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 and there is no r such that $q , then U(q) + 1 = U(p). THEOREM 2. Suppose $U(p) and $\xi_1 is a maximal descending chain in the fundamental order with ξ κ = [ 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 ∈ 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