Given an ordinal $\alpha$, I define $F_{n}(\alpha)$ as follows:

  • $F_0(\alpha)=\alpha$
  • $F_{n+1}(\alpha)$ is the smallest $\beta$ such that no first-order $\phi$ in the language of $\{\in\}$ has $(\mathrm{V}\models\phi(S,F_0(\alpha),F_1(\alpha)...F_{n}(\alpha)))\Leftrightarrow S=\beta$
  • $F_\omega(\alpha)=\mathrm{sup}\{F_{n}(\alpha):n<\omega\}$

The existance of Reinhardt cardinals implies that $F_1(\alpha)$ exists for every sufficiently small $\alpha$. In fact, every Reinhardt cardinal is $F_1(\alpha)$ for some ordinal $\alpha$. (Yes, even though Reinhardt cardinals are not consistent with AC, they are all ordinals.)

However, this is the only information I could find on this. Could somebody else tell me what this is called if it is already named?

Is it inconsistent to say $F_n(\alpha)$ exists for $n\leq\omega$? Which $n$ is it inconsistent for?

Information gathered post-question:

  1. If there is a first-order $\phi$ such that $(\mathrm{V}\models\phi(S))\Leftrightarrow S=\alpha$, then $F_n(\alpha)=F_n(0)$ for every $n$. Here are some $\alpha$ for which this is true:
    • The smallest Erdős initial ordinal, Mahlo cardinal, Ramsey, Rowbottom, Jonsonn, Inaccessible, Strongly Inaccessible, Limit, Weakly compact, ethereal, subtle (assuming each of these exists)
    • Every other minimum initial ordinal for a first-order large cardinal axiom
    • The second of each of those axioms, the third, the $\alpha$-th for $\alpha$ on this list (assuming they exist)
    • Every $\alpha<\omega_1^{CK}$ (hint for proving this: Kleene's $\mathcal{O}$ is first-order definable)
    • Every $\aleph_{\alpha}<\aleph_{\omega_1^{CK}}$ and $\beth_{\alpha}<\beth_{\omega_1^{CK}}$(a corrolary from above)
  2. $F_\alpha(\beta)$ is countable or $\omega_1$ (assuming ZF)

Honestly I think this is the first time $\beth_{\omega_1^{CK}}$ has been used in a theorem ever. If I'm wrong, please tell me.

  • 2
    $\begingroup$ I guess that is we assume that $V_\beta \prec V$, then $F_n(\alpha)$ exists for every $n$ and every $\alpha < \beta$ (the function $F_n$ is even definable from $\beta$ in this case). If we assume further that there is a class of cardinals $\beta$ such that $V_{\beta} \prec V$ (which is equiconsistent with ZFC), we should get that $F_n(\alpha)$ exists for all $\alpha$ and $n$. Also, it is not clear to me why the existence of $F_n(\alpha)$ for all $n$ imply the existence of $F_{\omega}(\alpha)$, even if we assume that $V$ is well founded. $\endgroup$
    – Yair Hayut
    Sep 18, 2017 at 9:57
  • $\begingroup$ Aren't these just the indiscernibles for a model containing $\alpha$? $\endgroup$
    – Not Mike
    Sep 18, 2017 at 13:09
  • $\begingroup$ You're right, it does not imply the existance of $F_\omega(\alpha).$ $\endgroup$
    – Zetapology
    Sep 18, 2017 at 22:00
  • 1
    $\begingroup$ Although it's not an issue now, since a moderator has deleted the MSE question, for future reference: on the question page for any question that you have asked, below the list of tags, where you would see just "share cite edit flag" for someone else's question, you will see "share cite edit close delete flag" for your own question. Possibly you'll see some different set of options, but 'delete' will definitely be one of them. Clicking that prompts to delete the question. $\endgroup$
    – LSpice
    Sep 18, 2017 at 22:56
  • $\begingroup$ What does it mean to say that something holds "for every sufficiently small [ordinal] $\alpha$"? Does $\alpha=0$ hold for every sufficiently small ordinal $\alpha$? $\endgroup$
    – bof
    Feb 18, 2018 at 10:23

3 Answers 3


For the definition of $F_{n+1}$ to make sense, we need, in addition to the usual axiomatic apparatus of ZFC, a notion of "satisfaction of formulas in $V$." If we have this additional notion and if we allow it to occur in replacement axioms, then we can prove that $F_n(\alpha)$ exists and is countable for every $n$ and $\alpha$, and therefore $F_\omega(\alpha)$ also exists and is countable. The argument is essentially as in Zetapology's answer, with the additional apparatus replacing the "Clearly" claim (which isn't justified without some definability and some use of replacement axioms).

