Prove that any countably incomplete ultraproduct of a collection of models is $\aleph_1$-saturated

Question

I'm using this article for the proof. everything sounds well, but I don't think I have a proper comprehension on some (especially the final) parts. For example, what is the function $f$ doing at the last stages and how is it working? please give me some intuition on what is going on in the proof.

Answer 1

Recall that an ultraproduct consists of the following data: An indexing set, an $L$-structure for each $i\in I$, and an ultrafilter. Then, we can think of $\prod_D A_i = \{f:I\to \prod_{i\in I} A_i | f(i) \in A_i\}/$~ where the elements of an ultraproduct are actually equivalence classes of functions from the indexing set to the cartesian product which pick out exactly one element from each structure.

So, we need to show that $\prod_D A_i$ is $\aleph_1$ saturated. This is equivalent to demonstrating that for every countable collection of elements $K=\{f_i\}_{i\in \aleph_0} \subset \prod_D A_i$, every complete type in $S_1(K)$ is realized in $\prod_D A_i$. So, the proof uses some combinatorial data about the ultrafilter to show this is true. We construct the element $f_D \in \prod_D A_i $ to satisfy any arbitrary given 1-type in $S_1(K)$.

If the cause of the confusion is more on the technical side, I'd be happy to add more to my explanation.