Show that $\{\alpha<\omega_1 : L_\alpha \prec L_{\omega_1}\}$ is closed unbounded in $\omega_1$.

Question

I was doing this exercise and there is a hint to consider the Skolem functions for $L_{\omega_1}$. However, I did not find any general definition of what a Skolem function may be in Kunen (1980), and I do not know how to proceed with this hint. Any help in clarification or hint in another directions would be appreciated. Thank you.

Answer 1

You can use Skolem functions to do this, but they're not necessary. Here's an argument I like:

First, I claim that it's enough to show unboundedness. We get this by applying two results, namely the coherence of elementary embeddings ($A\preccurlyeq C$, $B\preccurlyeq C$, and $A\subseteq B$ implies $A\preccurlyeq B$), and the nice behavior of elementary chains (if $A_i\preccurlyeq A_j\preccurlyeq C$ for $i<j<\lambda$ with $\lambda$ a limit ordinal then $A_i\preccurlyeq \bigcup_{i<\lambda}A_i\preccurlyeq C$ for all $i<\lambda$). Putting this all together we get that if $\alpha$ is a limit of $\{\beta: L_\beta\preccurlyeq L_{\omega_1}\}$ then $\alpha$ is in fact in that set.

Moreover, unboundedness itself is really all about constructing a single example with a parameter: it's enough to show that for each $a\in L_{\omega_1}$ there is an $\alpha$ such that $a\in L_\alpha\preccurlyeq L_{\omega_1}$. (Do you see why?)

So fix $a\in L_{\omega_1}$. By downward Lowenheim-Skolem we get an $M\preccurlyeq L_{\omega_1}$ with $a\in M$. Of course $M$ probably is not a level of $L$, so we aren't done yet. Instead we're going to build a sequence of structures:

• $M_0=M$.

• $M_{2i+1}=L_{M_{2i}\cap \mathsf{Ord}}$.

• $M_{2i+2}$ is any countable elementary substructure $N$ of $L_{\omega_1}$ such that $M_{2i+1}\subseteq N$.

Basically, at odd steps we "fill in the gaps" and at even steps we "re-attain elementarity." Crucially, the even steps require the observation that $L_\alpha$ is countable whenever $\alpha$ is; this is where this argument would break down if we tried to run it for the $V$-hierarchy instead. It's a good exercise to check that no $V_\alpha$ is a proper elementary substructure of $V_{\omega_1}$:

Already in $V_{\omega+1}$ we have elements "coding" all the countable ordinals, so $V_{\omega_1}$ will be the first level of the $V$-hierarchy after $\omega+1$ to satisfy "every countable well-order is isomorphic to some ordinal."

Now let $M_\omega=\bigcup_{i\in\omega}M_i$. We clearly have that $M_\omega$ is a countable substructure of $L_{\omega_1}$ containing $a$ as an element; moreover, as a union of levels of $L$ it is itself a level of $L$, and as a union of elementary substructures of $L_{\omega_1}$ it is itself an elementary substructure of $L_{\omega_1}$. So we're done!

Answer 2

For a route more similar to what Kunen is suggesting, proceed like the proof of the reflection theorem.

Let $\alpha < \omega_1.$ For a formula $\varphi(x,\vec y)$ and $\vec y \in L_\alpha,$ let $F(\alpha, \varphi, \vec y)$ be the least ordinal $\beta >\alpha$ such that there is an $x\in L_\beta$ with $L_{\omega_1}\models \varphi(x,\vec y)$ if there is any such $x\in L_{\omega_1},$ otherwise let $F(\alpha, \varphi, \vec y)=0.$ Define $$\alpha' = \sup\{F(\alpha, \varphi,\vec y): \varphi\in Form, \vec y\in L_\alpha\}.$$ Since there are countably many $\vec y$ and $\varphi,$ we have $\alpha'<\omega_1.$

Then, given an arbitrary $\alpha< \omega_1,$ let $\alpha_0 = \alpha$ and recursively let $\alpha_{n+1}= \alpha_n'$ and $\alpha^* = \lim_n \alpha_n.$

By the Tarski-Vaught test, $L_{\alpha^*}\prec L_{\omega_1}.$ Thus, the set in question is unbounded. Closure follows easily, as Noah mentions in his answer, since an increasing union of elementary submodels is an elementary submodel.