HOMEWORK WEEK 1 1. Let Δ Be an Ordinal with Uncountable Cofinality

HOMEWORK WEEK 1 1. Let δ be an ordinal with uncountable cofinality and A ⊆ δ be an unbounded nonstationary subset of δ. Show that there is a function f : A → δ satisfying the following three conditions: (a) f is regressive, that is, f(η) < η for all η ∈ A; (b) f is monotonic, that is, η < η′ =⇒ f(η) ≤ f(η′) whenever η, η′ ∈ A (it is not required that f(η) <f(η′) here !); (c) f is unbounded in δ, that is, rng(f) is an unbounded subset of δ, or equivalently (∀ξ<δ)(∃η ∈ A)(f(η) > ξ). Hint. Recall that you have a club C ⊆ δ such that A ∩ C = ∅. 2. Let µ<κ be cardinals and µ be regular. Let κ Sµ = {ξ<κ | cf(ξ)= µ}. Prove the following: κ (a) If cf(κ) ≤ µ then Sµ is a nonstationary subset of κ. κ (b) If hαξ | ξ < µi is a strictly increasing sequence of elements of Sµ and κ κ α is its supremum then α<κ =⇒ α ∈ Sµ . We briefly say that Sµ is closed under limits of sequences of length µ. (Notice that this terminology involves some sloppiness.) Also: is it necessary to require that the ordinals κ αξ above are in Sµ ? (c) If cf(κ) > µ and S ⊆ κ is unbounded and closed under limits of sequences of length µ then S is stationary in κ. κ (d) If cf(κ) > µ then Sµ is a stationary subset of κ. κ (e) If cf(κ) ≤ µ then Sµ is an unbounded subset of κ. 3. Let κ be a regular cardinal. Prove: κ+ + + (a) Sκ is a stationary subset of κ and no α<κ is its reflection point. κ κ (b) If µ<λ<κ are regular then every α ∈ Sλ ∩ lim(Sµ ) is a reflection point κ of Sµ . 4. Given a function f : δ → δ, an ordinal ξ<δ is a fixpoint of f if and only if f(ξ)= ξ. Assume cf(δ) >ω. 1 (a) If f : δ → δ is a monotonic continuous function then the set of all fixpoints of f is a closed unbounded subset of δ. (b) If S ⊆ δ is stationary and f : S → δ is monotonic and continuous on S then N = {ξ ∈ S | ξ is not a fixpoint of f} is a nonstationary subset of δ. 5. Let κ be regular. Given a sequence hxξ | ξ<κi where xξ ⊆ κ, recall that α ∈ △ xξ ⇐⇒ α ∈ xξ. ξ<κ ξ<α\ Thus, the diagonal intersection △xξ is defined for sequences hxξ | ξ<κi, that is, its value depends on the order in which we list the sets xξ. This implies that there is some ambiguity in the definition of the diagonal intersection. However, this ambiguity is inessential, as the diagonal intersections of two different listings of the same family of sets differ only on “small set of points”: Prove that if ′ hxξ | ξ<κi and hxξ | ξ<κi are two enumerations of the same family of subsets of κ, that is, {xξ | ξ<κ} = {xξ | ξ<κ}, then there are only nonstationarily ′ many α<κ on which the diagonal intersections △xξ and △xξ disagree. In fact, ′ A = {α<κ | α ∈ ( △ xξ) ∩ ( △ xξ)} ξ<κ ξ<κ contains a closed unbounded subset of κ. However, notice that it is not clear that A itself is closed unbounded. Prove that the analogous conclusion holds for diagonal unions ▽ xξ. ξ<κ 6. The previous exercise tells us that the diagonal intersection and union should be considered as operations on the quotient Boolean algebra P(κ)/NSκ. Prove that this is true, that is, the operations {[xξ] | ξ<κ} 7→ [ △ xξ] ξ<κ and {[xξ] | ξ<κ} 7→ [ ▽ xξ] ξ<κ are well-defined operations on P(κ)/NSκ in the sense that they do not depend on the choice of representatives xξ and on the choice of the order in which we list the equivalence classes [xξ]. Also prove that the following is true in the Boolean algebra P(κ)/NSκ, which means that this algebra is κ+-complete: [ △ xξ]= [xξ] and [ △ xξ]= [xξ] ξ<κ ξ<κ^ ξ<κ ξ<κ_ 2 Notice that the conclusions of Exercises 5 and 6 are easy to generalize for any normal uniform ideal ℑ on κ in place of the nonstationary ideal NSκ. 7. Let κ be regular and S be a stationary subset of κ. Prove that all but nonstationarily many elements of S are limit points of S. 8. Let κ be regular. Prove that ther is a stationary set S ⊆ κ such that its complement κ − S is also stationary. Hint. It is not possible to give a definition of such a set explicitly, as it is known that this may false if AC fails. Use the theory from the lecture to give an indirect argument. 