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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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    $\begingroup$ 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
    Commented Sep 7, 2020 at 14:35
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    $\begingroup$ If $X$ is a totally ordered set of uncountable cofinality, define (with choice) by transfinite induction $x_\alpha$ as some element $>x_\beta$ for all $\beta<\alpha$, for $\alpha\in\omega_1$. $\endgroup$
    – YCor
    Commented Sep 7, 2020 at 15:25
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    $\begingroup$ You seem to be misreading cofinality as cardinality. $\endgroup$
    – Emil Jeřábek
    Commented Sep 7, 2020 at 15:51
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    $\begingroup$ By "$X$ has uncountable cofinality" I mean "for every countable subset $Y\subset X$ there exists $x\in X$ such that $\forall y\in Y$, $y<x$". $\endgroup$
    – YCor
    Commented Sep 7, 2020 at 15:59
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    $\begingroup$ Let $(u_n)$ be a sequence of such sequences. I guess something like $u(m)=\max_{k\le m}\lceil 2^{-k}u_k(m)\rceil$ should work. $\endgroup$
    – YCor
    Commented Sep 7, 2020 at 16:52

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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? $\endgroup$
    – Alex Kruckman
    Commented Sep 16, 2020 at 16:47
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    $\begingroup$ @AlexKruckman HC means CH (as in "hypothèse du continu" and probably similar things in other languages). $\endgroup$
    – Andreas Blass
    Commented Sep 16, 2020 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
    Commented Sep 16, 2020 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
    Commented Sep 16, 2020 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
    Commented Sep 16, 2020 at 18:32

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