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Chapter 5. Infinitary Combinatorics ∗ Chapter 5. Infinitary combinatorics ∗ 5.1. Club and stationary sets. Definition. Let λ be a limit ordinal. A set D ⊆ λ is unbounded in λ if ∀α < λ ∃β ∈ D α < β. A set C ⊆ λ is closed in λ if for all limit ordinals δ < λ, C ∩ δ is unbounded in δ implies that sup(C ∩ δ) ∈ C. A set that is closed and unbounded in λ is often called “club” Examples. {γ < λ | γ is a limit ordinal} and {γ < λ | γ is a limit of limit ordinals} are closed in λ. And they are unbounded if λ is a cardinal. You should think of club sets as being “large” in some sense. Proposition. Let κ be an uncountable regular cardinal and F a set of finitary funtions on κ with F < κ (i.e., if f ∈ F we have that f : n(f)κ −→ κ for some n(f) < ω.) Then, C = {γ < κ | ∀f ∈ F f“ n(f)γ ⊆ γ } is closed and unbounded. Proof. It is clear that C is closed: if λ < κ is a limit ordinal and C ∩ λ is unbounded, f ∈ F, n(f) n(f) f : κ −→ κ and (α0, . , αn−1) ∈ λ then there is some γ ∈ (C ∩ λ) \ max({α0, . , αn−1 }), and we have f(α0, . , αn−1) < γ < λ. S To see C is unbounded, define G0(X) = X, Gn+1(X) = Gn(X) ∪ {β | ∃α0, . , αm−1 ∈ Gn(X) S ∃f ∈ F f(α0, . , αm−1) = β }, for each, n < ω and G(X) = n∈ω Gn(X) for each subset X of κ. If ξ < κ it is clear that ξ ⊆ G(ξ) ⊆ κ and G(ξ) < κ. Because κ is regular we can choose g(ξ) = min(κ \ sup(G(ξ))). Let gn+1(ξ) = g(gn(ξ)) some each ξ < κ, and let gω(ξ) = sup{gn(ξ) | n ∈ ω }. Then we have that gω(ξ) ∈ C and ξ < gω(ξ). Definition. Let g : κ −→ C enumerate a club set C in increasing order. Let C0 = g“{ωγ | γ < κ}. C0, ⊆ C, is the set of limit points of C. Lemma. Let C be a club subset of a regular cardinal κ and define f : κ −→ κ by f(γ) = min(C\(γ+1)). Then C0 = {γ < κ | f“γ ⊆ γ }, and hence C0 is club. Proof. It is clear that C0 = {γ < κ | f“γ ⊆ γ }, and thus by the previous proposition, is closed and unbounded. Proposition. Let κ be an uncountable regular cardinal, µ < κ, and let Cα be closed and unbounded T in κ for each α < µ. Then D = {Cα | α < µ} is closed and unbounded in κ. Proof. It is easy to see that D is closed. Now let fα(γ) = min(Cα \ γ + 1) for γ < κ and for all α < µ. D is unbounded because, by the previous proposition, {γ < κ | ∀α < µ fα“γ ⊆ γ } is unbounded (and closed), and is a subset of D by the lemma. In fact it is easy to see that one can prove similar results for singular cardinals κ with ω < cf (κ). For T example, if µ < cf (κ) and Cα be closed and unbounded in κ for each α < µ then D = {Cα | α < µ} is closed and unbounded in κ. First of all prove (by hand) that the intersection of any two club sets is club. Next, take a cofinal function f : cf (κ) −→ κ which is closed at limits, that is such that f(λ) = S{f(α) | α < λ} for ∗ version 1.0, 9/2/06 1 limit λ < cf (κ), and observe that f“ cf (κ) is a club, say C, in κ. Finally, if D ⊆ κ let D0 = D ∩ C and note that f −1“D0 is club in cf (κ). So to prove results about, for example, intersections of club subsets of κ (or the diagonal intersection lemma below) simply ‘slide’ the club sets down to cf (κ) using f −1, use the results for regular cardinals applied to the resulting sets and then use f to slide the result back up to κ Proposition. (Diagonal intersections) Let κ be a regular cardinal and let Cα be closed and T unbounded in κ for each α < κ. Then D = {β | ∀α < β (β ∈ Cα)} (= {β | β ∈ α<β Cα)}) is closed and unbounded in κ. Proof. It is easy to see that D is closed: if λ is a limit ordinal and D ∩ λ is unbounded in λ we have that ∀α < λ ∃β (α ≤ β < λ & ∀γ < β β ∈ Cγ ). Hence ∀α < λ ∃β (α ≤ β < λ & β ∈ Cα). T Thus Cα ∩ λ is unbounded in λ for all α < λ. Thus λ ∈ {Cα | α < λ}. T In order to see that D is unbounded, define g(γ) = min( α<γ Cα \ γ) for all γ < κ. (Note that this intersection is unbounded by