# Some questions regarding (relative) constructibility and the condensation lemma

I've got a question regarding the constructible universe and I'm a bit confused about the Condensation Lemma for the universe constructible from some set $A$. Help will be greatly appreciated:

Let's assume that we have $(M,E)$ a model of ZF (or maybe of ZFC) but not necessarily standard or transitive. Since it's a model of ZF we can create its constructible universe $L^{(M,E)}$. What can we say about this model? I think it is a model of ZF both inside $(M,E)$ and outside of it. Is this correct?

What about the continuum hypothesis and the axiom of choice? To show that the actual $L$ satisfies GCH we use the Condensation Lemma which is based on the fact that $V=L$ is satisfied in every $L_\gamma$. Furthermore to show the consistency of the axiom of constructibility and of its consequences we need to show $(V=L)^L$ which is done through absoluteness. Can we do something similar in the case of the arbitrary model? For example can we say that $L^{(M,E)}$ is an actual model of ZFC+GCH? Or it is so only inside $(M,E)$?

Regarding relative constructibility: I can see how $L[A]$ is a model of ZFC and I can see how for an inner model $M$ of ZF that has $A\cap M\in M$ we have that $L[A]\subset M$ through the Gödel operations. Furthermore it appears to me that since satisfiability even if we add a set as a predicate is absolute, we can through this show the absoluteness of "$x$ is relative constructible" and thus show both that $L[A]$ is the least inner model (under the restriction mentioned above) and $(\exists X\quad V=L[X])^{L[A]}$.

My problem is with the generalized continuum hypothesis and the condensation lemma. In Jech's book it states that GCH is true in $L[A]$ above some ordinal $\alpha$. But then the Condensation Lemma is stated as:

If $\mathcal{M}\prec(L_\delta[A],\in,A\cap L_\delta[A])$ where $\delta$ is a limit ordinal, then the transitive collapse of $\mathcal{M}$ is $L_\gamma[A]$ for some $\gamma\leq\delta$.

If this is indeed how the condensation lemma generalizes then why can't we prove the GCH much like we do in the case of $L$? For every $\alpha$, taking $X\in\mathcal{P}^{L[A]}(\omega_\alpha)$ there is of course some $\delta$ such that $X\in L_\delta[A]$. Then taking the Skolem Hull of $\omega_\alpha\cup\{X\}$ we get a model $\mathcal{M}\prec(L_\delta[A],\in,A\cap L_\delta[A])$ with $|\mathcal{M}|=|\omega_\alpha|$. Its transitive collapse would be some $L_\gamma[A]$ and since $|L_\gamma[A]|=|\gamma|$ we would have that $\gamma<\omega_{\alpha+1}$. I can't find any gap in this syllogism. What am I missing?

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Apostolos: About the first question, yes. The point is $(*)$: If $(M,\dot\in)$ is a model of set theory, $\phi$ is a sentence, and $(M,\dot\in)\models$"$(N,E)$ is a model of $\phi$", then $(N,E)\models\phi$.

More precisely, $(N,E)^*\models\phi$, where $(N,E)^*$ is the model that $M$ thinks $(N,E)$ is. Here, $(N,E)^*$ has universe $\{a\in M\mid (M,\dot\in)\models a\dot\in N\}$ and its relation is $\{(a,b)\in M\times M\mid (M,\dot\in)\models a,b\dot\in N\land aE b\}$.

$(*)$ is easily established by an induction on the complexity of formulas. Now, if $\phi$ is an axiom of ZFC, then $(M,\dot\in)$ thinks that $\phi$ is an axiom of ZFC (more precisely, if $n$ is a Gödel number for $\phi$, then in $M$, the numeral corresponding to $n$ is a Gödel number for a formula, and that formula is precisely $\phi$), and therefore, $(N,E)^*$ satisfies all ZFC axioms.

Note that $M$ may fail to be an $\omega$-model, in which case there are "natural numbers" in $M$ that code what $M$ believes are ZFC axioms, and $M$ may believe that $(N,E)$ satisfies them. We do not care about these "fake formulas". Similarly, there may be an $(N,E)$ in $M$ that $M$ thinks is not a model of ZFC but $(N,E)^*$ is, in fact, a ZFC model. The reason is similar: $M$ may think that $(N,E)$ does not satisfy one of the fake axioms, but this does not matter.

In the above, it doesn't matter whether $(N,E)$ is a proper class or a set in the sense of $M$.

The argument by induction in the complexity of formulas is perfectly general, so in fact $((L,\in)^M)^*$ is a model of every $\phi$ that follows from ZFC$+V=L$. In particular, GCH, diamonds, squares, etc, hold in $(L^M)^*$. Of course, there are also additional properties this model has that are not provable from ZFC$+V=L$.

About the second question: Suppose that CH fails, and let $A$ be a subset of $\omega_2$ that codes an injective $\omega_2$-sequence of reals, say the $\alpha$-th real is coded in $A\cap(\omega\cdot\alpha,\omega\cdot(\alpha+1))$.

Then $L[A]$ cannot possibly be a model of GCH, because $A\in L[A]$ so $L[A]$ sees at last $\omega_2^V\ge\omega_2^{L[A]}$ many different reals.

As you see, the usual proof of CH breaks down because we cannot ensure that for every real there is a countable $\alpha$ such that the real belongs to $L_\alpha[A]$. In fact, if we are careful we could even have a situation where $\omega_1=\omega_1^L$, ${\mathfrak c}=\omega_2$, $A$ codes all the reals, and $L_{\omega_1}[A]=L_{\omega_1}$, i.e., none of the "new" reals are visible in $L_{\omega_1}[A]$.

The issue is that $A$ may not collapse correctly, meaning, what you do is take $\delta$ large and a countable elementary $X\prec L_\delta[A]$ that contains $r$. Then you collapse $X$. Its collapse may not be $L_\alpha[A]$ for any $\alpha$, because the collapse of $A$ needs not be $A$. Of course, an initial segment of $A$ coincides with an initial segment of its collapse, but its collapse may "code information faster", in particular, it will code $r$ by a countable stage.

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(Hope this helps. I need to run, but I can expand details later if you feel something needs clarifying.) –  Andres Caicedo Mar 8 '11 at 16:24
Thank you very much. To be sure that I got it, collapsing $\mathcal{M}$ with $\pi$ would make it some $L_\gamma[\pi(A)]$? So to get the whole condensation lemma we need to be certain that $A\cap L_\delta[A]$ will collapse to the proper part of it, namely to $A\cap L_\gamma[A]$? And in the specific case of CH that would mean that for some $X\subset\omega$ the collapse might need to contain a part of $A$ "longer than" $\omega_1$ and so the size of the collapse will not be countable? –  Apostolos Mar 9 '11 at 5:52
Yes, I think that's a way of explaining the idea. If you take $X$ sufficiently large so it contains all of $A$, you can run the argument as usual, so you obtain GCH in $L[A]$, but only from some point on. –  Andres Caicedo Mar 9 '11 at 5:56
@Apostolos: An additional note is that there is significant work in ensuring that the predicates $A$ one uses "collapse correctly". One usually starts with $A$ and produces a new $A^*$ with this property, and the process is typically referred to as "reshaping". Jensen, Sy Friedman, Shelah-Stanley, Schindler, Kanovei, and Bagaria-Kanovei are some of the authors that have worked on this area. –  Andres Caicedo Mar 12 '11 at 7:22