Paradox with elementary submodels of the constructible tower

[Amir Livne]

This is an argument I thought up after a class on combinatrical properties of the model \textbf{L}. Our course is about set theory, not logic, so this paradox desn't seem relevant in its context. Can you help me figure out where I got it wrong?

The constructible heirarchy of sets is a series L_{\alpha} that is defined for all ordinal numbers \alpha. The important properties for my argument are:

1. L_{\alpha} is transitive for every \alpha
2. If \alpha < \beta, then L_{\alpha}\subset L_{\beta}
3. The transitive collapse (aka Montowski collapse) of every elementary submodel M \prec L_{\alpha} is L_{\beta} for some \beta
4. L_{\omega_{1}} satisfies "every set is countable" and L_{\omega_{2}} does not
5. L_{\alpha} is coutable iff \alpha is countable

So, we take an countable elementary submodel (CESM) M_{1} \prec L_{\omega_{1}}, and look at its transitive collapse, L_{\alpha_{1}} for some countable \alpha_{1}. We then take an CESM M_{2} \prec L_{\omega_{2}} that contains L_{\alpha_{2}}, and collapse it to get L_{\alpha_{2}} with countable \alpha_{2}. Then the same procedure yields a model L_{\alpha_{3}} \supset L_{\alpha_{2}} that has an elementary embedding into L_{\omega_{1}}. We generate an infinite series, switching between modelling L_{\omega_{1}} and L_{\omega_{2}}.

The limit L_{\alpha}=L_{\lim_{n<\omega}\alpha_{n}}=\bigcup_{n<\omega}L_{\alpha_{n}} is then the union of both subseries \{L_{\alpha_{n}}\}_{n=1,3,\ldots} and \{L_{\alpha_{n}}\}_{n=2,4,\ldots}. But a union of a series of elementary submodels is itself an elementary submodel, since it is a direct limit. In particular L_{\alpha} should be elementary equivalent to both L_{\omega_{1}} and L_{\omega_{2}}. This is impossible because of property (4), namely there is a statement true in one and not in another.

Where did I go wrong in my reasoning? All kinds of tips are appreciated...

 

[AKG]

"The union of elementary submodels is itself an elementary submodels" is only true when those submodels are elementary substructures of one another. This means more than just the fact that L_{\alpha_n} \subset L_{\alpha_{n+2}} and L_{\alpha_n} \equiv L_{\alpha_{n+2}}, it requires that the inclusion map is the elementary embedding. This would be the case if the L_{\alpha_n} were elementary substructures of their respective L_{\omega_i}, but they're generally not. You do get that they're elementarily equivalent subsets of their respective L_{\omega_i}, but the inclusion maps are not typically the elementary embeddings. The embeddings here are the inverses of the Mostowski collapses.

Nothing I said above proves that the inclusion maps aren't elementary embeddings, although nothing you said proves that they are. But here's why they're definitely not. Let's take for example L_{\omega_2}. For the same that GCH holds in L, we know that L_{\omega_2} thinks there is exactly one cardinal after \omega, and that this cardinal is \omega_1. If L_{\alpha} \equiv L_{\omega_2} with \alpha countable, then this structure will also think there's a unique cardinal after \omega, but it won't think \omega_1 is it.

 

[Amir Livne]

I see what you mean...

So, it seems that for every ordinal \alpha, the set \{\delta < \omega_{1} \mid L_{\delta} \prec L_{\alpha}\} is closed w.r.t taking limits. I thought about it some more and it's not hard to see this set is unbounded for \alpha = \omega_{1}, since for each \beta < \omega_{1} you can find find some L_{\beta'} with \beta<\beta'<\omega_{1} that is closed w.r.t Skolem functions of L_{\omega_{1}}. This natually means that the set of δs can't be unbounded for |\alpha|>\aleph_{1}.

 

[AKG]

I see what you mean...

So, it seems that for every ordinal \alpha, the set \{\delta < \omega_{1} \mid L_{\delta} \prec L_{\alpha}\} is closed w.r.t taking limits. I thought about it some more and it's not hard to see this set is unbounded for \alpha = \omega_{1}, since for each \beta < \omega_{1} you can find find some L_{\beta'} with \beta<\beta'<\omega_{1} that is closed w.r.t Skolem functions of L_{\omega_{1}}.

Correct. Another simple argument is to do the following construction:

• X_0 = Hull(\{\beta\},L_{\alpha})
• \beta_0 = \min\{\gamma : X_0 \subset L_\gamma\}
• X_{n+1} = Hull(L_{\beta_n},L_\alpha)
• \beta_{n+1} = \min\{\gamma : X_n \subset L_\gamma\}
• L_{\beta'} = \bigcup X_n

Here Hull(X,M) denotes some Skolem Hull of the set X in the structure M. Since we're working with L, there's a canonical choice for such hull.

This natually means that the set of δs can't be unbounded for |\alpha|>\aleph_{1}.

Indeed, the set of such \delta's is empty!