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Maximal chains in the fundamental order
Published online by Cambridge University Press: 12 March 2014
Abstract
Suppose T is superstable. Let ≤ denote the fundamental order on complete types, [p] the class of the bound of p, and U(—) Lascar's foundation rank (see [LP]). We prove
Theorem 1. If q < p and there is no r such that q < r < p, then U(q) + 1 = U(p).
Theorem 2. Suppose U(p) < ω and ξ1 < … < ξk is a maximal descending chain in the fundamental order with ξk = [p]. Then k = U(p).
That the finiteness of U(p) in Theorem 2 is necessary follows from
Theorem 3. There is an ω-stable theory with a type p ϵ S 1(ϕ) such that
(1) U(p) = ω + 1, and
(2) there is a maximal descending chain of proper extensions of [p] which has order type ω.
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- Copyright © Association for Symbolic Logic 1986
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