# End elementary extension in infinitary logic of some $L_\alpha$ producing a $L_\beta$

Let $$L_\alpha$$ be some admissible level of the constructible hierarchy and $$M \supseteq L_\alpha$$ an extension of $$L_\alpha$$. I am looking for conditions under which $$M \simeq L_\beta$$. It is not enough to ask that $$M$$ be an end elementary extension of $$L_\alpha$$ as $$M$$ may not be well-founded.

However I suspect that this may be provable when $$M$$ is an $$\mathcal{A}$$-elementary end extension of $$L_\alpha$$ for some fragment $$\mathcal{A}$$ of infinitary logic $$L_{\omega_1 \omega}$$. In the general case, well-orderdness is not definable in $$L_{\omega_1 \omega}$$ but here we also have the structure of $$L_\alpha$$ to help us.

The idea is to do as follows to show that $$M$$ is well-founded:

1. In $$M$$ we can define ordinals as hereditarly transitive sets and they are well-founded by foundation.
2. Then we can define by $$\Sigma$$-recusion (here we need $$\alpha$$ to be admissible and likely even $$L_\alpha$$ to be a model for infinitary $$KP$$) a predicate $$L(x, \alpha)$$ defined as: \begin{align*} L(x, \alpha+1) &\longleftrightarrow \bigvee_{n \in \omega} \bigvee_{\varphi} \exists p_1, \ldots, p_n \, \forall y \in x (L(y, \alpha) \wedge \varphi(y, p_1, \ldots, p_n)) \\ L(x, \alpha) &\longleftrightarrow \exists \beta \in \alpha \, L(x, \beta) \text{ for } \alpha \text{ limit} \end{align*} With it, we define $$L_\alpha = \left\{ x \mid L(x, \alpha) \right\}$$.
3. With this definition, the two following infinitary sentences are true in $$L_\alpha$$ (observe that the second sentence is not usually true when $$L(x, \alpha)$$ is defined in the finite setting using rudimentary functions instead of the definibality predicate) : \begin{align*} \begin{cases} \forall x \, \exists \alpha \, x \in L_\alpha \\ \forall \alpha \, (x \in L_\alpha \implies \forall y \in x \, \exists \beta \in\alpha \, y \in L_\beta) \end{cases} \end{align*}
4. Those two sentences can be reflectd in $$M$$ by $$\mathcal{A}$$-elementarity and with those, any infinite decreasing sequence $$a_0 \ni^M a_1 \ni^M a_2 \ldots$$ would yield an infinite decreasing sequence of ordinals $$\alpha_0 \ni^M \alpha_1 \ni^M \alpha_2 \ldots$$ which in turn would contradict the fact that ordinals are well-founded w.r.t. $$\in^M$$.

Now: do we need $$L_\alpha$$ to be a model for $$KP$$ for infinitary logic for the definition by recursion? And is there a more direct argument to show this?

• If $\mathcal A$ is a countable fragment and $\alpha$ is a countable ordinal, then by $\mathbf{\Sigma}^1_1$-boundedness, either there is an illfounded $\mathcal A$-elementary end extension of $L_\alpha$ or the $\beta < \omega_1$ such that $L_\beta$ is an $\mathcal A$-elementary extensions of $L_\alpha$ are bounded below $\omega_1$. So it seems there must be something wrong with your argument (or mine, I guess). I haven't looked closely at your argument though. Commented May 18, 2023 at 16:10
• Should the definition of $L(x,\alpha+1)$ instead be something along the lines of $L(x,\alpha+1)\leftrightarrow\bigvee_{n\in\omega}\bigvee_{\varphi}\exists p_1,\ldots,p_n \, \forall y(y\in x\leftrightarrow (L(y,\alpha)\land\varphi(y,p_1,\ldots,p_n)))$, with a $L(p_1,\alpha)\land\ldots\land L(p_n,\alpha)$ restriction and comprehension for $y$? As is currently written, if $L(x,\alpha+1)$ there does not seem to be anything forbidding $L(x',\alpha+1)$ where $x'$ is an arbitrary subset of $x$.
– C7X
Commented Feb 15 at 10:25
• Yes you're right. Which makes the definition actually $\Sigma_2$ and so we can't use $\Sigma$-recursion. Thanks, that explains the contradiction with Gabe Goldberg comment. Commented Feb 21 at 18:16
• @Johan I was also going to write an answer mentioning that well-foundedness is not expressible in $L_{\omega_1\omega}$ (according to this comment it is not even expressible in $L_{\infty\omega}$) and attempting to produce an $M$ whose ordinals are non-well-founded, however if $\Sigma$-recursion does not apply, is it still worth writing?
– C7X
Commented Mar 4 at 3:13
• Hi @C7X, sorry I forgot to answer your last comment. I think the error you pointed out plus the theoretical limitation that Gabe Goldbard suggested already builds a good case against what I was trying to do. Maybe you could add their observation in your answer for the sake of completeness? Commented Mar 21 at 10:09

The definition of $$L(x,\alpha+1)$$ was wrong, instead it should have been $$L(x,\alpha+1)\leftrightarrow\bigvee_{n\in\omega}\bigvee_{\varphi}\exists p_1,\ldots,p_n \, \forall y(y\in x\leftrightarrow (L(y,\alpha)\land\varphi(y,p_1,\ldots,p_n)))$$, and as this is not $$\Sigma_1$$, $$\Sigma$$-recursion is not applicable with the corrected definition.
Additionally, from step #1 on, the argument is dependent on stating an axiom of foundation in $$\mathcal L_{\omega_1\omega}$$ such that $$M$$ satisfying the axiom means that even just the ordinals of $$M$$ are externally well-ordered. But Karp ("Finite-Quantifier Equivalence", in The Theory of Models, J. Addison, L. Henkin, and A. Tarski (eds.), 1965) and Lopez-Escobar ("On Defining Well-Orderings, Fundamenta Mathematicae vol. 59, 1966) showed that external well-orderedness is not definable in $$\mathcal L_{\omega_1\omega}$$ nor in $$\mathcal L_{\kappa\omega}$$ for any cardinal $$\kappa$$.