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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 attempted to leave hidden avoid spoilers, if anyone knows how to fix it please do!):

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 $\varphi$).

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 $\varphi$).

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 attempted to leave hidden avoid spoilers, if anyone knows how to fix it please do!):

! 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 defintion ($\varphi$ is true in $L_{\omega_1^L}$ iff the unique element of $X$ contains $\varphi$).

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 attempted to leave hidden avoid spoilers, if anyone knows how to fix it please do!):

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 $\varphi$).

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 $\beta\in S$.

3. There is an unbounded set $X\subseteq \omega_1$ such that $L_\alpha\prec L_\beta$ for all $\beta\in S$.

In some sense, if $L_\alpha\not\prec L_{\omega_1}$, then there is a limit to how many times we can elementarily extend $L_\alpha$ within the $L$-hierarchy. (But you have to be careful here, since we can have an infinite ascending $\prec$-chain of countable $L$-levels not elementarily embedded into $L_{\omega_1}$.)

The implication from 1 to 2 is via dLS+Condensation directly (just build the club) together with the coherence property of $\prec$, and the implication from 2 to 3 is trivial since clubs by definition are unbounded. The direction 3 to 1 is a bit trickier. Note that we cannot directly use elementary chains, since given $X$ appropriate there is no guarantee that $L_\beta\prec L_\gamma$ for $\beta<\gamma$ elements of $X$. Instead we have to use definable Skolem functions, which each level of the $L$ hierarchy has (uniformly, even).

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!). 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 defintion ($\varphi$ is true in $L_{\omega_1^L}$ iff the unique element of $X$ contains $\varphi$).

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 attempted to leave hidden avoid spoilers, if anyone knows how to fix it please do!):

! 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 defintion ($\varphi$ is true in $L_{\omega_1^L}$ iff the unique element of $X$ contains $\varphi$).