Do these ordinals exist?

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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.