2
$\begingroup$

I'm trying to prove Lindström's theorem, which says that a countable inductive first order theory $T$ which is $\kappa$-categorical for some $\kappa\geq \aleph_{0}$ is model-complete.

The strategy I follow is proving each model has an existentially-closed extension of any bigger cardinality, which I proved using transfinite induction. Now, I try to prove an analogous result for non-e.c models - if $T$ has non-e.c. model, then it has a such a model of any infinite size. Using downwards Löwenheim-skolem, we can assume that $M\models T$ is a countable model which is not e.c.

So there is another model $N$ which thinks a certain existential sentence with parameters in $M$ that $M$ does not think. I tried to elementarily extend both $M$ and $N$ in a manner that preserves the inclusion in each step, but it doesn't seem to work.

Once I know this two facts, we obtain 2 models of cardinality $\kappa$ - one of which is e.c. and the other is not, contradicting the fact they are isomorphic. I'd like to know how to fill in this specific part.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Consider the two-sorted structure $(M, N)$ with constants representing elements in $M$ and an added function representing the inclusion map $M \to N$. Let $T$ be the theory of this structure. Add $\kappa$ many constants of the first sort to the language and add axioms stating these constants take distinct values. By compactness, we conclude that this new theory admits a model $(M’, N’)$. Clearly $|M’| \geq \kappa$. Using downward Löwenheim-Skolem if necessary, we may assume $M’$ and $N’$ are both of cardinality exactly $\kappa$. Clearly $M’$ elementarily extend $M$ and $N’ \supset M’$. $N’$ is also elementarily equivalent to $N$ with parameters in $M$. Thus, these structures satisfy the properties you want.

$\endgroup$
11
  • $\begingroup$ I'm not familiar with multiple-sorted structures, can this be proven without such reasoning? $\endgroup$
    – Oria
    Commented Mar 27 at 19:49
  • 1
    $\begingroup$ @Oria You can always pretend it’s a single-sorted structure by taking the disjoint union and add a predicate indicating which sort the element is. I’m unable to come up with a proof that avoids this kind of reasoning altogether at the moment though. $\endgroup$
    – David Gao
    Commented Mar 27 at 19:55
  • $\begingroup$ I see! The part I missed was using a predicate to distinguish between elements of $M$ and elements of $N$. I'll try to formalize it, thank you! $\endgroup$
    – Oria
    Commented Mar 27 at 20:05
  • $\begingroup$ I started writing down a first-order theory, but I seem to get stuck - Say I take the elementary diagram of $N$, and add the sentences $\left\{ c_{i}\neq c_{j}\;\vert\;i\neq j<\lambda\right\} \cup\left\{ P\left(c_{i}\right)\;\vert\;i<\lambda\right\}$ where all the $c$'s are new constants. How can I force $P$ to realize an elementary extension of $M$ inside of the new model? $\endgroup$
    – Oria
    Commented Mar 27 at 20:39
  • $\begingroup$ @Oria Add the axioms saying all elements of $M$ satisfy $P$ and none of the elements of $N \setminus M$ satisfies $P$. Then for any sentence $\varphi$ with parameters from $M$, clearly in the original structure $\varphi^P$ is true iff $\varphi$ is true in $M$, where $\varphi^P$ replaces all quantifiers in $\varphi$ by quantifiers that only range over elements satisfying $P$. (Since in the original structure satisfying $P$ and being in $M$ are equivalent.) $\endgroup$
    – David Gao
    Commented Mar 27 at 21:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .