Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks in advance,

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
(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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.