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Googology Wiki says this, concerning the relation between fast-growing hierarchies defined for all countable ordinals, and the existence of a system of assigning a canonical fundamental sequence to each countable limit ordinal:

It is an open problem whether an ordinal hierarchy can be defined without using fundamental sequences. More explicitly, the problem concerns whether there is a model of $ZF$ such that there exists an $F : \omega_1 \rightarrow (\mathbb{N} \rightarrow \mathbb{N})$ where for all $\alpha > \beta, F(\alpha)$ eventually outgrows $F(\beta)$, but there does not exist an $S: \omega_1 \cap \text{Lim} \rightarrow (\mathbb{N} \rightarrow \omega_1)$ such that for all $\alpha, \sup(R) = \alpha$ where $R$ is the range of $S(\alpha)$.

My question is, is it true that the question of whether there exists such a model of $ZF$ is an open problem?

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  • $\begingroup$ Can't you just define F in L? If F is valid in L, I think it would also be valid in V, right? $\endgroup$
    – Christopher King
    Commented Nov 28, 2019 at 21:43
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    $\begingroup$ @PyRulez (half a year late, sorry!) Defining $F$ in $L$, the constructible universe, needn't work, because $\omega_1^L$, what $L$ believes to be the least uncountable ordinal, might be countable in $V$. So this would only define a hierarchy up to $\omega_1^L$ which needn't be equal to $\omega_1$. $\endgroup$
    – Wojowu
    Commented Jun 13, 2020 at 13:53

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