I have found two versions of the famous Lowenheim-Skolem Theorem (LST) which appear outwardly to be saying different things. In Enderton's classic logic manual he states

LST (Enderton): (a) Let $\Gamma$ be a satisfiable set of formulas in a language of cardinality $\lambda$. Then $\Gamma$ is satisfiable in some structure of size $\le \lambda$. (b) Let $\Sigma$ be a set of sentences in a language of cardinality $\lambda$. If $\Sigma$ has a model, then it has a model of cardinality $\le \lambda$.

Whereas Jech in his classic tome 'Set Theory' states

LST (Jech): Every infinite model for a countable language has a countable elementary submodel. [my italics].

Now the problem I have is with Jech's use of elementary submodel. It seems to me that Jech is saying something much stronger. Enderton's version says that if a countable $\Sigma$ has a model $V$ then it has a countable model $U$. Jech on the other hand not only asserts this, but in addition states that $U$ is an elementary submodel of $V$. Surely this is a much stronger statement?


No, it isn't. And here's why! Skolem functions!

Suppose that $\cal L$ is a countable language. Then there are only countably many formulas in the language to begin with. So if $M$ is a structure in the language $\cal L$, we may assume that $\Sigma$ is the theory of $M$.

Now we close $\cal L$ under Skolem functions. Namely, for every existential formula $\exists x\varphi(x,\bar b)$ we add a function $f_\varphi(\bar b)$ and the axiom that $\exists x\varphi(x,\bar b)\rightarrow\varphi(f_\varphi(\bar b),\bar b)$. Convince yourself that this extended language is again countable, and that there is a natural way of interpreting this extended language over $M$ in such way that we do not change the way we interpreted $\cal L$ itself.

[Note: the above paragraph relies heavily on the axiom of choice, and for a good reason, too.]

Next, use the Tarski–Vaught criteria to convince yourself that a countable model of the theory of $M$ in this expanded language has an elementary embedding into $M$.

Finally, crack open a cold beer and celebrate a well-made proof.


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.