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Let $\mathcal{F}$ be a non-principal ultrafilter on $\omega$. Let $^*\mathbb{N}$ = $\mathbb{N}^\omega/\mathcal{F}$ be an ultrapower. Let $\{n_\alpha\}_{\alpha\in\omega_1}$ be a strictly increasing sequence in $^*\mathbb{N}$.

Assuming ZFC, would this sequence be unbounded in $^*\mathbb{N}$, i.e. $\forall n \in {^*\mathbb{N}}\ \exists \alpha \in \omega_1\ n_\alpha > n $? Does it depend on CH?

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  • $\begingroup$ I don't think (2) will play any role (if $u$ satisfies (1), it's easy to produce $u'$ satisfying (1) and (2), and such that $u$ is bounded iff $u'$ is bounded). $\endgroup$
    – YCor
    Sep 7 '20 at 14:29
  • $\begingroup$ The set of $n\in{^*\mathbb{N}}$ such that $\lim_{m\to\omega} n(m)/m=0$ is convex but has uncountable cofinality. So there exists $(n_\alpha)$ satisfying (1) such that each $n_\alpha$ is less than the sequence $m\mapsto m$. $\endgroup$
    – YCor
    Sep 7 '20 at 14:35
  • $\begingroup$ The question seems to be whether every sequence is unbounded, which is purportedly negatively answered here. Isn't the question rather whether some sequence is unbounded (which would better fit the title)? $\endgroup$
    – YCor
    Sep 7 '20 at 14:39
  • $\begingroup$ @Ycor thank you for your first note. I removed the redundant condition. But how uncountable cofinality implies the existence of strictly increasing $\omega_1$ sequence? $\endgroup$ Sep 7 '20 at 15:23
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    $\begingroup$ You seem to be misreading cofinality as cardinality. $\endgroup$ Sep 7 '20 at 15:51
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If HC (continuum hypothesis in French) holds, then some of those sequences are cofinal whereas some are not.

Indeed, HC implies that the corresponding ultrapower$\ ^*\mathbb{R}$ of $\mathbb{R}$ is a saturated ordered field with cardinal $\omega_1$. This is the same for the field $\mathbf{No}(\omega_1)$ of surreal numbers with countable birthday. So they are isomorphic.

In $\mathbf{No}(\omega_1)$, there is a strictly increasing and cofinal embedding $x \mapsto \omega^x: \mathbf{No}(\omega_1) \rightarrow \mathbf{No}(\omega_1)^{>0}$ which satisfies in particular $\forall x,y(0\leq x<y\Longrightarrow \omega^x+1<\omega^y)$. So taking integer parts in $^*\mathbb{N}$, we obtain a cofinal order embedding $ x \mapsto \left\lfloor \omega^x \right\rfloor: \mathbf{No}(\omega_1)^{\geq 0} \longrightarrow\ ^*\mathbb{N}$. Since there are copies of $\omega_1$ in $\mathbf{No}(\omega_1)^{\geq 0}$ which are cofinal, and others which are bounded, this yields cofinal and bounded $\omega_1$-sequences in$\ ^*\mathbb{N}$.

In ZFC, I think (but I am not sure) that it is consistent that the cofinality of$\ ^*\mathbb{N}$ be $\omega_2$, meaning that each $\omega_1$-sequence would be bounded.

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  • $\begingroup$ What does it mean to say HC holds? $\endgroup$ Sep 16 '20 at 16:47
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    $\begingroup$ @AlexKruckman HC means CH (as in "hypothèse du continu" and probably similar things in other languages). $\endgroup$ Sep 16 '20 at 16:53
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    $\begingroup$ @AlexKruckman I didn't even realize I wrote HC, which is indeed the french acronym for the continuum hypothesis. $\endgroup$
    – nombre
    Sep 16 '20 at 17:46
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    $\begingroup$ "each $\omega_1$-sequence is bounded" literally means that the cofinality of $^*\mathbf{N}$ is $\ge\omega_2$. $\endgroup$
    – YCor
    Sep 16 '20 at 18:04
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    $\begingroup$ You might get information about cofinality of $^*\mathbf{N}$ from projecteuclid.org/download/pdf_1/euclid.ndjfl/1093635237 $\endgroup$
    – YCor
    Sep 16 '20 at 18:32

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