NOTES ON SQUARES, SCALES, AND REFLECTION

NOTES ON SQUARES, SCALES, AND REFLECTION MAXWELL LEVINE 1. Some Basic Facts Proposition 1. If κ <κ = κ, then κ holds. Proof. We define C α : α < κ + by considering the cases of cf α < κ and cf α = κ separately. If cf α < κ, then let C α = {C α : C club in α, C < κ, cf β < κ β C}. Since α κ and κ <κ = κ, we have C α κ as required (the C α s are also clearly nonempty). We can t use the same trick in the cf α = κ case because κ κ > κ. Instead we choose a club C in α of order type κ and let C α = {C}. Observe that if β lim C for C C α, then C β C β : If cf α < κ, then this is immediate since we guarantee that cf β < κ. If cf α = κ and β lim(c) for C C α, then o. t.(c β) < κ, so C β C β. Proposition 2. If SCH κ fails (where κ is a singular strong limit such that 2 κ > κ + ) then Refl(cf κ, κ + ) fails. Proof. Assume without loss of generality that cf κ = ω, and suppose for contradiction that Refl(ω, κ + ) holds while SCH κ fails. Decompose κ + into κ-many disjoint stationary sets S α : α < κ. This can be done by appealing to Solovay s Theorem or using the canonical stationary sets for regular cardinals less than κ. Let f : κ ω κ + by given by mapping α n : n < ω to the point below κ + where S αn : n < ω simultaneously reflects. There is a set X κ ω of cardinality at least κ ++ on which f is constant with value γ: this follows from regularity of κ ++ and the fact that κ ω = 2 κ κ ++. We claim that there is a sequence S βi : i < κ of stationary sets in γ. This suffices for our proof because if C γ is a club and o. t. C = cf γ < κ (noting that κ is singular, so any ordinal below κ + has cofinality strictly less than κ), then C cannot possibly meet all of the S βi s and thus we obtain a contradiction. Define β i : i < κ as follows: Pick an arbitrary sequence in X and let β 0 be its first element. If we have already defined β i : i < δ, then observe that P ({β i : i < δ}) < κ because κ X and κ is a strong limit. Hence there is a sequence s X \ P ({β i : i < δ}), so we can pick β δ s \ {β i : i < δ}. The S βi s are disjoint by construction, and S βi γ is stationary for all i < κ because β i s X. Proposition 3. κ κ + I[κ + ], i.e. AP κ + holds. Proof. Let C = C α : α < κ + be a weak square sequence such that α < κ +, C C α such that o. t. C = cf α (this property is equivalent to the usual weak square) and let M = H θ,, < θ, C for θ large enough. We claim that all points in the club (κ, κ + ) are approachable with regard to M. Say α (κ, κ + ) and N = Sk M (α). Since M contains an enumeration of C α and κ N, we can assume C α N by elementarity. Pick C C α with o. t. C = cf α, 1

2 MAXWELL LEVINE which will witness approachability of α: For all β < α, C β C β by definition of the square sequence and so C β N. 2. Forcing and Approachability Theorem 1. If P is countably closed and S κ cf(ω) is stationary in V, then S is stationary in V P. Proof. Let C V P be a club in κ and let p, Ċ witness this. Let M = H θ, p, P, Ċ, κ (where H θ is all sets with transitive closure of size less than θ, and θ is large enough for p, P, Ċ, κ H θ). It s easy enough to show that D = {x P κ (θ) : x = dom(n), N M} and D = {sup(x κ) : x D} are both clubs in P κ (θ) and κ respectively. Therefore there is N M with α = sup(n κ) S. Since α S cf(ω), we can write α = sup n<ω α n with α n : n < ω N by elementarily. Construct a decreasing sequence p n : n < ω N and an increasing β n : n < ω N as follows: Using that p Ċ is unbounded in κ and p n p, find p n+1, β n+1 N such that p n+1 β n+1 Ċ and β n+1 β n, α n (again, using elementarily). Note that β n+1 < α for all n since sup(n κ) = α. Use countable closure to find q p n for all n < ω. Then since sup β n = sup α n = α (by interleaving) and q Ċ unbounded, α n Ċ n < ω, we have q p with q α Ċ. We showed that {q p : q Ċ S } is dense, so we are done. Let I[κ] be the approachability ideal. Theorem 2. If P is µ + -closed, S κ cf(µ), and S I[κ], then V P = S is stationary. Proof. Let C V P be a club in κ with p Ċ club in κ. Since S I[κ], there is a = a β : β < κ and a club D such that every point in D S is approachable with regard to a. Let M = H θ, κ, µ, p, P, Ċ, a. Then E = {x H θ (κ) : x = dom(n), N M} and E = {sup(x κ) : x E} are both clubs. If γ E S D, then γ then there is some N M such that sup(n κ) = γ. Let A γ witness the approachability of γ with regard to a so o. t. A = µ and {A β : β < γ} {a β : β < γ}. Write A = {α ξ : ξ < µ}. Define an increasing p ξ : ξ < µ N and a decreasing β ξ : ξ < µ N as follows: For ξ = 0, just pick p 0 p and β 0 such that p 0 β 0 Ċ. For ξ > 0, since sup(n κ) = γ > α ξ, there is α ξ β < γ, β N. M = x a, sup x < β (because A β = a β for some β < γ), so N satisfies the same statement, witnessed by some α. Pick p ξ p ν ( η < ξ, using µ + -closure) such that p ξ β ξ Ċ, β ξ > α. Finally use µ + closure to find q p ξ for all ξ < µ. Then sup A = sup α ξ = sup β ξ = γ, and so q γ S Ċ. We showed {q p : q Ċ S } and so we are done by density. 3. Supercompactness versus Squares and Scales Let κ be supercompact, let λ κ, and let j : V M be a λ-supercompact embedding. Proposition 4. If s is a sequence of length less that κ then j(s) = j s. Proof. By elementarity, x s implies j(x) j(s); hence j s j(s). If s = x ξ : ξ < τ and τ < κ then j(s) = α ξ : ξ < j(τ) = τ. V = α ξ is the ξ th element

NOTES ON SQUARES, SCALES, AND REFLECTION 3 of s, so M = j(α ξ ) is the j(ξ) = ξ th element of s. Hence j(α ξ ) = αξ and so j(s) j s. Proposition 5. If ρ = sup j λ and C ρ is a club in M, then C = {α : j(α) C} is a < κ-club in λ. Proof. κ-closure: suppose s = α ξ : ξ < τ C and τ < κ and β = sup s. Then j(β) = j(sup s) = sup j(s) = sup j s C, so β C. unboundedness: Let α < λ. Pick β 0 C such that j(α) < β 0 < ρ and γ 0 < λ such that β 0 < j(γ 0 ). In general pick β n+1 C with j(γ n ) < β n+1 and γ n < λ with β n < j(γ n to define s = β n : n < ω and t = γ n : n < ω such that s and j t are interleaving. If γ = sup t then j(γ) = j(sup t) = sup j(t) = sup j t = sup s C, so γ C. Proposition 6. If ρ = sup j λ, then ρ < j(λ). Proof. This is because ρ = [x sup x] (if f : x sup x then (jf)(j λ) = f(j λ) = [f]) and j(λ) = [x λ]. Since sup x < λ for all x P κ (λ) the result follows from Los. Theorem 3. If κ is supercompact and λ κ then λ fails. Proof. Let j : V M be a λ + -supercompact embedding and C = C α : α < λ +, lim(α) be a λ sequence. Then j(c) = Cα : α < j(λ + ) is a j(λ) sequence by elementarity. Let C = {α : j(α) Cρ} where ρ = sup j λ +. C is a < κ-club in λ +. We will argue that o.t.(c α) < λ for α < λ +. This implies that C has no λ th element (for one) and therefore cannot be cofinal in λ +. Fix α < λ +. Pick β lim(c) such that α < β < λ + and cf β < κ. If γ C β then j(λ) < j(β) and j(γ) Cρ. Furthermore Cρ j(β) = Cj(β) = j(c β), and so j(γ) j(c β ), implying γ C β byi elementarity. We have demonstrated that C α C β C β. Because cf β < κ λ it follows that o.t.c β < λ and therefore that o.t.(c α) < λ. Proposition 7. Let f α : α < κ be a scale in a singular cardinal κ and suppose α < κ + and cf α > cf κ. Then the following are equivalent definitions of α being a good point: (1) There exists an unbounded A α, i < cf κ, such that β, γ A, j i, β < γ implies f β (j) < f γ (j). (2) There exists an exact upper bound h for f β : β < α and j < cf κ such that cf h(i) = cf(α) for j i < cf κ. (3) There exists a pointwise increasing sequence h γ : γ < cf α cofinally interleaved modulo I with f β : β < α. 1 2. Given A and i define h(j) = sup β A f β (j) for j i and 0 otherwise. Without loss of generality A = cf α, so cf h(j) cf α for j i. The fact that f β (j) : β A is strictly increasing tells us that cf h(j) cf α. To show that h is an upper bound for the sequence, suppose γ < α. There is some β A with γ < β. Since f γ < I f β, j such that i j implies f γ (i) < f β (i) < sup{f δ (i) : δ A} = h(i), i.e. f γ < I h. We must show that f β : β < α is cofinal in h before we are done. Suppose g < I h. There is j such that j i implies g(i) < h(i). Pick β i A such that g(i) < f βi (i), and let γ = sup i<cf κ β i < α (because cf α > cf κ). Then g < I f γ.

