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Let $\mathcal L$ be a language with only a binary predicate $E$, and let $T$ be a theory of structures in which $E$ is an equivalence relation which partitions the structure into two infinite sections. It is easy to see that $T$ is $\aleph_0$-categorical but not $\aleph_\alpha$-categorical when $\alpha>0$.

My question: can there be a $\kappa \not \le \aleph_0$ such that $T$ has a model of cardinality $\kappa$ and is $\kappa$-categorical? What about $\kappa$ is infinite but Dedekind-finite?

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    @ChrisEagle Yes, thank you.2012-11-28
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    By $\kappa$-categorical, do you mean $T$ has (up to isomorphism) *exactly* one model of cardinality $\kappa$, or *at most* one? If $\kappa$ is amorphous, $T$ has no models of cardinality $\kappa$ at all.2012-11-28
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    @ChrisEagle Yes, I have corrected my post.2012-11-28
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    @AsafKaragila Okay, thanks to you.2012-11-28

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If $M$ is a model of $T$ let $M_0$ and $M_1$ denote the two parts defined by $E$.

Suppose that $A$ is an infinite Dedekind-finite set. If $A$ is amorphous then it cannot be a model of $T$ to begin with because it cannot be split into two infinite sets.

Suppose that $A$ can be split into $A_0$ and $A_1$. If $|A_0|=|A_1|+\frak a$ for $\frak a$ nonzero, then taking any $B\subseteq A_0$ such that $0<|B|<\frak a$ defines a non-isomorphic model by $A'_0=A_0\setminus B$ and $A'_1=A_1\cup B$. To see that this partition is not isomorphic to the previous partition note that it could only happen if $|A'_1|=|A_0|$ (because we remove points from $A_0$ and its cardinality is strictly smaller now) but this means there is a bijection $f\colon A_0\to A_1\cup B$ which is false because $|A_0|=|A_1|+|B|=|A_1|+\frak a$ and therefore $|B|=\frak a$, in contradiction to the way we chose $B$.

If $A$ can be split into two incomparable parts, then any exchange of points will keep it incomparable by a similar argument.


However it is possible to get categoricity in $\aleph_1$-amorphous cardinals. Namely cardinals of sets that cannot be split into two disjoint uncountable sets.

Suppose that $A$ is $\aleph_1$-amorphous, and it has no infinite Dedekind-finite subset, then any partition of $A$ into two parts would have to have one part of size $\aleph_0$ and another part of size $|A|$. Therefore categoricity follows for any $\aleph_1$-amorphous cardinal.


Examining this proof shows that $T$ is categorical in $\kappa$ if and only whenever $\kappa=\frak a+b$ then we can guarantee that either $\frak a$ or $\frak b$ is $\aleph_0$, and the other is Dedekind-infinite.

This happens if and only if $A$ cannot be split into two uncountable sets, has no infinite Dedekind-finite subsets, and if $B\subseteq A$ is infinite then $|B|=|A|$ or $|B|=\aleph_0$.

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    I have to admit that I wrote my initial comment with an elaborate idea I had in mind, but as the details began to unfold it became increasingly difficult to prove, and I decided to pick a simpler example for now. If I finish the proof I will definitely post it.2012-11-28
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    I am under the impression that OP meant to ask about models of this particular theory.2012-11-29
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    @tomasz: You know... I think you're right. :-) Let me modify my answer to address the actual question! Thanks!!2012-11-29
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    Er...I'm not sure. Assume $\kappa=\lambda+\aleph_0$ be an $\aleph_1$-amorphous cardinal. It is Dedekind-infinite, but Does it exist some cases $\lambda<\kappa$?2012-12-01
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    @Popopo: If $\kappa$ is $\aleph_1$-amorphous then $\lambda=\kappa$.2012-12-01
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    It need to show that if $\lambda$ is Dedekind-infinite then $\lambda+\aleph_0=\lambda$ and if $\lambda$ is Dedekind-finite then $\kappa$ is not $\aleph_1$-amorphous. But it seems non-trivial. Could you please give me a hint?2012-12-01
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    @Popopo: Err, no... of course I was wrong. Take any amorphous set $A$ then $A\cup\omega$ is $\aleph_1$-amorphous. However if we also add the requirement that every D-finite subset of our $\aleph_1$-amorphous is finite then the claim is true. Because this means that if $|A|=|B|+\aleph_0$ then $\aleph_0\leq|B|$ and then $|B|+\aleph_0=|B|$. To see that the last claim is true note that if $\omega\leq\alpha<\aleph(A)$ then $A+\alpha\sim A$ (where $\aleph(A)$ is the Hartogs number of $A$, and $\sim$ means same cardinality).2012-12-01
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    Note that in the last part I covered that case. Namely if $A$ is an $\aleph_1$-amorphous set created by taking the union of an amorphous set and a countable set then we can split it into two parts where one is infinite but not Dedekind-infinite. This is a contradiction to the requirement about $\frak a+b$ written in the last part.2012-12-01
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    Okay. that all things is clear now. Thank you very much.2012-12-02