I asked my friend (a Set Theorist) this question and he said that every model of ZFC thinks it is the Standard Model. But, I'm not sure it is so simple. First, because I don't know how a Universe could test whether or not it is the Standard Model. And second, because I think that various models of ZFC could have different opinions about whether they are the standard model.

For a model which does think it is standard, I'm thinking of a model which has enough evidence to believe it is the standard model, perhaps one obtained after many generic extensions, or one which has "nice" properties, perhaps because PD (projective determinacy) holds there. But, I don't have the knowledge or vocabulary to put this in precise language yet.

I can imagine a universe which does not think that it is the standard model, one which is "close" to its generic extensions, perhaps one where generic filters exist (a universe where MA holds for example).

What do you think? Is there any way to tell, within a given model, whether it has enough evidence to conclude that it is the standard model? Or, is it as simple as my friend suggests and every model thinks it is standard? Or, is it simple because no model thinks it is standard since there are always generic extensions of any given model?

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    $\begingroup$ I think that, instead of "does every model of ZFC think it is the standard model?", you really want "does every model of ZFC think it is the `real world'?" And the answer to this question can reasonably be said to be Yes. Given a model $M$ of ZFC, its theory $Th(M)$ is a completion of ZFC that fully describes the world of sets (e.g. CH, existence of supercompacts, etc. are all settled one way or the other). In this sense, $M$ makes an exact claim as to what $V$ looks like. But I also might not be getting what you are after. $\endgroup$
    – Ed Dean
    Dec 31, 2011 at 1:12
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    $\begingroup$ I believe I'm pretty standard, or at least should be considered so. However, at a wedding in the late 1980's, another graduate student asked me what it was like to take LSD. I pointed out that I had never taken LSD. He said "You mean you're like this all the time?" $\endgroup$
    – Will Jagy
    Dec 31, 2011 at 1:15
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    $\begingroup$ With Ed, I might not be getting what you are after, but I don't see a real question here. Peppered throughout the question are words like "think", "believe", "opinions" as actions ascribed to a model, and I don't know how else to interpret those words except in the sort of tautologous way given by Ed. (I might feel more lenient if the setting were more relative -- what sort of 'opinions' a model might have about an inner model or something -- but not here where a model is asked to opine about itself, as pointed out more wittily by Will.) $\endgroup$
    – Todd Trimble
    Dec 31, 2011 at 2:03

1 Answer 1


One of the main lessons of set theory is that many of our familiar set-theoretic concepts, such as countability, uncountability, well-orderedness, ill-foundedness, even finiteness, depend on the set-theoretic context in which they are considered. We know that different models of set theory can disagree about whether a set, the same set, is countable or uncountable, about whether a relation is well-founded or ill-founded and even about whether a set is finite or infinite.

First, of course, one can play all kinds of games using the compactness theorem or ultrapowers to make models with vastly different beliefs about the same set. For example, by starting in a universe $V$ and constructing a nonprincipal ultrapower $W=V^\omega/\mu$, one has nonstandard finite sets of natural numbers in $W$ that $W$ thinks are finite, but $V$ thinks have size continuum.

But second, the method of forcing allows us to construct new universes that stay close to the original universe, the ground model, in some ways while differing in others, which we can precisely control. They have the same ordinals, for example, and thus agree on the well-foundedness of the relations they have in common. But meanwhile, we can still make them disagree on other matters, such as the power set operation, since a forcing extension can have additional set of natural numbers and thus additional reals. So the concept of the "set of reals" is not absolute between models of set theory, even when they agree on the ordinals.

The issue of background dependence is sometimes thought not to apply to arithmetic, since we can prove that there is only one standard model of arithmetic. But one can object to this slight-of-hand: the uniqueness of $\mathbb{N}$ is proved in set theory, and so each model of set theory has a unique concept of natural number, but different models of set theory can have different non-isomorphic versions of $\mathbb{N}$, as my ultrapower example above shows.

Because of this pervasive dependency of set-theoretic concepts on the set-theoretic background in which they are considered, some set theorists, perhaps like your friend, are pushed to the view that every model of set theory provides its own concept of standardness, and that the notion of an absolute set-theoretic background is illusory. Every model of set theory look upon itself as the entire universe, a standard world by its own lights. This is a sense in which the notion of standardness itself is relative-to-a-model.

Other set-theorists retain the idea that there is an absolute sense of standardness, that we are living in the real universe of sets, and that it makes sense to judge models as being correct or incorrect about their judgements of well-foundedness. As you know, such questions were precisely the topic of my recent course on the philosophy of set theory, and the debate between these two perspectives is the thrust of my article on the set-theoretic multiverse, along with the other readings we had in the course.

But let me close by mentioning one case often mentioned by set-theorists, where a model of set theory can begin to have an inkling that something is seriously awry with its notions of standardness. Specifically, if ZFC is consistent, then we know that $\text{ZFC}+\neg\text{Con}(\text{ZFC})$ is also consistent. What would it be like to live in a model of this latter theory? In that universe, we would see that all the ZFC axioms are true, but also we would think that the ZFC axioms are formally inconsistent, admitting a proof of a contradiction.

  • $\begingroup$ Thanks Joel. Your first point is very interesting - to see a difference in perspective so vast that a model would have a finite set of natural numbers but whose ultrapower thinks such set has size continuum! I hope to learn more. In response to your second point, I wonder if (since elements of power sets of finite numbers as small as 2 are real numbers which are not natural numbers) adding reals to a model would affect the size of the power set of finite numbers in that model? $\endgroup$
    – user10290
    Dec 31, 2011 at 6:11
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    $\begingroup$ @Joel: If I lived in a model of ZFC + $\neg$Con(ZFC), I don't think I "would see that all the ZFC axioms are true." I'd have a nonstandard notion of what "all the ZFC axioms" means. I could check the truth of each individual one of the "genuine" ZFC axioms (but the general notion of "genuine" here would make no sense to me). $\endgroup$ Dec 31, 2011 at 16:17
  • $\begingroup$ Yes, I agree with that, and your comment is usually also part of my response to those who advance this argument. Nevertheless, it is true that if we lived in that universe, then we would think separately of each of the ZFC axioms that they are true and also that the theory ZFC is inconsistent. There is no way to express "All of ZFC is true" since we can have no internal first-order definition of truth. $\endgroup$ Dec 31, 2011 at 20:32
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    $\begingroup$ > I'd have a nonstandard notion of what "all the ZFC axioms" means. Why not a nonstandard notion of what inconsistency means? Perhaps on account of proofs that have nonstandard length. For example, Robinson's Q is finitely axiomatized but rich enough for the Godel incompleteness theorems. So wouldn't a model of Q + ¬Con(Q) have the same axiom notion of all the axioms as any other model of Q? $\endgroup$ Apr 8, 2012 at 3:48

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