Let $\mathfrak{t}$ be the least ordinal such that $L_{\mathfrak{t}}$ has undefinable ordinals; i.e. there is an $\alpha<\mathfrak{t}$ such that $L_{\mathfrak{t}}$ cannot define $\alpha$.

This ordinal is quite large, but may have countable bounds under certain conditions. Because this ordinal is at most $\omega_1$ ($L_{\omega_1}$ is an uncountable structure and therefore must have undefinable sets in every uncountable subset, like $\omega_1$), it is definitely definable because its definition only quantifies over formulae bounded in $L_{\alpha}$, which are overall bounded in $L_{\omega_1}$.

Here's some data on it:

  • It must be a limit ordinal; $L_{\alpha+1}$ defines $\alpha$ (as the largest ordinal) and thus $L_\alpha$ allowing for statements like "$\varphi^{L_\alpha}(\beta)$". Therefore, if $\beta\in\alpha$ is definable in $L_\alpha$ by the formula $\varphi$, then $\beta$ is definable by $\varphi^{L_\alpha}$.
  • It must be larger than $\omega$ because all finite numbers are clearly definable in $L_\omega=V_\omega$
  • It must be larger than $\zeta$ (the supremum of eventually writable ordinals) because:
    1. Every turing machine is definably encoded as a natural number in $L_\omega$
    2. If $\alpha>\omega$, then $L_\alpha$ can define $L_\omega$
    3. Klev's $\mathcal{O}^{++}$ is therefore definable in $L_\alpha$ for $\alpha>\omega$ on any $n$ such that $\mathcal{O}^{++}(n)\in L_\alpha$
    4. If every ordinal below $\mathfrak{t}$ is eventually writable, then for every $\alpha<\mathfrak{t}$, $\alpha$ can be defined in $L_{\mathfrak{t}}$ as $\mathcal{O}^{++}(n)$ for some $n$
    5. That's a contradiction, so there is some ordinal below $\mathfrak{t}$ which is not eventually writable, one of which must therefore be $\zeta$
  • If there is some countable $\alpha$ such that $\mathcal{P}(L_\alpha)\cap L\models\text{MK}$, for example an $L$-inaccessible $\alpha$, then $\alpha$ is an upper bound to $\mathfrak{t}$ (that is, $L_\alpha$ has some undefinable ordinals). This is because:
    1. If $\mathcal{P}(L_\alpha)\cap L\models\text{MK}$, then $\mathcal{P}^{L}(L_\alpha)\models\text{MK}$. Therefore $L\models\mathcal{P}(L_\alpha)\models\text{MK}$.
    2. (Working in $L$ for this bullet) Because $\text{MK}$ proves that $V$ has undefinables in every uncountable subclass, $\mathcal{P}(L_\alpha)$ also satisfies that $V$ has undefinables in every uncountable subclass.
    3. (Still working in $L$) $V^{\mathcal{P}(L_\alpha)}=L_\alpha$, and therefore in every $L_\alpha$-uncountable subset of $L_\alpha$, for example $\omega_1^{L_\alpha}$, there are $L_\alpha$-undefinables.
    4. (No longer working in $L$) $L_\alpha^L=L_\alpha$, so $\omega_1^{L_\alpha}$ has some $L_\alpha$-undefinables, and therefore $\mathfrak{t}$ is smaller than $\alpha$.

EDIT: Thanks to user Miha Habič (you should check out some of his brilliant work if you haven't already), we know that this ordinal is for sure countable. Any countable ordinal $\alpha$ which is larger than an $L_{\omega_1}$-undefinable ordinal with $L_\alpha\prec L_{\omega_1}$ is clearly an upper bound.

So here's the question, just how big is it?

Is this ordinal admissible? Computably inaccessible? Computably mahlo? Just how big is it? Is it larger than $\Sigma$, the least ordinal which is not accidentally writable?

  • 9
    $\begingroup$ Your ordinal will definitely be countable. If $\alpha$ is an ordinal not definable in $L_{\omega_1}$, then you can find a countable $\beta>\alpha$ such that $L_\beta\prec L_{\omega_1}$. It follows that $\alpha$ is already not definable in $L_\beta$. $\endgroup$ Aug 31, 2018 at 11:47
  • $\begingroup$ Oh, duh. I should have thought about that. Also, I should edit to remove "indiscernables" from the question because that is wrong. $\endgroup$ Aug 31, 2018 at 14:24
  • $\begingroup$ Miha, you should post your comment as an answer. $\endgroup$ Sep 1, 2018 at 1:54
  • $\begingroup$ Perhaps $\mathfrak{t}$ is the least $\beta$ such that $L_\beta\prec L_{\omega_1}$ where $\beta>\alpha$ for some $L_{\omega_1}$-undefinable ordinal. $\endgroup$ Sep 1, 2018 at 2:37
  • $\begingroup$ Somewhat related: mathoverflow.net/q/278033/17064 $\endgroup$
    – Gro-Tsen
    Sep 4, 2018 at 12:16

1 Answer 1


${\mathfrak t}$ is the least $\beta$ such that there is a $\gamma<\beta$ with $L_\gamma \prec L_\beta$. That ${\mathfrak t} \leq$ the least such $\beta$ is obvious. On the other hand, if $X \subset L_{\mathfrak t}$ is $\subseteq$-least with $X \prec L_{\mathfrak t}$, then $X \not= L_{\mathfrak t}$; hence if $\sigma \colon L_\gamma \cong X$, then either $\sigma$ is the identity and thus $\gamma<{\mathfrak t}$ (as desired) or $\sigma$ is not the identity, in which case $L_{{\rm crit}(\sigma)} \prec L_{{\sigma}({\rm crit}(\sigma))}$, so that ${\mathfrak t} \leq \sigma({\rm crit}(\sigma)) < {\mathfrak t}$ by the other direction (giving a contradiction). In particular, $L_{\mathfrak t}$ is a model of ZFC${}^-$ (ZFC w/o the power set axiom), in fact of ZFC${}^-$ plus ``every set is at most countable,'' but ${\mathfrak t}$ is not the least $\beta$ such that $L_\beta$ is a model of ZFC${}^-$ and it is much bigger than $\Sigma$, the supremum of the accidentally writable ordinals.

  • 1
    $\begingroup$ By $\prec$, do you mean "elementary submodel" or "$\Sigma_1$-elementary submodel"? $\endgroup$
    – Gro-Tsen
    Sep 4, 2018 at 12:15
  • 2
    $\begingroup$ "fully elementary" $\endgroup$ Sep 4, 2018 at 12:16
  • $\begingroup$ Well. The least $\gamma$ for which $L_\gamma$ is a model of ZFC also satisfies that it is a pointwise definable model. So certainly all the ordinals there are definable... $\endgroup$
    – Asaf Karagila
    Sep 4, 2018 at 12:30
  • $\begingroup$ Do you mean "assuming there are models of ZFC" because $L_{\mathcal{t}}$ exists under ZFC in every model, even in pointwise definable models. $\endgroup$ Sep 5, 2018 at 2:30
  • $\begingroup$ Thanks, Keith, this was a typo. I meant ZFC${}^-$ rather than ZFC, I edited and clarified. $\endgroup$ Sep 5, 2018 at 8:45

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