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Let $\mathbb{A} = (A, \leq^\mathbb{A})$ be a linear order without a maximal element and let $\mathbb{B} = \mathbb{A}^\mathbb{N} / \mathcal{U}$ be its ultrapower with resprect to some ultrafilter. I must prove that for every (countable) infinite sequence $(a_1, a_2, \ldots)$ of the ultrapower's elements (meaning that $a_i \in B$, not $A$) there is another element $b \in B$ such that $b \geq a_i$ for all $i$.

I generally understand ultraproduct tricks, but here I'm at a loss. My first idea was to create a set of sentences "$\forall_{x_1, x_2, \ldots, x_k} \exists_y y \geq x_1 \wedge \ldots \wedge y \geq x_k$" with increasing $k$'s and use Łoś theorem, but I don't really see how we could 'jump' with this into infinite sequences (unless the compactness theorem comes into play in a way that I can't see). My second attempt was the counting argument: there are uncountably many elements in the whole ultrapower, but only countably many in the sequence $(a_1, a_2, \ldots)$. But in fact that doesn't prove anything.

Do you have any suggestions how to tackle this problem?

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    $\begingroup$ For this to work you need that $\mathcal U$ is non-principal. In my answer I assumed that this is the case. $\endgroup$ – Stefan Mesken Feb 18 '18 at 15:51
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Let $(a_n \mid n \in \mathbb N)$ be a countable sequence in $\mathbb B$. We define $$ b \colon \mathbb N \to \mathbb A, i \mapsto \max_{\le_{\mathbb A}}\{ a_j(i) \mid j \le i \}. $$ Verify that this is a well-defined element of $\mathbb B$ and -- using Łoś $(\dagger)$ -- show that for all $n \in \mathbb N$ $$ a_n \le_{\mathbb B} b. $$


$(\dagger)$ It's here that you need $\mathcal U$ to be non-principal. (And it may be useful to note that $\mathcal U$ is non-principal if and only if it doesn't contain a finite subset of $\mathbb N$.)

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  • $\begingroup$ Thank you! Every elegant, and there I was with my silly ideas. ;) $\endgroup$ – Mike Feb 18 '18 at 16:03
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    $\begingroup$ @Mike You're very welcome. This sort of diagonal argument is useful to keep in mind when it comes to ultrapowers. Variations of the construction above are used all the time. $\endgroup$ – Stefan Mesken Feb 18 '18 at 16:05

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