.e. j < i, is O.K. Next, as S
j≤i
A
j⊆ A, the set
B
i:= \
j≤i
\
β∈Aj