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Maximal chains in the fundamental order

Published online by Cambridge University Press:  12 March 2014

Steven Buechler*
Affiliation:
Department of Mathematics, University of Wisconsin/Milwaukee, Milwaukee, Wisconsin 53201
*
Current address: Department of Mathematics, University of California, Berkeley, California 94618

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.

Type
Research Article
Copyright
Copyright © Association for Symbolic Logic 1986

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References

REFERENCES

[B] Buechler, S., Locally modular theories of finite rank, Annals of Pure and Applied Logic (to appear).Google Scholar
[L] Lascar, D., Ranks and definability in superstable theories, Israel Journal of Mathematics, vol. 23 (1976), pp. 5387.CrossRefGoogle Scholar
[LP] Lascar, D. and Poizat, B., An introduction to forking, this Journal, vol. 44 (1979), pp. 330350.Google Scholar
[M] Makkai, M., A survey of basic stability theory, with particular emphasis on orthogonality and regular types, Israel Journal of Mathematics, vol. 49 (1984), pp. 181238.CrossRefGoogle Scholar
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