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I was flipping through Baldwin's Stability Theory book and saw an example that has me confused...

The example is a 1st-order theory $T$ of refining equivalence relations $E_i(x, y), i< ω$, where

  • $E_j$ refines $E_i$ when $i < j$
  • each $E_{i+1}$ relation splits each $E_i$-equivalence class into two $E_{i+1}$-equivalence classes.

Here's what I'm picturing: $E_0$ splits the model into two classes, call them $a_0/E_0$ and $a_1/E_0$. Then $E_1$ splits $a_0/E_0$ into two pieces $a_{00}/E_1$ and $a_{01}/E_1$, and $E_1$ splits the class $a_1/E_0$ into two pieces $a_{10}/E_1$ and $a_{11}/E_1$, ad infinitum.

So the classes are arranged like the tree ${}^{<ω}2$ of finitely long $0-1$ sequences (the tree ordering is sequence extension), where the $n$th level of the tree has the $E_n$ classes as nodes.

My confusion:

  1. Each infinite branch of the tree (there are $2^\omega$ of them) should correspond to an incomplete type of an element, right? And those types are pairwise contradictory. So, if $M$ is a countable model of $T$, there are $2^ω$ types over $M$. If all that is right, then $T$ is $\omega$-unstable. $T$ is supposed to be (some kind of) stable however.
  2. But can't the type space still blow up over uncountable sets of parameters for cardinals of countable cofinality? Is it that such a T is stable at some cardinals, but doesn't “stay stable”?

Sorry if this is boring or obtuse! -M

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  • $\begingroup$ I edited your post to use more standard markup. For future reference, try to typeset math using TeX and break lines using just two regular line breaks (it makes the bullet and enumeration markup work right, and makes the code easier to edit for others). Also, try to always enclose math expressions in dollar signs, even if you don't need any special markup and you just use variable name. It makes the text more readable. $\endgroup$ – tomasz May 26 '13 at 17:23
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This theory is stable in all cardinals that are at least $\mathfrak c$. To see this, notice that a type is determined entirely by which class of each $E_n$ it belongs in (of which there are $\mathfrak c$ many possible combinations in total) and which elements it is (not) equal to.

Such a theory is called superstable.

There are theories which are stable, but not superstable, and do explode precisely for parameters of size $\kappa$ such that $\kappa^\omega>\kappa$. For example, the theory of (countably) infinitely many independent equivalence relations with infinitely many classes. With $\kappa$ parameters you can have each relation have $\kappa$ many classes, yielding $\kappa^\omega$ types.

(Edit: the theory I mentioned is of infinitely many independent relations with infinitely many classes; the theory of infinitely many independent equivalence relations with two classes is, I believe, bi-interpretable with the one you described in your question.)

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  • $\begingroup$ Nice, thanks Tomasz! I wonder how hard it would be to find an uncountable set of parameters X, with |S(X)| > X, if one assumes ~CH... I might fool with that. $\endgroup$ – Marco May 26 '13 at 3:13
  • $\begingroup$ Why would it be hard? Just take a suitably large model. Any model consists of set of points covering some dense subset of infinite branches, with arbitrarily many points covering each branch. $\endgroup$ – tomasz May 26 '13 at 9:37
  • $\begingroup$ Yeah, I guess there's not much interesting there after all. How about this though: $\endgroup$ – Marco May 26 '13 at 14:50
  • $\begingroup$ Modifying T can make it unsuperstable: Change T so that there are ωω "many" E_i's. Then E_ω has infinitely many equivalence classes. In general, for each n, E_ωn has infinitely many classes. Then for each n, the set of formulas that say "there are κ many E_ωn-inequivalent elements" is consistent with T. <br /><br />Find a model M, |M| = κ, cf(κ) = ω. The types over M are determined by possible paths through the E_ωn's, of which there are κ^ω > κ. <br /><br />This looks like a special case of what you were pointing out, eh Tomasz? $\endgroup$ – Marco May 26 '13 at 15:02
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    $\begingroup$ @Marco: It seems as though you have a couple different account hanging around here. See this page for information about having them merged. $\endgroup$ – user642796 May 26 '13 at 17:57

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