Two equivalent statements of Lowenheim-Skolem 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?
 A: 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.
