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My definition of $\omega-$categorical: a theory $T$ is $\omega-$categorical if any two countable models of $T$ are isomorphic. In particular, if $T$ doesn't have a countable model, then $T$ is $\omega-$categorical.

I am solving some exercises from Hodges shorter model theory, and while solving Ex 6.3.1, I wanted to figure out the following:

I either want to find an example or prove that such example cannot exist, of the following:

Find an example of a language $L$ and an infinite $L$ structure $M$, such that there is a finite tuple $\vec{m}\in M$ such that precisely one of the following two theories is $\omega-$categorical: $Th(M)$ and $Th(M,\vec{m})$.

If $L$ is countable, then by Ryll-Nardzewski, using the fact that "$S_n(Th(M))$ is finite for all $n$ iff $S_n(Th(M,\vec{m}))$ is finite for all $n$", we can see that such an example cannot exist.

Thanks for any help!

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  • $\begingroup$ I'm confused about what exactly you're asking. It seems like you know that when $L$ is countable, and $\overline{m}$ is a finite tuple from an $L$-structure $M$, then $\mathrm{Th}(M)$ is $\omega$-categorical if and only if $\mathrm{Th}(M,\overline{m})$ is $\omega$-categorical. So are you asking about whether this can fail when $L$ is uncountable? $\endgroup$
    – Alex Kruckman
    Commented Nov 17, 2023 at 16:26
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    $\begingroup$ by modifying the example here I think it is not hard to find examples of theories with many non-isomorphic countable models, but such that adding a single constant can preclude the existence of countable models. more precisely, if $M$ is any uncountable model of the theory in his post, then there is some non-standard $m\in M$, and then the theory of $(M,m)$ will have no countable models for the reason he describes $\endgroup$
    – Atticus Stonestrom
    Commented Nov 17, 2023 at 18:49
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    $\begingroup$ so eg you can take a two-sorted theory, where the first sort is just the theory Andreas describes in the post, and the second sort is your favorite non-$\omega$-categorical theory, and there are no relation or function symbols between the two sorts. then taking $M$ to be any uncountable model will let you get an example where adding a constant makes the theory vacuously $\omega$-categorical (as it has no countable models) even though $Th(M)$ is not $\omega$-categorical $\endgroup$
    – Atticus Stonestrom
    Commented Nov 17, 2023 at 18:51
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    $\begingroup$ however this is not so satisfying as the only reason the expanded theory is $\omega$-categorical is that it has no countable models. I think a non-vacuous example would be much nicer $\endgroup$
    – Atticus Stonestrom
    Commented Nov 17, 2023 at 18:53
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    $\begingroup$ You can think of a two-sorted structure in the following way: two predicates $A,B$ partition the universe; for the theory that you want to be referring to the first sort, "relativize the quantifiers to $A$" the same way as in set theory; similarly for $B$. This should work generally to simulate any finite number of sorts in a single-sorted structure. $\endgroup$
    – Tesla Daybreak
    Commented Nov 18, 2023 at 4:10

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