Lascar inequalities (assume T simple, a, b finite) I. SU(a/Ab) + SU(b/A) ≤ SU(ab/A) ≤ SU(ab/A) + SU(a/Ab) Proof: By induction on a: SU(b/A) ≥ a then SU(ab/A) ≥ SU(ab/A) + a. a=0: SU(ab/A) ≥ SU(a/A) ≥ SU(a/Ab). Limit: By ind hyp. (since a on right-hand-side) SU(b/A) → ∞ so ∃ c st. SU(a/Ac) ≥ a, b & c. WMA c ≥ a. ⇒ SU(Ab) = SU(A/bc) By ind hyp, SU(ab/Ac) ≥ SU(a/Ac) + a = SU(a/Ab) + a. since b ≤ c ⇒ SU(ab/A) ≥ SU(ab/Ac) + 1 ≥ SU(a/Ab) + a + 1. Now other inequality, we prove by induction. ∀ x if SU(ab/A) ≥ x ≥ SU(a/Ab) + SU(b/A). Again x=0, limit ✓. Assume SU(ab/A) ≥ x + 1. Then ∃ c st. b ≤ c ≤ SU(ab/Ac) ≥ x. ⇒ SU(a/bc) + SU(b/Ac) ≥ x by inc. Either b ≤ c ≤ a or a ≤ c ≤ b (otherwise by trans we get a ≤ SU(ab/A) + SU(b/A) ≤ SU(ab/Ac) + SU(b/Ac) ≤ x + 1). In either case SU(ab/A) ≥ SU(a/Ab) + SU(b/A) ≥ x + 1. II. If a ∪ b = b ⇒ SU(ab/A) = SU(a/A) + SU(b/A). Assume SU(a/A) + SU(b/A) ≥ x + 1. So wlog assume ∃ p, q st. SU(a/A) ≥ p + 1, and SU(b/A) ≥ q ≥ 0, p + q ≥ x. ∃ c a ≤ c ≤ b SU(a/Ac) ≥ p. WMA c ∪ b = b ⇒ ac ∪ b = b ⇒ a ≤ b. ⇒ SU(ab/Ac) ≥ p + q ≥ x. Explicitly SU(ab/Ac) ≥ SU(a/Ac) + SU(b/Ac) ≥ p + q ≥ x + 1. So we have equality. I. Assume that p ∈ S(A) has nonsplitting extensions to every set B ≥ A. Then p is Lascar strong. Proof. Let N ≥ A be (|A| + |T|)-saturated. Let g ∈ S(N) be a nonsplitting extension of p. Let a ∈ b ∈ p. We need: a = b. We may assume that a, b ∈ N (realize tp(a,b/A) in N). By induction on i ∈ ω, find ci ∈ N st. ci ∈ A-abaci. Then each a ∈ {co, c1, c2, ...} and b ∈ {co, c1, c2, ...} is indiscernible since nonsplitting and induction. (Since a ∈ {co, c1, c2, ...} has same type / A (p), ...) ⇒ a = b. Corollary (T stable). Every type over a Lascar strong hyper A in a T/t-saturated model is Lascar strong. (1st order: don't need T stable or T/t-saturated). Defn A class of sets (in the universal domain) S is cofinal ∀B ∃A ∈ S, A ≥ B. Eg Let MITIR = {I | I is T-saturated models} Then MITIR is cofinal. Prop Assume that for every finite tuple n, for every increasing sequence (Ai: i < |T|) in S, ∃j < |T| st. a ∈ Ai. Then T is simple. Proof Assume T is not simple ⇒ ∃a, bi: i < |T| st. a ∈ bi, ti. Let pi = tp(a,bi). Find (Ai: i < |T|) increasing in S and ci: i < |T| and st. ∃b, ci ∈ Ai st. if a ∈ b = ci then a ∈ ci ∀i. (contradiction) Construction: i = 0: A0 anything in A. i limit: anything containing Uj Aj. i+1: Ai,r ≥ Ai, ci. Now choose ci: We have Ai, ci, ci; want to find ci. Since a ∈ bi ∃ bi indiscernible sequence (di,j: j ∈ ω) witnessing it. Since by ind hyp, ci ∈ bi, find (ei,j: j ∈ ω) st. ei,j = ci = di,bi. By extension/extraction, we may assume (ei,j: j ∈ ω) is Ai-indiscernible. Ci ∈ ei,0. (1) ci = bei (since di,0 = bi). Let p(x, y, z) := tp(a, bi, ci). Then ∧p(xi, bi, ci) is inconsistent and the sequence (xi,j: j ∈ ω) is Ai-indiscernible ⇒ it ⊢ ¬p(xi, ci, ci) ⇒ ci ∈ Ai. Now we can prove the main theorem

18.996A Simplicity Theory: Lecture 10 Notes

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