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Even though $\mathbb{N}$ cannot be defined by first order means, it can be defined by second order means. Anyway: it can be defined, and there is no doubt, which abstract structure $\mathbb{N}$ represents.

Accordingly $\mathbb{Z}$, $\mathbb{Q}$, and $\mathbb{R}$ can be characterized unambiguously, even as purely linear orders (without addition and multiplication).

Until today I am not sure, if and how some model of some set theory can be uniquely characterized. Is for example $V = L$ enough to unambiguously distinguish one specific and unique structure among myriads of possible models of the ZF(C) axioms?

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    $\begingroup$ No. V=L settles surprisingly many questions, but not everything. And the incompleteness theorem renders the search for such first order axioms than pin down set theory completely hopeless. $\endgroup$
    – Michael Greinecker
    Aug 31, 2012 at 23:43
  • $\begingroup$ No restriction to first order axioms: anything goes! With the common number systems it works, but what about set theory? $\endgroup$
    – Hans-Peter Stricker
    Aug 31, 2012 at 23:46
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    $\begingroup$ But Hans, second order already assumes [philosophically] that you have some notion of set. $\endgroup$
    – Asaf Karagila ♦
    Aug 31, 2012 at 23:47
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    $\begingroup$ The above been said, Joel D. Hamkins recently proved a theorem stemming from a question on this very site that [quantifier free] bi-embeddability isomorphism classes of countable models of ZFC are linearly ordered, in fact well-ordered. In this aspect you can consider $V=L$ of the least member of this class. $\endgroup$
    – Asaf Karagila ♦
    Aug 31, 2012 at 23:50
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    $\begingroup$ Timothy Chow in "Forcing for dummies" (www-math.mit.edu/~tchow/mathstuff/forcingdum): "strictly speaking, nobody has ever exhibited even one model of ZFC". $\endgroup$
    – Hans-Peter Stricker
    Sep 3, 2012 at 17:39

1 Answer 1

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I gathered my comments into an answer, because there are more and more of them piling in my head. I have added a summary at the bottom as well.

  1. Note that $\mathbb Q$ is the unique countable model of DLO, an $\aleph_0$-categorical theory, and $\mathbb R$ is the unique Dedekind-completion of the unique $\mathbb Q$. This means, amongst other things that we can "canonicalize" some model. This canonical model, however, is defined within ZFC as a meta-theory. Namely, we use ZFC to prove the existence and uniqueness (up to isomorphism) of such models.

    ZFC, on the other hand, is very very far from being complete. Even if we add $V=L$, and even more axioms deciding things independent of $V=L$, this is either going to stay incomplete -- or become an "ugly" (in sense of effective enumerations) theory. With respect to the last remark, I should add that it seems less likely to canonicalize a model of ZFC when your meta-theory is ZFC itself.

  2. Allowing ourselves to extend to stronger logics, there is a philosophical obstacle. We want to canonicalize a model for ZFC without the need for a universe of ZFC to exist from the start (otherwise we have some sort of circularity issue here, see point 6).

    In the case of high-order logic we presuppose that we have a notion of a set. We already have that in mind, and talking about second-order (or more) set-theory as what you begin with seems quite peculiar to me, for this very reason.

  3. The above been said, Joel Hamkins proved a theorem stemming from a question on this site, that if we consider quantifier-free embeddings between countable models of ZFC, then the isomorphism classes are well-ordered in ordertype $\omega+1$

    We can therefore canonicalize a model by talking about the least or the last model. If one should do so, one should probably be wary of the following points:

    1. Hamkins himself is an advocate of the multiverse approach to set theory, namely there is no canonical model. Instead there are models for most, if not every, possible consistent extension of ZFC.

      Using his work to go against the multiverse approach is not very nice... :-)

    2. If we observe closely, we'll see that the embeddings are quantifier-free, and while that is something, it's not enough to say that there is some canonical model -- at least from my perspective. I feel that way because ZFC is supposed to be a foundational theory, and as such we expect some quantified formula to be passed to the canonical model from any model.