4 MAXWELL LEVINE 2 3. There is g i : cf α h(i) cofinal and increasing for all i < cf κ. Let h γ (i) = g i (γ) for every γ < cf α. There h γ s are pointwise increasing because the g i s are increasing. Since h γ < I h and since h is an exact upper bound for f β : β < α, there is β < α with h γ < I f β. We still need to show that for every β < α, there is γ < cf α such that f β < h γ. Fixing β, the fact that f β < I h implies there is a j such that f β (i) < h(i) for every i j. Suppose towards a contradiction that γ k i k such that h γ (k) f β (i). Then if k = j the assignment γ i is regressive for γ > cf κ (again because cf α > cf κ), so by Fodor s Theorem there is a stationary S cf α and i 0 j on which h γ (j) f β (i 0 ) for i 0 j and all γ S. Then g j (γ) f β (i 0 ) < h(j) for unboundedly many γ < cf α, contradicting the fact that g j is cofinal in h(j). It follows that there is some γ such that f β < I h γ. 3 1. Define β δ : δ < cf α and γ δ < cf α such that f βδ < I < h βδ < I f βδ+1, etc. For every δ < cf α there is i δ such that j i δ implies f βδ (j) < h βδ (j). The function δ iδ is regressive for δ > cf κ, so there is a stationary A cf α on which j i 0 implies f βδ (j) < h βδ (j). Since the h s are pointwise increasing, it follows that f γ (j) : γ A is strictly increasing for j i 0. Theorem 4. If κ is supercompact, κ < λ, and cf λ < κ, then every scale on λ is bad. Proof. Let j : V M be a λ + -supercompact embedding and λ, f be a scale on the ideal I cf λ where λ = λ i : i < cf λ and f = f α : α < λ +. Let ρ = sup j λ +. Note that j(f) = fα : α < j(λ + ). We claim that ρ is a bad point for j(f). It follows that if C λ + is a club then M = j(c) contains a bad point of j(f) (because ρ j(c)); thus V = {α < λ + : α is a bad point of f} is stationary. A key fact is that since M λ+ M it follows that (cf α) M = (cf α) V if cf α λ +. Hence ρ < j(λ + ) because cf ρ = λ +. By the above proposition it is enough to dispay an exact upper bound h for fα : α < ρ such that cf h(i) : i < cf λ is bounded below λ +. Let h be given by h(i) = sup j λ i. h is an upper bound of fα : α < ρ : for all α < ρ there is some β < λ + such that α < j(β), and so fα < j(i) fj(β) = j(f β) < h by elementarity and the fact that f(i) < λ i for all i < cf κ. Also, cf h(i) = λ i < λ + for all i. 4. Some Relevant Facts on Forcing Theorem 5. The following are equivalent definitions of a κ-distributive poset P (assuming that P is separative): (1) For all λ < κ, if D α : α < λ is a sequence of open dense sets in P, then α<λ D α is open dense. (2) If λ < κ, f : λ V, and f V P, then f V. (3) If λ < κ, f : λ ON, and f V P, then f V. 2 1. Let λ < κ and let D α : α < λ be a sequence of open sense sets. It s rather simple to see that α<λ D α is open, so our task is to show that the intersection is dense. For each α < λ let A α D α be a maximal antichain and let f = { α, ˇq, q : q A α }. Fix p P and take p G to be P-generic over V. By our hypothesis f G V, so there is some g V with g(α) = q α A α such that g = f G. Let r G be such that

NOTES ON SQUARES, SCALES, AND REFLECTION 5 r p and r f = ǧ. We claim that r q α for every α < λ and hence r D α for all α < λ using openness. Suppose s q α. Then by separativity there is some t s such that t q α. Let t G be generic. Then there is some q G A α such that q q α (because G must intersect every maximal antichain but cannot contain incompatible elements) and therefore f G (α) = q q α. Therefore s f = ǧ. It follows that r q α for all α. 1 2. Let f : λ V be in V [G] where G is P-generic. There is some p G and A V such that p f : λ A. Our goal is to show that D = {q p : g V, r ǧ = f} is dense below p. Then there is some q G D and so f = g V. The set D α = {q P : a A, q f(α) = a} = {q P : q f(α)} is open dense. Let q p, q α<λ D α. Then we claim that f(α) is the unique a A such that q f(α) = a. Since q D α, there is such an a, and moreover it is unique since q cannot force contradictory information. 