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I had this question up on Math stackexchange: https://math.stackexchange.com/questions/1349247/number-of-non-isomorphic-models/1350763#1350763 . While it was answered partially there, I'm posting here in the hope that I might get a more detailed answer.

Let $C$ be the class of cardinals. Define by recursion $C_0 = C$, $C_\alpha = C_\beta\cup P(C_\beta)$ if $\alpha=\beta+1$ and $C_\alpha = \bigcup_{\beta<\alpha}{C_\beta}$ for limit $\alpha$ (Here $P(C_\beta)$ is the class of subsets of $C_\beta$).

We say that a complete theory $T$ has an invariant system of rank $\alpha$ iff there is some (class) function $f$ associating to each model of $T$ an element of $C_\alpha$ such that given models $A, B$ of $T$; $f(A)=f(B)$ iff A is isomorphic to B. And a theory $T$ is classifiable iff it has a invariant system of some rank $\alpha$.

I recently read that having $2^\lambda$ models for each uncountable cardinal $\lambda$ meant that you could not have an invariant system of the above style. I would like to know the reason why.

I know that classification theory is very complicated and technical, I would be quite happy with an intuitive answer that doesn't have all the details for this question.

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Under the global choice principle (which asserts that there is a class well-ordering of the set-theoretic universe), then every theory $T$ has rank $0$ as you define it, regardless of the number of models of the various uncountable sizes. The reason is that the well-ordering allows us to select a unique representative from each isomorphism class, and these representatives are themselves well-ordered in type at most Ord, which is bijective with $C_0$ via $\alpha\mapsto\aleph_\alpha$. So we can map the isomorphism classes injectively into $C_0$.

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  • $\begingroup$ Perhaps the OP had meant to impose another requirement on $f$ that connected the size of the model $A$ to the value of $f(A)$? $\endgroup$
    – Joel David Hamkins
    Commented Jul 6, 2015 at 16:03
  • $\begingroup$ I found this here: books.google.com/… $\endgroup$
    – user75685
    Commented Jul 6, 2015 at 16:50
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    $\begingroup$ I don't have any particular fix in mind; rather, I was merely speculating that Shelah had omitted condition on $f$ that would insist that $f(A)$ is chosen from the part of $C_\alpha$ using only cardinals up to and including $|A|$, or something like that. It seems plausible to me in that case that you couldn't find such a function for a fixed $\alpha$, since there might not be enough room. $\endgroup$
    – Joel David Hamkins
    Commented Jul 6, 2015 at 17:49
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    $\begingroup$ When I said in my math.stackexchange answer that "it doesn't seem obviously inconsistent to me that there could be a theory $T$ with the maximal number of models ($2^\lambda$) in all uncountable $\lambda$ with an invariant system of rank $0$," I actually had this global choice argument in mind, but I hadn't really stopped to think about it. I'm glad you've confirmed that it works! Note that this definition is not due to Shelah, it's an interpretation of his work in the book by Marcja and Toffalori. $\endgroup$
    – Alex Kruckman
    Commented Jul 6, 2015 at 18:10
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    $\begingroup$ The cardinals which appear in the dimension invariants in Classification Theory are necessarily bounded above by the cardinality of $A$, just as you suggest. $\endgroup$
    – Alex Kruckman
    Commented Jul 6, 2015 at 18:11

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