the proposition that the intersection of fewer than κ club sets is club, thus g(γ) is well-defined.) Now let β < κ, β0 = β, βn+1 = g(βn) for n < ω, and βω = sup{βn | n < ω }. Then βω ∈ D. Definition. Let κ be a cardinal with ω < cf (κ). A set S ⊆ κ is stationary if S ∩ C 6= ∅ for all C which are closed and unbounded in κ. A set X ⊆ κ is non-stationary if it is not stationary. Stationary sets are “medium-sized” as subsets of κ and non-stationary sets are “small” in this sense – even though they can have cardinality κ. The proposition that any intersection of fewer than cf (κ) many club sets is club in κ says that any union of fewer than cf (κ) non-stationary sets is non-stationary. Examples. Let µ < cf (κ) be a regular cardinal. Then {α < κ | cf (α) = µ} is stationary, because if C is closed and unbounded in κ the µth element in the increasing enumeration of C has cofinality µ. For example, both {α < ω2 | cf (α) = ω } and {α < ω2 | cf (α) = ω1 } are both stationary in ω2. Proposition. (Fodor’s Lemma) Let κ be a regular cardinal, ω < κ, let S be a subset stationary of κ and let f : S −→ κ be such that f(α) ∈ α for each α ∈ S. Then there is some γ < κ such that {α ∈ S | f(α) = γ } is stationary in κ. Proof. If the conclusion fails, then choose for each γ ∈ κ a set Cγ which is closed and unbounded −1 in κ and such that Cγ ∩ f “{γ } = ∅. Let D = {β | ∀γ < β (β ∈ Cγ )}. The diagonal intersections proposition says that D is closed and unbounded in κ. But D ∩ S = ∅, because if β ∈ D we have f(β) 6= α for each α < β. This is a contradition to the supposition that S is stationary. An illustration that stationary sets are not as “fat” as club sets are is given by the following proposition. + Proposition. Let κ = µ and let S be stationary in κ. Then there is some Sα for α < κ with S S = {Sα | α < κ} and Sα ∩ Sβ = ∅ for each pair α, β < κ with α 6= β. ξ Proof. Let fγ : γ −→ µ be an injection for each γ < κ. Define Sα = {γ | α < γ < κ & ξ ξ fγ (α) = ξ } ⊆ κ. Then Sα ∩ Sβ = ∅ if α 6= β for each ξ < µ because each fγ is a injection. Also S ξ we have that ξ<µ Sα = {γ | α < γ < κ} is stationary in κ. Because the union of fewer than κ non-stationary subsets of κ is non-stationary, it must be that for each α there is some ξα < µ ξα such that Sα is stationary. Since µ < κ there is some ξ < κ such that {α < κ | ξ = ξα } = κ. ξ Let A = {α < κ | ξ = ξα } and α0 = min(A). Define Sα = Sα for α ∈ A \{α0 }, and define S = S \ S{Sξ | α ∈ A \{α }} (since Sξ ⊆ S ). We have that S ⊆ S is stationary for each α0 α 0 α0 α0 α α ∈ A and Sα 6= Sβ where α, β ∈ A and α 6= β. 2 Fact. With a little more work one can show the proposition for all regular cardinals and not only for successor cardinals as is done above. 5.2 Silver’s Theorem. Now we are going to use the concepts of closed and unbounded and of stationary sets to prove Silver theorem about exponentiation of singular cardinals of uncountable cofinality that was mentioned at the end of §4.4. It is useful to have one more definition. ω1 Definition. F ⊆ ωω1 is called (an) eventually different (family) if ∀f, g ∈ F (f 6= g −→ {α | f(α) = g(α)} ≤ ω. ℵα ℵω1 Theorem. (Silver, 1975) If 2 = ℵα+1 for all α < ω1, then 2 = ℵω1+1. α Proof. By the hypotheses of the theorem, supposethatP(ωα) = {Xβ | β < ωα+1 } for all α < ω1. α Then for every X ⊆ ωω1 and α < ω1 there is some β < ωα+1 such that X ∩ ωα = Xβ . So for each α X ⊆ ωω1 we can define a function fX : ω1 −→ ωω1 by fX (α) = β(< ωα+1) if X ∩ ωα = Xβ . Note that F = {fX | X ∈ P(ωω1 )} is eventually different because X 6= Y implies that there is some α < ω1 such that fX (α) 6= fY (α), and fX (α) 6= fY (α) implies fX (γ) 6= fY (γ) for every γ ≥ α. Now partially order F by f < g if and only if {α < ω1 | g(α) ≤ f(α)} is a non-stationary subset.
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