I'm using this article for the proof. I thought some parts are extra and tried to make a new shorter proof. Here goes:

  • Let $\Delta(x)$ be a set of formulas (with one free-variable $x$) in the language $\mathcal L$. It suffices to show that if each finite subset of $\Delta(x)$ is realized in the ultraproduct $\prod \mathcal M _ {i}$ over the ultrafilter $D$, then $\Delta(x)$ is also realized in $\prod \mathcal M _ {i}$.

  • Now, since every finite subset of $\Delta(x)$ is realized in $\prod \mathcal M _ {i}$, we have that $\prod \mathcal M _ {i} \models \exists x \delta_{n}(x)$ for any $n\in N$.

  • So by Łoś's Theorem, for any $n\in N$ we have $X=\{i\in I:\mathcal M _ {i} \models \exists x \delta_{n}(x)\}\in D$. So, for all $i$ in $X$ we have that $\mathcal M_i \models \delta_{n}(f_i)$ for some $f_i \in \mathcal M_i$.
  • Again by Łoś, we would have $\prod \mathcal M _ {i} \models \delta_{n}(f_D)$ such that $f_D(i)=f_i$, for all $n\in N$.


If it is true (and I doubt it is!), it means that we don't need our language to be countable, or ultrafilter be countable incomplete.

Now someone please explain what's wrong with me (or the original proof!).

  • $\begingroup$ Long time ago, but if you were satisfied with my answer from below, it would be nice if you accept it (that's the common practice here on stackexchange)... $\endgroup$ – russoo Feb 25 '16 at 18:00

First of all, the assumption that the ultrafilter is countably incomplete is indeed necessary. For example, let $D$ be a principal ultrafilter on a set $I$. Then, $D$ is not countably incomplete. But if $\mathcal{M}$ is some model which is not $\aleph_1$-saturated, the ultrapower $\mathcal{M}^I/D$ cannot be $\aleph_1$-saturated since $\mathcal{M}^I/D$ is isomorphic to $\mathcal{M}$. This shows, as you expected, that your argument must be flawed.

Now, what is wrong with your argument? The mistake is in the fourth step. The problem is that the element $f_D$ depends on $n$ and the reason is that the set $X$ from step 3 already depends on $n$. So, what you can conculde from step 3 is

$\bullet$ for every $n$, there is an element $f_D$ in the ultraproduct with $\prod \mathcal{M}_i \models \delta_n(f_D)$

(Note that this can even be inferred from the assumption that $\Delta(x)$ is finitely realized). However, you conclude that for a fixed element $f_D$, we have

$\bullet$ $\prod \mathcal{M}_i \models \delta_n(f_D)$ [...] for all $n$

which is not valid in this situation.

  • $\begingroup$ Thank you. I got the flaw in my proof. However, I still believe I have no proper comprehension of the proof. For example, I still don't understand what the function $f$ is doing here (I know its final purpose, but not how it works!). May you please clarify more on what's going on in the proof? $\endgroup$ – Ak9 Dec 4 '15 at 8:31
  • $\begingroup$ Besides, by definition, to prove that the ultraproduct $\mathcal M'$ is $\aleph_1$-saturated , we must consider finite realizablity of $\Delta(x)$ on $Th(\mathcal M'_a)$ for an extended model $ \mathcal M'_a$ of $\mathcal M'$ over any countable subset $A$ of $M'$. Now why such an extended model is not considered in the proof? $\endgroup$ – Ak9 Dec 4 '15 at 8:40
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    $\begingroup$ Answer to your 2nd comment: The statments "$\Delta(x)$ is finitely realized in $\mathcal{M}'$" is equivalent to the statment "the theory $\Delta(x) \cup Th(\mathcal{M}'_a)$ is consistent (or equivalently, finitely satisfiable)". It is therefore sufficient to consider finite realizability in $\mathcal{M}'$. $\endgroup$ – russoo Dec 4 '15 at 12:31
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    $\begingroup$ Answer to your 1st comment: I think the best idea is that you make a new question where you explain in detail what parts of the proof in the paper you do not understand. If I have time and understand the proof in the paper, I can try to help you. Otherwise, I think there are a lot of people here on stackexchange who can help you. $\endgroup$ – russoo Dec 4 '15 at 12:37
  • $\begingroup$ Just (due to your first answer) if those statements are equivalent (I guess because $Th(\mathcal M'_a) = Th(\mathcal M')$), what is worthy about the choice of $A$? It won't make a difference how to choose $A$. What do you think? $\endgroup$ – Ak9 Dec 4 '15 at 13:35

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