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There are many equivalent definitions for stability, one of them being that being unstable is equivalent to the existence of a formula having the order property.

While intuitively it makes sense that having a lot of types (which is the definition that Shelah uses) would lead to a lot of models , the order property isn't as intuitive. Where the examples/partial results floating around that would have indicated that it was likely? The only result I'm aware of is Ehrenfeucht result, which said there were at least two models (for uncountable cardinals).

Edit1: I'm unsure why this question was tagged as unclear. If someone could explain why so, I can edit it.

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    $\begingroup$ Having an order also implies having lots of types. Different cuts give you different types and you can engineer orders with lots of cuts. $\endgroup$
    – Levon Haykazyan
    Commented Jul 30, 2015 at 21:52
  • $\begingroup$ @Levon : It isn't very intuitive though and the engineering of these orders is not intuitive either (at least they are not for me). $\endgroup$
    – user75685
    Commented Jul 30, 2015 at 22:10
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    $\begingroup$ I would post the following as an answer, but unfortunately the question has been closed. The order property was first defined by Shelah, but its origins in connection counting types and counting models dates back further to the work of Ehrenfeucht and Morley. Here is an excerpt from Morley's paper "Categoricity in Power": "Suppose $A$ is a model of $T$, $R$ a relation of degree $n$ of $A$, $X\subseteq |A|$, and $S_n$ the permutation group on $(0,\dots,n-1)$. Following Ehrenfeucht, we define $R$ to be connected over $X$ if for every sequence of $n$ distinct elements $x_0,\dots,x_{n-1}$..." $\endgroup$
    – Alex Kruckman
    Commented Aug 1, 2015 at 20:31
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    $\begingroup$ "... there is an $s\in S_n$ such that $A\models R(x_{s(0)},\dots,x_{s(n-1)})$. $R$ is anti-symmetric over $X$ if for every sequence of $n$ distinct elements $x_0,\dots,x_{n-1}$ of $X$, there is $s\in S_n$ such that $A\models \lnot R(x_{s(0)},\dots,x_{s(n-1)})$. Theorem 3.9: If $T$ is totally transcendental and $A$ a model of $T$, then no relation of $A$ is connected and anti-symmetric over any infinite $X\subseteq |A|$." $\endgroup$
    – Alex Kruckman
    Commented Aug 1, 2015 at 20:34
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    $\begingroup$ The notion "connected and anti-symmetric over infinite $X$" is a clear ancestor of the order property. Morley proved that every theory categorical in an uncountable cardinality is totally transcendental, so Theorem 3.9 shows that no such theory has an "unstable formula" in this sense. Morley also notes "For the case where $T$ is categorical in $2^\kappa$, this result was obtained by Ehrenfeucht ["On theories categorical in power"]. Dana Scott (unpublished), by a different and simpler proof, extended the result to theories categorical in power $\kappa^{\aleph_0}$." So the ideas were in the air! $\endgroup$
    – Alex Kruckman
    Commented Aug 1, 2015 at 20:40

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I disagree with the claim that the order property isn't intuitive.

When I took my first model theory class, types were introduced to me as a generalization of Dedekind cuts. So the very first examples we looked at were linear orders. Moreover, the construction of theories with exactly $n$ distinct countable models (for $n\in\mathbb{N}-\{2\}$) was done via linear orders with additional structure; I believe this is standard. So we were already used to counting the models of theories with linear orders.

Given this, it was very intuitive to me to believe that having an ordering meant having lots of types. Now, that's not actually true - DLO has the order property obviously, but only $n!$-many $n$-types - but it's a first approximation of a true fact: that having an ordering means having lots of models. So for me this was the progression of ideas.

Although I describe above the way I gained an intuition for the order property, Shelah et al were certainly aware of all of the above and more. Finally, there were already at the beginning of stability theory lots of set-theoretic results about the different kinds of linear orders, so there was already a strong sense that $\{$linear orders$\}$ is an extremely rich class. So I think the order property was actually a very intuitive thing to define in the context of counting uncountable models. The fact that it corresponds exactly to instability is less intuitive of course, but the order property as a thing to study in the context of stability theory seems natural to me.

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  • $\begingroup$ The $n!$ count omits those $n$-types that say that two or more of the $n$ elements are equal. (Of course that doesn't affect the point you were making.) $\endgroup$
    – Andreas Blass
    Commented Jul 31, 2015 at 1:19
  • $\begingroup$ Could you please add a couple of the results about linear orders that you were thinking about to the answer? $\endgroup$
    – user75685
    Commented Jul 31, 2015 at 1:49