# Chain of elementary submodels of $L_{\omega_2}$

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 $$\leq_L$$?

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

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.