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By the downward Lowenheim-Skölem theorem we can find two countable ordinals $\alpha < \beta$ such that $L_\alpha \prec L_{\omega_1}$ and $L_\beta \prec L_{\omega_1}$. That is, $L_\alpha$ and $L_\beta$ are elementary submodels of $L_{\omega_1}$. Consequently, $L_\alpha \prec L_\beta$.

Hence my question, the other way around:

Given countable ordinals $\alpha$ and $\beta$ such that $L_\alpha \prec L_\beta$, do we always have $L_\alpha \prec L_{\omega_1}$ ? If not, how do construct/prove the existence of such a $\alpha$ and $\beta$?

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3 Answers 3

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Very clearly not. Take some countable elementary submodel $M_0$ of $L_{\omega_2}$, and take $M_1$ to be another one, but with $M_1$ a end extension of $M_0$. We can find such models by first finding two uncountable $\gamma<\delta$ such that $L_\gamma\prec L_\delta\prec L_{\omega_2}$, and then taking $M_1$ be a countable elementary submodel of $L_\delta$ with $L_\gamma$ added as a predicate to the language, and then letting $M_0$ be $M_1\cap L_\gamma$.

Now, since both are well-founded models of enough set theory and $V=L$, their transitive collapses are some $L_\alpha$ and $L_\beta$ which are countable, and the collapses agree on $M_0$.

Consequently we have that $L_\alpha\prec L_\beta$. But it is also clear that $L_\alpha$ (and $L_\beta$) think that they have an uncountable ordinal, whereas $L_{\omega_1}$ thinks that all the ordinals are countable.

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    $\begingroup$ Since $M_1$ has $M_0$ as an element and knows that it isn't transitive, it will have new elements witnessing that, and so the Mostowski collapse of $M_1$ won't agree with the collapse of $M_0$. So you'll only get an elementary embedding $L_\alpha\to L_\beta$, not elementary substructure. Right? For example, the order type of $M_0$ will be a new countable ordinal in $M_1$ but not in $M_0$, and this is less than $\omega_1^L\in M_0$. $\endgroup$ Commented Jan 11, 2023 at 17:42
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    $\begingroup$ To make an argument like this work, you need to find $M_0\prec M_1\prec L_{\omega_2}$ with $M_1$ being an end extension of $M_0$. $\endgroup$ Commented Jan 11, 2023 at 17:49
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    $\begingroup$ Ugh. That was my first instinct, but I hedged and changed it to the one written now. I'll edit it later when I'm back with a keyboard. $\endgroup$
    – Asaf Karagila
    Commented Jan 11, 2023 at 19:07
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    $\begingroup$ You can find end-extension models as needed by first taking an uncountable $\alpha$ with $L_\alpha\prec L_\beta\prec L_{\omega_2}$, and then taking a countable elementary substructure $M_1$ of $\langle L_\beta,\in,L_\alpha\rangle$, letting $M_0$ be the part of $M_1$ below $\alpha$. These will now collapse as desired. $\endgroup$ Commented Jan 12, 2023 at 0:25
  • $\begingroup$ Thanks, Joel. I've edited. $\endgroup$
    – Asaf Karagila
    Commented Jan 12, 2023 at 10:21
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No, there are many instances of $L_\alpha\prec L_\beta$ without $L_\alpha\prec L_{\omega_1}$.

Here is one easy way to construct one. Consider the smallest $\alpha$ that has an elementary extension $L_\alpha\prec L_\beta$. There is indeed such a countable ordinal $\alpha$ as you observed, and this is visible inside $L_{\omega_1}$. So if $L_\alpha\prec L_{\omega_1}$, then there would have to be one below $\alpha$, contradicting minimality.

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    $\begingroup$ 12 seconds. Usain bolt ran a 100 meter dash between our answers, and had time to catch his breath afterwards, too! $\endgroup$
    – Asaf Karagila
    Commented Jan 11, 2023 at 15:58
  • $\begingroup$ Yes, indeed, you beat me by at least 150 meters! But we have different methods, so it is interesting... $\endgroup$ Commented Jan 11, 2023 at 15:59
  • $\begingroup$ Yes, I was glad to see that we took fairly different approaches. $\endgroup$
    – Asaf Karagila
    Commented Jan 11, 2023 at 16:00
  • $\begingroup$ Thank you for those quick answers! I guess those 12 seconds mean that I'll accept Karagila's answer. $\endgroup$
    – Johan
    Commented Jan 11, 2023 at 16:19
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As Asaf and Joel have observed, the answer to your question is negative. However, there is a sense in which being an elementary submodel of $L_{\omega_1}$ is the only way to "persistently" get elementary submodelhood relations.

Specifically, the following are equivalent:

  1. $L_\alpha\prec L_{\omega_1}$.

  2. There is a club $S\subseteq\omega_1$ such that $L_\alpha\prec L_\beta$ for all $\beta\in S$.


But on the other other hand, if $V=L$ then there is an unbounded $U\subseteq \omega_1$ such that for all $\alpha,\beta\in U$ we have $L_\alpha\equiv L_\beta\not\equiv L_{\omega_1}$ (note that this can't be proved using just a counting argument or forcing + absoluteness). This is a beautiful short application of Tarski's undefinability theorem due to Hjorth, answering question 10.4 of A. Miller. Hjorth's argument, with minor formatting edits from me, is copied below (which I've left hidden to avoid spoilers):

Let $X$ be the set of complete theories that satisfy "everything is countable" and have unboundedly many $\alpha<\omega_1^L$ with $L_\alpha$ realising them. The theory of $L_{\omega_1^L}$ is one such theory, and we will be done if we prove that there are some others. Now $X$ is a definable class in $L_{\omega_1^L}$, and so it must have some other elements or else $L_{\omega_1^L}$ would admit a truth definition ($\varphi$ is true in $L_{\omega_1^L}$ iff the unique element of $X$ contains $\varphi$).

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  • $\begingroup$ This first result might help me in some way. Do you know a reference for it? $\endgroup$
    – Johan
    Commented Jan 13, 2023 at 18:08
  • $\begingroup$ @Johan Not offhand, but maybe it's an exercise in Kunen or Jech. $\endgroup$ Commented Jan 13, 2023 at 18:39
  • $\begingroup$ Thanks, I'll have a look in those. $\endgroup$
    – Johan
    Commented Jan 13, 2023 at 18:48

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