6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ You're welcome. Where did you find this question? It seems like a nice fit for my students in two weeks. $\endgroup$
    – Asaf Karagila
    May 25 '14 at 21:38
  • $\begingroup$ I actually haven't seen this question anywhere else. I am currently studying model theory and did a research project last quarter on set theory; so the question just came to mind. $\endgroup$ May 25 '14 at 21:42
  • $\begingroup$ It's a nice question. I might consider using it at some point, if that's okay. $\endgroup$
    – Asaf Karagila
    May 25 '14 at 21:46
  • $\begingroup$ That's absolutely fine. $\endgroup$ May 25 '14 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.