  • $\begingroup$ Just to be clear, the reason $F_\omega(\alpha)$ is countable is because $\aleph_1$ has cofinality $\omega_1$, meaning that the set of all $F_n(\alpha)$ for finite $n$ (which has cardinality $\aleph_0$) is not cofinal in $\omega_1$, and therefore does not have supremum $\omega_1$. Is this correct? $\endgroup$
    – Zetapology
    Sep 18, 2017 at 22:39
  • $\begingroup$ Yes, the supremum $F_\omega(\alpha)$ of a countable set $\{F_n(\alpha):n\in\omega\}$ of countable ordinals is countable. This uses the axiom of choice. I'm also using the availability of replacement for formulas involving the satisfaction predicate; without that, I couldn't guarantee that $\{F_n(\alpha):n\in\omega\}$ is a set. $\endgroup$ Sep 19, 2017 at 1:37

Let $V$ be a transitive model of $ZFC$. Without special assumptions about the model $V$, it is possible that $F_0(0)$ does not exist (in other words - it is possible that every ordinal is definable without parameters). In fact, every model of $ZFC + V=HOD$ has an elementary submodel such that all its elements are definable without parameters - for example, the Skolem closure of the empty set, using the definable Skolem functions. These models are called "Pointwise definable models".

The situation in which $F_0(0)$ exists but $F_n(0)$ does not exist for some $n$ may occur as well. For example, let $V$ be a pointwise definable model of $ZFC + V = HOD$ + there is a measurable cardinal (this is a vast overkill). Let $\kappa$ be a measurable cardinal in $V$ and let $j \colon V \to M$ be the ultrapower embedding by a normal measure on $\kappa$.

Let us claim that $\kappa$ is the first undefinable ordinal in $M$. Indeed, if $\varphi$ was a definition for $\kappa$ in $M$ then, by elementarity, $\varphi$ defines an unique ordinal in $V$, $\gamma$. But this implies that $j(\gamma) = \kappa$ which is impossible.

Every element in $M$ is of the form $j(f)(\kappa)$ for some $f\in V$. Since $f$ is definable without parameters in $V$, $j(f)$ is definable (with the same definition) in $M$. In particular, every element in $M$ is definable from the parameter $\kappa$. Thus, $F_0(0) = \kappa$ and $F_1(0)$ doesn't exist.

One can iterate this process in order to get for every $n < \omega$ a model in which $F_n(0)$ exists while $F_{n+1}(0)$ doesn't exists.

  • $\begingroup$ I thought pointwise definable models have to satisfy something like V=OD. I mean, add a Cohen real to L, then what would be a pointwise definable elementary submodel of that? $\endgroup$
    – Asaf Karagila
    Sep 19, 2017 at 17:40
  • $\begingroup$ You're right, I need to assume Global Choice in order to get that the Skolem function are definable. $\endgroup$
    – Yair Hayut
    Sep 19, 2017 at 18:12
  • $\begingroup$ Ah, so it's not probable from ZFC whether or not they exist. $\endgroup$
    – Zetapology
    Sep 19, 2017 at 22:28
  • $\begingroup$ I meant provable, this won't let me edit it for some reason $\endgroup$
    – Zetapology
    Sep 20, 2017 at 5:27

You can guarantee $F_n(\alpha)$ is countable.

Assume the contrary. There is a first-order formula for every countable ordinal $\phi$ such that $(V\models\phi(S,F_1(\alpha)...))\Leftrightarrow S=\alpha$

Clearly there is a bijection between the set of all these formulas and $\omega_1$. But using Godel numbers, this set is countable. Therefore we have a Contradiction.

  • 3
    $\begingroup$ This argument is flawed since being definable is not first order definable. See Pointwise definable models of ZFC. $\endgroup$
    – Yair Hayut
    Sep 18, 2017 at 14:26
  • $\begingroup$ I didn't say that it was though $\endgroup$
    – Zetapology
    Sep 18, 2017 at 14:27

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