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One of the ways to define the Morley rank of a definable set is with respect to a model, say $M$, i.e. a set has rank $\alpha+1$ if there are infinitely many definable subsets with parameters in $M$ of rank $\alpha$ (and a similar clause for limit ordinals). One then shows that once $M$ is $\aleph_0$-saturated then considering definable subsets with parameters in elementary extensions of $M$ doesn't change the rank.

However, in some theories, e.g. algebraically closed fields of fixed characteristic, any model will do, i.e. Cantor-Bendxson rank and Morley rank coincide.

What is known about this phenomenon, i.e. in which theories Cantor-Bendixson equals Morley rank in any (not necessary $\aleph_0$-saturated) model?

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  • $\begingroup$ Dima, I will add to your question the following remark: the fact that this is true in an algebraically closed field of any characteristic has nothing to do with the field and everything to do with the strong minimality of the theory. When the theory is strongly minimal, then Morley rank is definable in families, so working in an existentially closed model would suffice. $\endgroup$
    – James Freitag
    Commented Jul 27, 2012 at 2:55
  • $\begingroup$ Dear James, how is definability of Morley rank related to the fact that one can compute Morley rank using parameters from existentially closed models? $\endgroup$
    – Dima Sustretov
    Commented Jul 30, 2012 at 12:47
  • $\begingroup$ Does this mean that in $DCF_0$, say, there are examples of sets such that their Cantor-Bendixson rank (wrt some model) is not the same as their Morley rank? $\endgroup$
    – Dima Sustretov
    Commented Dec 11, 2012 at 20:45

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The distinction between Morley rank as defined by arbitrary formulas and by definable families of formulas is essential. $\aleph_1$- categoricity in particular implies the rank can defined by definable families. Since my 1973 [?] article in the transactions AMS or Shelah's book or say Pillay's geometric model theory book

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    $\begingroup$ Welcome to MathOverflow, John! $\endgroup$
    – Joel David Hamkins
    Commented Feb 11, 2013 at 12:26
  • $\begingroup$ By the way, you can use simple tex here, just write it with dollar signs and slashes as usual. $\endgroup$
    – Joel David Hamkins
    Commented Feb 11, 2013 at 14:53
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Maybe the following remark will help you:

For any strongly minimal forumla $\varphi (x)$ and any formula $\psi(x_1,..,x_n,y)$ which implies $\varphi (x_i)$ for all i, the class $ \lbrace b : MR(\psi(x_1,..,x_n,b))=k \rbrace$ is definable for every k.

This is $26.4$ of Tent and Ziegler's book "A course in model theory"

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  • $\begingroup$ Dear Tim, I do not get it how some kind of definability in families might be helpful. Let our definable set contain infinitely many mutually disjoint definable sets of rank $\alpha$. They might be defined all by different formulas $\psi_i(x, b_i)$. We need to prove that there exist infinitely many such formulas with parameters in a given model, probably by changing the parameters $b_i$. The trouble is not so much to control the rank but to ensure that no intersections occur when we change the parameters. $\endgroup$
    – Dima Sustretov
    Commented Dec 12, 2012 at 11:45
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In $ACf_p$ or $DCF_0$, one always assume that they are working in a monster model. So saturation still does the job. Note that they are both $\omega$-stable. So saturated model for any regular cardinal exists.

But maybe that is not what you are asking.

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