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First of all, I'm only starting to study independence results in set theory. And there is one obstacle that confuses me a lot. Probably such questions have already been asked, but I haven't found anything that clarifies it for me. I'm new to that subject, so please feel free to correct me if I'm using wrong terminology or reasoning.

Here is how I understand the reasoning behind proving that a statement $\varphi$ is not a theorem of a theory $T$. From mathematical logic we know that $T \not\vdash \varphi$ if and only if $\mbox{Con}(T + \neg \varphi)$. From Gödel's Completeness theorem it follows that $\mbox{Con}(T)$ if and only if $T$ has a model (that is a set with a collection of operations, relations and constants). So in order to prove something like $\mathsf{ZFC} \not\vdash \mathsf{CH}$ we should produce a model of $\mathsf{ZFC} + \neg\mathsf{CH}$. On the other hand, by Gödel's Incompleteness theorem this can't be done inside $\mathsf{ZFC}$. So all consistency results should be relative and be of the form $$\mbox{Con}(\mathsf{ZFC}) \rightarrow \mbox{Con}(\mathsf{ZFC} + \varphi).$$

As I've understood, when the method of forcing is used, we start with the assumption like $\mbox{Con}(\mathsf{ZFC})$ that gives us a (set) model $(M, \in)$ of $\mathsf{ZFC}$ from which (using Löwenheim–Skolem theorem) we can get a countable model of $\mathsf{ZFC}$. After that we extend this model to another model $(N, \in)$ that satisfies some desired statement $\varphi$ obtaining a model of $\mathsf{ZFC} + \varphi$. Such reasoning looks perfectly fine to me, since we are working with set models of $\mathsf{ZFC}$ whose existence is based on the assumption $\mbox{Con}(\mathsf{ZFC})$.

The question is about class models of $\mathsf{ZFC}$. For example, there is the Gödel's constructible universe $L$ that is a model for $\mathsf{ZFC} + V = L$ (and also for $\mathsf{AC}$ and $\mathsf{CH}$). My main question is

How can we formally deduce from the existence of such a (class) model that $$\mbox{Con}(\mathsf{ZFC} + V = L)?$$ In other words, how the existence of a class model for $T$ implies $\mbox{Con}(T)$?

Is there some kind of Gödel's completeness theorem for class models? Or again we are starting with some set model of $\mathsf{ZFC}$ and constructing $L$ inside it? If so, how this changes the proof of that $L$ is a model of $\mathsf{ZFC}$? It seems like I'm missing some basic fact (and commonly used technique) that allows to go from class models to set models. I would be very grateful if someone could explain me in details how this issue is resolved. Thanks in advance!

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    $\begingroup$ Answer: we can't. What we are actually doing is taking a set model $M$ of ZFC and then we prove that $L^M$, which is the constructible universe of $M$, is a model of $ZFC+V=L$. So this is still a result of the sort $Con(ZFC)\Rightarrow Con(ZFC+V=L)$. $\endgroup$ – Wojowu Jan 19 '16 at 14:25
  • $\begingroup$ This might be relevant, maybe even a duplicate: math.stackexchange.com/questions/557924/… $\endgroup$ – Asaf Karagila Jan 19 '16 at 14:35
  • $\begingroup$ I totally understand that all consistency results are relative (I explicitly wrote it in my question). My question is how to obtain such a result using a class model $L$? @Wojowu: Is this an obvious corollary of the fact that $L$ is a class model of $\mathsf{ZFC}$ or it requires different reasoning? $\endgroup$ – Random Jack Jan 19 '16 at 14:48
  • $\begingroup$ A class model is a set model in the meta-theory. $\endgroup$ – Asaf Karagila Jan 19 '16 at 15:22
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The theorem that $\operatorname{Con}\sf (ZF)$ implies $\operatorname{Con}\sf (ZF+\mathit{V=L})$ is a meta-theorem.

It is formulated in the meta-theory, and not internally. You are absolutely right that we can't quite formulate this internally without violating Godel's incompleteness theorem. But the nice thing is that what we can prove is that for every axiom $\varphi$ of $\sf ZFC$, $\sf ZF\vdash\varphi^L$, where $L$ is the class defined via the axiom of constructibility.

So while $\sf ZF$ does not prove that $L\models\sf ZFC$, it does prove that every axiom of $\sf ZF$ is relatively true. This tells us even a bit more than just the meta-theorem, which is great.

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  • $\begingroup$ So the statement "for every axiom of $\mathsf{ZFC}$, $\mathsf{ZF} \vdash \varphi^L$" has nothing to do with the relative consistency result "$\mbox{Con}(\mathsf{ZF})$ implies $\mbox{Con}(\mathsf{ZF} + V = L)$"? My question is how from the first statement (that every axiom of $\mathsf{ZF}$ is true relative to $L$) we can deduce the second? I've seen statements like $L$ is a model of $\mathsf{ZFC} + V = L$, hence as a corollary we have the relative consistency result. Is it an obvious corollary? $\endgroup$ – Random Jack Jan 19 '16 at 14:39
  • $\begingroup$ Well, it has something to do with it. If we assume in the meta-theory that $\sf ZF$ is consistent, we immediately get that the meta-theory cannot possibly think that $\sf ZF$ proves a false statement. So we get that $\sf ZFC+\it V=L$ is also consistent. This is stronger than saying that the proof happens entirely in the meta-theory. $\endgroup$ – Asaf Karagila Jan 19 '16 at 14:42
  • $\begingroup$ I'm sorry, may be I'm asking unclear questions, but my questions is: how do we prove that $\mathsf{ZFC} + V = L$ is also consistent? We know that there is a class model of it (namely, $L$). Assume that $\mbox{Con}(\mathsf{ZF})$, so there is a set model $(M, \in)$ of $\mathsf{ZF}$. How can we use $L$ and this model to produce a set model for $\mathsf{ZFC} + V = L$? $\endgroup$ – Random Jack Jan 19 '16 at 14:47
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    $\begingroup$ @Asaf: "If we assume in the meta-theory that ZF is consistent, we immediately get that the meta-theory cannot possibly think that ZF proves a false statement." Is that a specific property of ZF? For example, it doesn't seem to hold about ZFC: if our meta-theory is ZF+Con(ZFC)+not-AC, then the meta-theory thinks ZFC is consistent, but also thinks that ZFC proves (trivially) the statement AC which the meta-theory thinks is false ... $\endgroup$ – Henning Makholm Jan 19 '16 at 14:54
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    $\begingroup$ UGH! Everyone, calm down. I can only write so many comments at a single time! :-) $\endgroup$ – Asaf Karagila Jan 19 '16 at 14:55

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