Ultraproducts of models of ZFC Let $D$ be a non-principal ultrafilter over $\mathbb{N}$. Let $A_i$ for $i\in\mathbb{N}$ be (countable) models of ZFC such that $A_i\models \mathfrak{c}=\aleph_i$. Then, what is the size of the continuum in $\prod_D A_i$ ? More specifically, what does $\prod_D A_i$ 'think' the the size of the continuum is (or what can it think the size of the continuum is on our choice of $A_i$'s)?
Note: $\prod_D A_i \not \models \mathfrak{c}=\aleph_\omega$ since this would violate Kornig's lemma. We also have that $\prod_D A_i$ thinks that there exists an injection from $\mathfrak{c}$ to $\aleph_\omega$ since each model in our sequence asserts this claim.  
 A: That's a very nice question.
Note that for every $n\in\omega$ we have some $E\in D$ such that for all $i\in E$ we have $A_i\models\mathfrak c>\aleph_n$. Therefore in the resulting model $\frak c$ must be greater than $\aleph_n$ for every standard integer $n$.
But as you said, it is also true, every that $\frak c<\aleph_\omega$. So it is impossible that internally $\prod_DA_i$ will satisfy $\frak c>\aleph_\omega$.
Luckily there's a very cute solution here. We note that since $D$ is free, it is not $\sigma$-complete, and therefore the ultraproduct is not well-founded. In particular it adds new integers. So $\prod_D A_i$ would satisfy $\frak c=\aleph_\tau$ where $\tau$ is some non-standard integer which corresponds to the identity function in the product. So while this $\tau$ is larger than all the standard integers, it is strictly smaller than $\omega$ of the ultraproduct.
Note that this is exactly the same as considering the ultraproduct of $(\Bbb N,c)$ where the $n$-th copy satisfies $c=n$. In the ultraproduct, $c$ will be a non-standard integer.
