Let $\mathcal L = \{ E(\_,\_) \}$ and $T$ be the $\mathcal L$-theory that says that $E$ is an equivalence relation with an infinite number of infinite classes. (I find this statement not clear, because it does not seem to specify the number of finite classes. In the following I assume that all classes are infinite)

I'd like to know how many models with given cardinality there are. I think there is only one model of $T$ of cardinal $\aleph_0$, because there is $\aleph_0$ classes of cardinal $\aleph_0$, hence any two models would be isomorphic by simply identifying the classes. This shows that $T$ is $\aleph_0$-categorical and hence complete.

Now if I look at models $\mathcal M$ of cardinal $\aleph_1$, then the "isomorphism type" of $\mathcal M$ (I don't know if this is a valid terminology) is given by the number $\kappa$ of equivalence classes and for all $i<\kappa$ by the cardinal $\lambda_i$ of the $i$th class. Since $Card(\mathcal M) = \aleph_1$, we have that $\sum_{i<\kappa} \lambda_i = \aleph_1$. If all $\lambda$'s are equal, I know that this sum is $\kappa\times\lambda = \max\{\kappa,\lambda\}$ and there are 3 non-isomorphic models of $T$. But if the $\lambda$'s are different, I don't know how to evaluate the infinite sum of cardinals.

Could you help me finish this computation? Any reference for cardinal arithmetic would be appreciated too!

  • $\begingroup$ If I split the $\lambda$'s into those which are $\aleph_0$ and those which are $\aleph_1$, I get that there are $2^{\aleph_1}$ non-isomorphic models of cardinal $\aleph_1$, is that right? $\endgroup$ – zarathustra Oct 26 '13 at 8:59
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    $\begingroup$ That's too many. I only see countably many choices. $\endgroup$ – André Nicolas Oct 26 '13 at 9:03

Let $M$ be the underlying set of $\mathscr{M}$, and assume that $|M|=\omega_1$. For a partition $\mathscr{P}$ of $M$ into infinitely many infinite blocks let $\kappa(\mathscr{P})$ be the number of blocks of cardinality $\omega_1$, and let $\lambda(\mathscr{P})$ be the number of blocks of cardinality $\omega$.

Clearly $\kappa(\mathscr{P})$ can assume any value in $K=\{0,1,2,\ldots,\omega,\omega_1\}$.

  • If $\kappa(\mathscr{P})=0$, then clearly each block of $\mathscr{P}$ has cardinality $\omega$, so $\lambda(\mathscr{P})=\omega_1$.
  • If $\kappa(\mathscr{P})>0$, $\lambda(\mathscr{P})$ can assume any value in $K$.
  • The isomorphism type of $\mathscr{P}$ is completely determined by $\left\langle\kappa(\mathscr{P}),\lambda(\mathscr{P})\right\rangle$.
  • $\begingroup$ So there is one model with type $(0, \omega_1)$ and then for the $\omega$ other different values of $\kappa(\mathscr{P})$, there is $\omega$ choices for $\lambda(\mathscr{P})$, whence a total number of $\omega$ models? $\endgroup$ – zarathustra Oct 26 '13 at 9:12
  • $\begingroup$ @Antoine: That’s right. $\endgroup$ – Brian M. Scott Oct 26 '13 at 9:13
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    $\begingroup$ Congrats on the 200k, Brian. $\endgroup$ – Asaf Karagila Oct 26 '13 at 9:29
  • $\begingroup$ Congrats 200k man !!!!!!!! $\endgroup$ – ILoveMath Oct 26 '13 at 12:08
  • $\begingroup$ @DonAnselmo: Thanks! $\endgroup$ – Brian M. Scott Oct 26 '13 at 13:13

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