What's wrong with my proof on "any countably incomplete ultraproduct of a collection of models is $\aleph_1$-saturated"? 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 $n\in N$.


$\dashv$
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!).
 A: 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.
