I have a chain of elementary submodels defined inductively in the following way.

Pick $N_0\prec L_{\omega_2}$ s.t $|N_0|=\aleph_0$ and $N_0$ is minimal in $\leq_L$. For every $\alpha<\omega_1$ let $N_{\alpha+1}\prec L_{\omega_2}$ be the minimal (with respect to $\leq_L$) elementary submodel s.t $N_\alpha \in N_{\alpha+1}$ and $|N_{\alpha+1}|=\aleph_0$. For $\alpha<\omega_1$ a limit let $N_\alpha = \bigcup_{\beta<\alpha}N_\beta$.

I want to show $N_\alpha \subseteq N_{\alpha+1}$.

In a previous discussion done in this question I asked I understood the argument should be something similar to

Let $\alpha<\omega_1$ be some ordinal, from construction $|N_\alpha|=\aleph_0$ hence $\exists f:\omega \rightarrow N_\alpha$ a bijection using the $L$ global well order

$$L_{\omega_2}\models f \text{ is the} \leq_L \text{minimal bijection from } \omega \text{ to } N_\alpha$$

$N_{\alpha+1} \prec L_{\omega_2}$ and $N_\alpha,\omega\in N_{\alpha+1}$ thus

$$N_{\alpha+1}\models f \text{ is the} \leq_L \text{minimal bijection from } \omega \text{ to } N_\alpha$$

Hence $\forall n<\omega,\ f(n)\in N_{\alpha+1}$, $f$ is bijective giving $N_\alpha \subseteq N_{\alpha+1}$.

Doesn't the transition to $N_{\alpha+1}$ affects the chosen function? Is it because $L_{\omega_2}$ thinks $N_\alpha$ is $\aleph_0$ so $N_{\alpha+1}$ thinks it too? Or $f$ is described uniquely using a single ordinal via $\leq_L$?

  • $\begingroup$ Yes, the transition does not affect $f$. The reason is that $f$ is definable and $N_{\alpha+1}$ is an elementary submodel of $L_{\omega_2}$. $\endgroup$
    – Hanul Jeon
    Jan 24, 2021 at 1:54

1 Answer 1


This is really an instance of a more general fact.

Suppose $\kappa$ is uncountable and regular and $M\prec H_\kappa$ is an elementary submodel. If $p\in M$ and $p$ is countable, then $p\subseteq M$.

To start, notice that for uncountable regular $\kappa$, $H_\kappa$ gets countability facts correctly. That is, if $p$ is countable, then $H_\kappa\vDash \exists f:\omega\to p \text{ surjective}$. Since $\omega, p\in M$, this sentence is true in $M$ as well. We now show that for any $q\in p$, we have also $q\in M$.

First fix $f_p\in M$ such that $M\vDash f_p:\omega\to p \text{ is surjective}$. Since $M$ thinks $f_p$ is a function, by elementary $H_\kappa$ thinks $f_p$ is a function, so $f_p$ is really a function. Now let $q\in p$ be arbitrary. We want to show that $q$ is an element of $M$. In $V$, we know that there is some $n$ such that $f_p(n)=q$. Also, for each $n\in\omega$, $n\in M$ by definability of the natural numbers. So we can ask for what $M$ thinks is $f_p(n)$. I claim that this object is $q$.

To see this, notice that $H_\kappa\vDash (\exists! x)(x\in p\wedge x=f_p(n))$. By elementarity, we have $M \vDash (\exists! x)(x\in p\wedge x=f_p(n))$. Now if $M$ is transitive then we are done: the unique $x$ is just $q$. But $M$ might not be transitive. We work around this obstacle by appealing to elementarity.

Fix the element $z\in M$ such that $M \vDash z \text{ is the unique element in } p \text{ such that } z=f_p(n))$. Passing this up to $H_\kappa$ by elementarity, we have $H_\kappa\vDash z \text{ is the unique element in } p \text{ such that } z=f_p(n))$. But then $H_\kappa$ also thinks that $q$ is the unique element satisfying the description. Hence $H_\kappa\vDash q=z$, and so $q$ and $z$ are really identical. This proves $q\in M$.

As to the question: notice that we have in general $(L_\kappa=H_\kappa)^L$. This means the general fact applies in this case, if we apply it in $L$. So just take $N_{\alpha+1}$ to be $M$ and take $p$ to be $N_\alpha$ and apply the fact.


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