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It is often said that unsuperstable theories do not admit a classification in the sense of Shelah. Why exactly is this so? And also what exactly does in the sense of Shelah mean? It is hand waved in a lot of places and it is hinted that the dividing line is having few models.

However the only definition that I know of that works (See here for a definition that fails Number of non-isomorphic models); having the maximal number of models would not exclude the possibility of a classification (see for example the definition in Hart, Hurshovski, Laskowski; http://arxiv.org/pdf/math/0007199.pdf of classifiable, i.e. superstable, has prime models over pairs, and does not have the dimensional order property).

So to re-iterate, what is "admits a classification in the sense of Shelah"? and why does it fail for unsuperstable theories? Is there no hope for recovering any sort of classification in this case? And is there any place where this is discussed in any depth?

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    $\begingroup$ Shelah's book itself discusses this. $\endgroup$
    – Andrés E. Caicedo
    Commented Jul 8, 2015 at 22:48
  • $\begingroup$ @Andres : I admit to not having read the entire book, but in the bits and pieces I have read, he just sort of outlines his plan, with few specifics and no true definition of what it means to be classifiable, which is what my question is about. If he does, then it would be at a part I skipped and I would be extremely grateful to you if you could point me to some places where he discuss this. $\endgroup$
    – user75685
    Commented Jul 9, 2015 at 0:16
  • $\begingroup$ Chapter 13 of Shelah's book is called "For Thomas the Doubter" and is entirely devoted to your question. $\endgroup$
    – Levon Haykazyan
    Commented Jul 9, 2015 at 13:10
  • $\begingroup$ The idea behind "admits a classification in the sense of Shelah" is that the class of models of T in a given cardinal \aleph_\alpha can be described by bounded many numercial invariance (generalizing the situation that a vector space is determined by its dimension). In the case of a countable first-order theory T, if T admits classification then the number of isomorphism types of models of card \aleph_\alpha is bounded by \beth_{\omega_1}(|\alpha|). When \alpha is such that \aleph_\alpha is much bigger than \alpha then \beth_{\omega_1}(|\alpha|)<< 2^{\aleph_\alpha}. $\endgroup$
    – Rami Grossberg
    Commented Jul 10, 2015 at 15:19

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The idea behind "admits a classification in the sense of Shelah" is that the class of models of $T$ in a given cardinal $\aleph_\alpha$ can be described by bounded many numercial invariance (generalizing the situation that a vector space is determined by its dimension). In the case of a countable first-order theory $T$, if $T$ admits classification then the number of isomorphism types of models of card $\aleph_\alpha$ is bounded by $\beth_{\omega_1}(|\alpha|)$. When $\alpha$ is such that $\aleph_\alpha$ is much bigger than $\alpha$ then $\beth_{\omega_1}(|\alpha|)<< 2^{\aleph_\alpha}$.

Thus $I(\lambda,T)=2^\lambda$ for all uncountable lambda implies that $T$ is not classifiable.

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    $\begingroup$ Rami, I added dollar signs in order that your TeX compiles. $\endgroup$
    – Joel David Hamkins
    Commented Jul 10, 2015 at 15:30

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