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:
- Every turing machine is definably encoded as a natural number in $L_\omega$
- If $\alpha>\omega$, then $L_\alpha$ can define $L_\omega$
- Klev's $\mathcal{O}^{++}$ is therefore definable in $L_\alpha$ for $\alpha>\omega$ on any $n$ such that $\mathcal{O}^{++}(n)\in L_\alpha$
- 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 $n$
- 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:
- If $\mathcal{P}(L_\alpha)\cap L\models\text{MK}$, then $\mathcal{P}^{L}(L_\alpha)\models\text{MK}$. Therefore $L\models\mathcal{P}(L_\alpha)\models\text{MK}$.
- (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.
- (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.
- (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?

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