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I have to answer the next questions:

  • What is the number of complete 1-types of the theory of atomless Boolean algebras?

  • What is the number of complete 2-types of the theory of atomless Boolean algebras?

I know that the theory of atomless Boolean algebras is countably categorical (has up to isomorphism only one countable model) and therefore has only finitely many n-types for each n. But what next?

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You could start by looking at the countable model and thinking about how its automorphisms act. – Chris Eagle Jan 31 '13 at 16:05

Hint: (elaborating a little on Chris Eagle's comment) if a theory is $\omega$-categorical, then its only countable model is necessarily saturated.

Now, if you have any saturated model $M$, then it realizes all $n$-types without parameters for each $n$ and it is strongly homogeneous. This means that not only can you find representatives for each $n$-type, but $n$-types correspond exactly to orbits of $\operatorname{Aut}(M)$ acting on $M^n$.

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Thanks on this, I think I have the right answers now. But a next question: How can you prove saturated implies strongly homogeneous? – natural Feb 5 '13 at 15:37
@natural: in general, $\kappa$-saturation implies $\kappa$-homogeneity (this is easy), and homogeneity in a model's own cardinality is equivalent to strong homogeneity in its own cardinality (which can be done using a standard back-and-forth argument). – tomasz Feb 5 '13 at 20:42
Thanks, I can follow the reasoning now. But I cannot get a grip on what the orbit of say an element $a$ (not 0 or 1) is. For we are dealing with a countably infinite BA. – natural Feb 6 '13 at 13:43
@natural: If you're fairly familiar with descriptive set theory, it might help to think of the countable boolean algebra as algebra of closed-open sets in Cantor set, and of homeomorphisms of the Cantor set. Or, more directly, about binary trees. I'm pretty sure you can take any element distinct from zero and unit to any other such element, and for pairs, you have to preserve zero, unit, containment and noncontaimnent, and that should be it (the case of two elements should follow directly from the case of one element). – tomasz Feb 7 '13 at 21:15

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