  4. Returning to the case at hand, there is work being done towards an "Ultimate L" model of ZFC, it should be a model for large cardinals and should decide a lot of the independence results. This work is not finished yet. It will have the same problems as adding $V=L$ I have pointed out in the my first point - it will not be enough to talk about a truly canonical model.

  5. If we have a canonical model of ZFC then we essentially solved all independence questions. We also eliminated the need for forcing (in general) and possibly much of the theory of large cardinals too.

    Note that if there is a large cardinal then there is a set model of ZFC in the universe. Therefore a canonical model of ZFC cannot really have large cardinals since those would imply that the canonical model house many non-canonical models too, and possibly some canonical models as well... how could that be canonical?

  6. Note that all those canonical models depend a lot on your universe of ZFC. Especially if we allow "arbitrary" properties to be taken. If there are enough large cardinals -- all projective sets of real numbers are deteremined; if not, then they are not determined.

    It means that a canonical model of ZFC would have to be a set in a larger universe of ZFC. Is the larger universe itself a canonical model? That should mean that the canonical model itself knows about a canonical model of ZFC, and that smaller canonical model ... somewhere we just hit well-foundedness quite hard.

  7. Lastly, for the layman mathematics is not based on set theory. Set theory is a mathematical theory which has a rich inner life of its own. Try to explain mechanical engineers what is the axiom of choice and they'll stare at you vacantly with expressions ranging from fear, to quizzical, to disbelief that someone actually cares about that.

    If a layman is sophisticated enough to ask me about a canonical model of ZFC then it means (usually) one of three things:

    1. This is not a layman, but rather someone that has enough mathematical knowledge to understand what a canonical model is.

    2. This is someone that read somewhere about the incompleteness theorem, or the term "canonical model", but doesn't grasp these ideas in full and I first have to explain that.

    3. This is someone that is just trying to piss set theorists off. I am sad to say that I have encountered several of these folks before.

Summary:

Canonical model requires a model, namely an object in the universe, to begin with. This means that the universe itself cannot be a model of ZFC, or we might get some sort of problem.

Furthermore, if a layman can discuss about set theory, or canonical models of set theory, then this was not a layman to begin with.

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  • $\begingroup$ RCF is real closed fields? The real closure closure of $\mathbb{Q}((x))$ (ordered with x positive infinitesimal) also has cardinality $2^{\aleph_0}$, doesn't it? It should be the field of Puiseaux series with real algebraic coefficients. As an ordered field it's not isomorphic to $\mathbb{R}$, because $\mathbb{N}$ is a bounded subset. $\endgroup$
    – user14972
    Sep 1, 2012 at 2:42
  • $\begingroup$ @Hurkyl: That field sounds countable to me. Why is $\mathbb Q((x))$ has size continuum? Edit: Hmm. You might be right, I'll edit once I get to a computer. $\endgroup$
    – Asaf Karagila ♦
    Sep 1, 2012 at 6:16
  • $\begingroup$ @Hurkyl: I've corrected this, thanks. I was confused because it was 3am and I was thinking about quantifier elimination and elementary equivalence. $\endgroup$
    – Asaf Karagila ♦
    Sep 1, 2012 at 6:46
  • $\begingroup$ @Asaf: Can you tell me more about the "ugly" theories (in the sense of effective enumerations)? $\endgroup$
    – Hans-Peter Stricker
    Sep 3, 2012 at 7:49
  • $\begingroup$ @Hans: It's simple. Godel's theorem tells us that every effectively enumerated theory which is "strong enough" is either incomplete or inconsistent. If you take any completion of ZFC (e.g. all sentences true in a particular model) then the theory is complete and strong enough, therefore we cannot effectively decide which sentences are axioms in this theory and which are not. $\endgroup$
    – Asaf Karagila ♦
    Sep 3, 2012 at 8:03

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