9. Let S ⊆ ω1 be stationary and α<ω1 be a countable limit ordinal. Prove that S contains a set of order type α. More generally, prove that the set Cα = {δ ∈ lim(S) | δ = sup(x) for some x ⊆ S with otp(x)= α} contains a club. Hint. First use Exercise 7 to get the result for α = ω. Then proceed by induction on α<ω1. This takes some work. 10. Let κ be regular. Define a binary relation ≺ on P(κ) by A ≺ B ⇐⇒ A ∩ β is stationary for all β ∈ B That is, A ≺ B if and only if every β ∈ B is a reflection point of A. (a) The relation ≺ is transitive. (b) The relation ≺ is well-founded. Hint. For (b), starting from an infinite descending sequence of sets under ≺ construct an infinite descending sequence of ordinals. Clause (a) should give you a clue how to pick these ordinals. 11. Prove that there is a ω-sequence. 12. ′ + ′ Let κ be a cardinal. A sequence hcα | α ∈ lim ∩(κ,κ )i is a κ-sequence if and only if the following are satisfied: ′ (a) Each cα is a closed subset of α. ′ (b) If cf(α) >ω then cα is unbounded in α. ′ ′ ′ (c) (Full coherency) Ifα ¯ ∈ cα then cα¯ = cα ∩ α¯. ′ (d) otp(cα) ≤ κ. 3 ′ Thus, the difference between a κ-sequence and κ-sequence is that in the case ′ of a κ-sequence we strengthen the coherency codition to full coherency, but relax the requirement on unboundedness. ′ Prove that for every cardinal κ, a κ-sequence exists if and only if a κ- sequence exists. Hint. ′ For the direction κ ⇒ κ: Look at sets of the form lim(cα). ′ Formulate the notion of a <λ-sequence and prove statement analogous to the above. + 13. Let κ be regular and hcα | α ∈ lim ∩(κ,κ )i be a κ-sequence. Prove that otp(cα)= κ if and only if cf(α)= κ for all α. 14. Let λ be regular and hcα | α ∈ Sing ∩ λi be a <λ-sequence. Let S ⊆ λ be a set such that S ∩ cα has at most one element. Let the unique ordinal in S ∩ cα if such an ordinal exists γα = 0 otherwise Then let ′ cα = cα − (γα + 1). ′ Prove that hcα | α ∈ Sing ∩ λi is again a <λ-sequence. 15. Combine the ideas from Exercises 12 and 14 to finish the proof of Proposi- tion 1.22 from the lecture. 16. Show that the following are equivalent: (i) There is a κ-sequence. (ii) There is a <κ+ -sequence. Hint. The nontrivial direction is from (ii) to (i). Let hcα | α ∈ Sing ∩ λi be + a <κ+ -sequence. Define a sequence hdα | α ∈ lim ∩(κ,κ )i by recursion on α as follows: Assuming dα¯ was constructed for allα ¯ < α, let δα = otp(cα). We have a unique normal function fα : δα → α enumerating cα. By assumption + δα < α, so dδα is defined. Let dα = fα[dδα ]. Show that hdα | α ∈ lim ∩(κ,κ )i is a κ-sequence. Remark. Notice that the above argument does not use the full length of + the <κ+ -sequence; it only uses the segment hcα | α ∈ lim ∩(κ,κ )i, because for α<κ we only need to know that otp(cα) <κ. So we could start the recursion ′ ′ ′ from the sequence hcα | α ∈ Sing ∩ λi where cα = α if α ≤ κ and cα = cα − κ if α>κ. 17. Let κ be a singular cardinal and hcα | α ∈ Sing∩λi be a κ-sequence. Show ′ ′ that there is a κ-sequence hcα | α ∈ Sing ∩ λi such that otp(cα) <κ for all α. Hint. Use an argument similar to that in the proof in Exercise 16. 4 18. Let κ be a regular cardinal and S ⊆ κ be a stationary set. Prove that the set S′ = {α ∈ S | α is not a reflection point of S} is stationary. (This is the lemma needed for the proof of splitting theorem at inaccessible κ.) Hint. Given a closed unbounded set C ⊆ κ, consider the first ordinal in S ∩ lim(C). 19. ω3 ω3 Prove that there are stationary sets S ⊆ Sω and T ⊆ Sω1 such that every α ∈ T is a non-reflection point of S. Hint. ω3 First, for each α ∈ Sω1 pick a closed unbounded subset cα of α with ω3 otp(cα)= ω1. Next split Sω into ω2 many disjoint stationary sets hAξ | ξ<ω2i. ω3 ∅ Notice that for each α ∈ Sω1 there is some ξ<ω2 such that Aξ ∩ cα = . Use this observation and the pigeonhole principle to obtain S and T . Remark. On the other hand, it is consistent with ZFC that every stationary ω2 ω2 S ⊆ Sω reflects at all but nonstationarily many α ∈ Sω1 .

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