2 3. This is immediate since ON V. 3 2. Let A ON be a code for im f. (That is, if B = tc(im f), and g : τ B is a bijection, and C = {(α, β) : g(α) g(β)}, and D = {g(α) : α B}, then A = Γ[D] where Γ is the canonical bijection ON ON ON.) Then h : λ A ON V, so f V. Theorem 6 (Easton). Let P be κ-c.c. and Q be κ-closed. Then Q is λ-distributive in V P. We first need two facts: Lemma 1. If Q is λ-distributive, H is Q-generic, and q ξ : ξ < µ is descending in Q for µ < λ, then there is q H such that q q ξ for all ξ < µ. Proof. Define D ξ = {p Q : p q ξ or p q ξ } which is open dense, so D := ξ<µ D ξ is open dense. Assume by induction that q ξ H for all ξ < µ and let q H Q. q q ξ because H is a filter, so it follows that q q ξ for all ξ < µ. Lemma 2. If P is κ-c.c. and Q is κ-closed, then V Q = P is κ-c.c. Proof. Suppose q Q and q f : κ A enumerates a maximal antichain. Then using κ-closure we can define a decreasing sequence q ξ : ξ < κ and p ξ : ξ < κ such that ξ < κ, q ξ f(ξ) = p ξ. All of the p ξ s are incomparible: if ξ < η < κ, then q ξ p ξ p η, which implies that V = p ξ p η. Hence p ξ : ξ < κ is an antichain of size κ in V. Proof of Easton s Theorem. Let G be P-generic and let H be Q-generic. Our goal is to show that if f V [G][H] and f : λ ON for λ < κ, then it follows that f V [G]. Without loss of generality assume that f is forced by the empty condition in P Q. Define D α = {p P : q H, (p, q) ϕ(α)} V [H], which is open dense (a fact that uses that we can define P Q-generic filter K in terms of G and H). Let A α D α be a maximal antichain for each α < λ. Note that P has the κ chain in V Q because Q does not add new κ-sized sets and an antichain A in V Q will remain

6 MAXWELL LEVINE an antichain in V. It follows that A α < κ and therefore A α V (again because of λ-closure) for all α. Let us then enumerate A α = {p α,β : β < κ α < κ}. Use the lemma to define a lexicographically ordered descending sequence q α,β : α < τ, β < κ α of conditions in H such that (p α,β, q α,β ) f(α). Let q q α,β for all indices. Then we claim that f(α) =!β s.t. p G, (p, q) f(α) = β. We know that p α,β G A α for some β and (p α,β, q) (p α,β, q α,β ), so such a p exists. Moreover, p won t force contradictory information. Since f is defined using G and q V, it follows that f V [G]. Definition 1. Given a poset P we define the game G α (P) for two players as follows: Suppose β α and we are given p ξ : ξ < β. If β is an even successor or a limit, then player I selects p β such that p β p ξ for ξ < β. If β is an oddtheorem successor then player II selects p β p ξ for ξ < β. P is κ-strategically closed if player II has a winning strategy for G α (P) for all α < κ. Theorem 7. If P is κ-strategically closed, then it is κ-distributive. Proof. Suppose λ < κ and f : λ ON is a function in V P. Define p ξ : ξ < λ as follows: If ξ+1 is an odd successor, then pick p ξ+1 p ξ such that p ξ+1 f(ξ), f(ξ+1). If ξ is an even successor or limit, than use the winning strategy of player II to select p ξ p η for η < ξ. Finally use the winning strategy of player II to choose q p ξ for all ξ < λ. Then if g : λ ON is defined such that g(ξ) is the unique β such that q f(ξ) = β, then g V and q g = f, so we are done by density.