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.



Number of non-isomorphic models
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 $C\_0$. 


