Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $T$ be a countable first-order theory, and assume that $T$ has exactly one atomic model up to isomorphism in every uncountable cardinality. (By "atomic" I mean a model which omits the non-principal types).

Now let $\mathfrak{M}$ be a countable transitive model of set theory, and assume that $T$ is also (countable) in $\mathfrak{M}$.

Is the above property preserved in $\mathfrak{M}$? i.e. does it hold in $\mathfrak{M}$ that there exists only one atomic model of $T$ in every uncountable cardinality in $\mathfrak{M}$?

share|improve this question
6  
I may be missing something obvious, but could the close voters explain what they think is wrong with the question? –  Emil Jeřábek Dec 28 '13 at 20:16
1  
Related question: mathoverflow.net/questions/20590 . –  Emil Jeřábek Dec 28 '13 at 20:31

1 Answer 1

up vote 14 down vote accepted

This is really a comment, but I need a bit more space.

If $\phi$ is a sentence of ${\cal L}_{\omega_1,\omega}$ that is $\aleph_0$-categorial, there is a complete first order theory $T$ in an expanded vocabulary such that the models of $\phi$ are exactly the reducts atomic models of $\phi$. The expansion is done in such a way that two structures will be isomorphic in the original language if and only if they are isomorphic in the expanded language.

So your question is really the same as: For $\phi$ a sentence of ${\cal}L_{\omega_1,\omega}$ is ``$\phi$ is $\kappa$-categorical for all infinite $\kappa$" absolute?

This is, as far as I know, still an open question. It is also open if ``$\phi$ is $\aleph_1$-categorical" is absolute.

John Baldwin in his paper ``Amalgamation, Absoluteness, and Categoricity" addresses some issues around this. Here is a link http://homepages.math.uic.edu/~jbaldwin/pub/singsep2010rev.pdf

share|improve this answer
6  
I think this is absolutely valuable as an answer, not a mere comment. –  Todd Trimble Dec 30 '13 at 19:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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