2
$\begingroup$

Let $\mathcal L=\{c_{i}:i<\omega\}$ be a language in first order logic, and:

$\Gamma =\{\forall x(x=x),\forall x\forall y(x=y\rightarrow y=x),\forall x\forall y\forall z(x=y\wedge y=z\rightarrow z=y)\}\cup\{c_{i}\neq c_{j}:i,j\in\omega ,i\neq j\}$

Show that $\Gamma$ is $\kappa $-categorical for $\kappa>\aleph_{0}$ , but not $\aleph_{0}$-categorical.

¿It can be an arbitrary equivalence relation $R$ , not necessarily $=$?

$\endgroup$
  • $\begingroup$ You shouldn't use "=" for anything other than equality. Please edit to make it clear what the sentences in $\Gamma$ are... $\endgroup$ – universalset Dec 13 '13 at 3:10
  • $\begingroup$ Hello Yesid.${}$ $\endgroup$ – Camilo Arosemena-Serrato May 3 '15 at 19:10
2
$\begingroup$

Your theory is $\kappa$-categorical for all uncountable $\kappa$, since every model of uncountable size $\kappa$ of the theory has the $\omega$ many distinct constants, plus $\kappa$ many other individuals, and all such models are isomorphic. But your theory is not $\aleph_0$-categorical, since your theory has a model with only the constants, and another with the constants plus another individual that is not a constant, or two others or $n$ others or $\omega$ many others, and these models are not isomorphic. So there are countably many distinct countable models up to isomorphism.

The result isn't true if you interpret $=$ only as an equivalence relation, since in this case, there can be non-isomorphic models of uncountable size $\kappa$, depending on the size of the equivalence class of each $c_i$.

$\endgroup$

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.