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In Chang & Keisler "Model Theory" it is claimed that the theory of a one-to-one function of A onto A with no finite cycles is $\omega_1$- categorical (page 140). Why is that, and is there a reference for this?

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up vote 6 down vote accepted

Let $A$ and $B$ be two models of size $\omega_1$. In both A and B, being in the same cycle is an equivalence relation. Each equivalence class has size $\omega$. So there are $\omega_1$ many equivalence classes in both A and B. Fix a bijection between the equivalence classes in A and B. You can use this to make an isomorphism between A and B because every single equivalence class in either A and B is isomorphic to every other.

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You left out important detail: because $f$ is onto, these equivalence classes are all doubly infinite paths. So they are all isomorphic to each other. – David Harris Apr 1 '11 at 22:13
- What do you mean by doubly infinite paths? aren't they isomorphic since all of them of size $ \omega $ ? - Can't we have in f_1 a cycle of size $ \omega_1 $ while in f_2 all cycles are of size $ \omega $? – Eran Apr 2 '11 at 14:28
David's point is just that cycle keeps going in the direction of repeated applications of f: $f(x), f(f(x)), f(f(f(x))), \ldots$ and in the direction of repeated applications of $f^{-1}$: $f^{-1}(x), f^{-1}(f^{-1}(x)), f^{-1}(f^{-1}(f^{-1}(x))), \ldots$ – Richard Dore Apr 2 '11 at 16:06
You can't have an uncountable cycle. Let f be your function, and x some point in your cycle. Then the cycle is exactly $\{ f^n (x) : n \in \mathbb{Z}\}$, which is a countable set. – Richard Dore Apr 2 '11 at 16:06
Do we need $f$ to be onto? if $f^{i}(x)$ is the i'st element in the cycle for i limit, then $f^{-1}(f^{i}(x))$ is an $f^{j}(x)$ for some j<i which is a contradiction since $f(f^{j}(x))$ is another element in between (so we can't have i limit). – Eran Apr 2 '11 at 21:46

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