0
$\begingroup$

I read on the Wikipedia page that "an equivalence relation with exactly two infinite equivalence classes is an easy example of a theory which is ω-categorical, but not categorical for any larger cardinal number." https://en.wikipedia.org/wiki/Equivalence_relation I understand the statement intuitively but I'm not sure if I'm on the right track proving it. Here is my approach to prove a theory with exactly two infinite equivalence classes is countably categorical but not uncountably categorical.

For every two countable models M1, M2 of the theory T of an equivalence relation with exactly two infinite equivalence classes. |M1|=|M2| since both equivalence classes have infinite elements. The domain of all models of T are countably infinite. Therefore there is an isomorphism between every two models of T.

For the second part, I'm trying to find a counterexample which shows two uncountable models are not isomorphic but I failed to find any. What I don't understand is, how can there be any uncountable model if the equivalence relation of a certain language has exactly two countably infinite equivalence classes?

$\endgroup$
  • $\begingroup$ At the very end of your question, you wrote "exactly two countably infinite equivalence classes" but the theory says only "exactly two infinite equivalence classes." Smuggling in the word "countable" was the source of your problem. $\endgroup$ – Andreas Blass Apr 4 '17 at 17:34
  • $\begingroup$ Sorry my bad. I'm going to edit the question. But I think the statement still holds true in the sense of countably categorical but not uncountably categorical. And I used the countable property in my proof as well. $\endgroup$ – user432451 Apr 4 '17 at 17:58
  • $\begingroup$ Even after the edit, the end of your question still says "exactly two countably infinite equivalence classes" and there's no reason for "countably" here. $\endgroup$ – Andreas Blass Apr 4 '17 at 18:31
  • $\begingroup$ Oh I just got what you mean. So such models could exist since there's no need to guarantee the equivalence classes are countable. Thank you so much. $\endgroup$ – user432451 Apr 4 '17 at 19:06
1
$\begingroup$

Recall cardinal arithmetic for infinite cardinal: $$\kappa+\lambda=\max\{\kappa,\lambda\}.$$

Try to play with the cardinality of one of the two classes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.