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In asking a question on this site about the unprovability on the Continuum Hypothesis, many people explained to me that for a given set of axioms, there are many different models that satisfy that set of axioms, a concept that I am very comfortable with now.

Hence, there exists a model for ZFC set theory that satisfies CH, and one that doesn't. (Correct me if I'm wrong about anything)

However, for practical purposes, it seems like there is model of set theory that is used almost universally, at least in branches of mathematics other than logic. It also seems like this model is "maximal" in the sense that if the existence of a set $A$ doesn't violate any of the axioms of ZFC, then $A$ exists in the universe of sets.

First off, am I even remotely correct in thinking this? Is there such a model? And if so, is it maximal in any sense?

If I am correct, then what's the big deal about the unprovability of the Continuum Hypothesis? It seems true of a lot of things. From reading the axioms, I don't see any reason why a set with the cardinality of the real numbers has to exist. And why be so squirrelly when talking about sets with cardinality between that of the reals and the natural numbers? If they exist within our maximal universe of sets, why not use them?

If I'm not correct, then does a maximal universe even exist? Or, is there a universe of sets, $U$, that satisfies ZFC, and all other models of ZFC are subsets of $U$?

If so, why wouldn't we use it? And my earlier questions still apply.

If not, why not? Is there a proof? And what model do we actually use, and why?

Sorry if this isn't a very well-asked question. It's kind of rambling and contains a lot of sub-questions, but I'm not sure how else to phrase this while still asking what I want to.

RothX
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  • "From reading the axioms, I don't see any reason why a set with the cardinality of the real numbers has to exist." Did you really intend to write this? $\mathbb{R}$ itself is a set ... – Noah Schweber Sep 15 '18 at 19:33
  • @NoahSchweber Yes, of course it's a set. My point is that I don't think you can prove $\mathbb{R}$ exists just from the ZFC axioms. Hence, some models don't include it. – RothX Sep 15 '18 at 19:38
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    I'm not sure what you mean - ZFC definitely proves that $\mathbb{R}$ exists (incidentally, it might be simpler to think of $\mathcal{P}(\omega)$ instead of $\mathbb{R}$). Of course, different models might disagree about what it is ... – Noah Schweber Sep 15 '18 at 19:39

3 Answers3

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EDITED in response to comments:

I think the issue you're facing is that there is actually no universal agreement on what the "mathematical universe" looks like, or whether it even exists, or in what sense.

This is one of the things a "foundational theory" like ZFC is for: it frees us from the constraint of having to commit to a philosophical perspective. An ultrafinitist, a formalist, a Platonist, and a multiversist (see below) can all agree with the statement "ZFC proves that the continuum does not have cofinality $\omega$" (for example). In my opinion, ZFC exists precisely because of the problems inherent to your question.

Specifically, your question isolates the following philosophical points:

  • Is there in fact a model of ZFC used by mathematicians, or at least the majority of mathematicians ("the model of set theory used almost universally")?

  • If the answer is yes, then do we have some mechanism for figuring things out about it, which goes beyond just using the ZFC axioms (and so in particular could decide CH) ("If they exist within our maximal universe of sets, why not use them?")?

Both of these questions are extremely controversial (and in fact I would answer "no" to the first right off the bat). In fact, consider the following weak claim:

  • Most mathematicians work in ZFC.

Even that is, in my opinion, of dubious truth - if you pick a mathematician at random and ask them to state the ZFC axioms without looking them up, with high probability they won't be able to. The ZFC axioms simply don't play a role in most mathematicians' activities. So in what sense can the point above even be claimed?

(Well, a Platonist commitment can get around it - "You're living in a model of ZFC even though you don't know it" - but that just takes us back to points $1$ and $2$ above.)

And this takes us to the strange paradox of philosophy in practice: that mathematicians are on the one hand empirically inconsistent (not just amongst each other, but even with ourselves) about these philosophical issues, and on the other hand nonetheless able to do mathematics (and even sometimes benefit from this "flexibility"). I have many opinions on this, but the two relevant points I want to make - at the risk of repeating myself a bit - are:

  • This "flexibility" poses a fundamental obstacle to giving a clear, consistent description of mathematical practice, especially if you want to avoid formalism; and in particular makes the role of anyone looking for a "model used in practice" almost impossible.

  • One of the important things ZFC does is to provide us with a "bulwark" against incomprehensibility: at the end of the day, I can always walk back a Platonist claim ("$2^{\aleph_0}$ has uncountable cofinality") to a formalist one based on a specific formal theory $T$ ("the specific theory $T$ proves that $2^{\aleph_0}$ has uncountable cofinality"), and by making the sociological agreement to treat ZFC as our "default $T$" we ensure that this fallback does not lead to "mathematical disintegration."

One question looming in the background, then, is: why ZFC? This is really twofold: on what basis do we justify the ZFC axioms (as true, or useful, or used, or ...), and why can't we justify anything more - or can we?

Both parts get us into interesting territory. I recommend the following articles:

Incidentally we can go even deeper, and question logic itself:

To clarify, I am not claiming agreement with them, but I do think they're good sources.

Martin Sleziak
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Noah Schweber
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  • Right, I think what I meant by "the model of set theory used almost universally" is the model of set theory used in fields of math other than logic and set theory, which I feel is usually the same model, but I could be wrong. – RothX Sep 15 '18 at 19:41
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    @RothX You're making two assumptions - that such a model exists (which is easy to justify via Platonism, if you're willing to adopt that philosophy, but which pretty much any other philosophical perspective would at least demand justification for and in my opinion probably oppose strongly) and that we can tell what it looks like. Even assuming that there is an "actual model" used by mathematicians - in fact, a single one we all share - what mechanism do you propose we use to figure out of CH is "truly true?" – Noah Schweber Sep 15 '18 at 19:42
  • Right, well my problem is that I wasn't sure if such a model existed or if we could tell what it looked like, hence my question. Also, I'm fully on board with the unprovability of CH. I was wondering whether it was true or not in our "standard model" of ZFC, if such a model exists. – RothX Sep 15 '18 at 19:45
  • @RothX That's exactly my point: that the existence of such a standard model is a point of contention, and that even if it weren't, there's no obvious way to "ask questions about it" - for that matter, why do we believe that ZFC is "true?" – Noah Schweber Sep 15 '18 at 19:46
  • Outside of logic and set theory, I would say that actually typically no models of set theory are used at all. See https://mathoverflow.net/questions/300753/circular-or-missing-definition-in-set-theory/301061#301061 for some discussion (and also Noah's answer on that post for some related discussion). – Eric Wofsey Sep 15 '18 at 19:46
  • To clarify: I agree with @EricWofsey, although there's some nuances there about how this plays with mathematical Platonism outside logic. – Noah Schweber Sep 15 '18 at 19:47
  • @NoahSchweber So are you saying, even given some model of ZFC, there's no way to know whether CH is true in that model or not? Because that would make a lot more sense. Also, I never said that I think ZFC is "true." It doesn't make any sense to ask whether a set of axioms is true or not. All I was asking about were models of ZFC. – RothX Sep 15 '18 at 19:49
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    @RothX It does make sense to ask whether a set of axioms is true of a specific structure - if you're claiming that there is a "set-theoretic universe" used by mathematicians, how do you know ZFC is true of that universe? Why couldn't it be the case that the "set-theoretic universe" mathematicians use is actually a model of NF, instead? – Noah Schweber Sep 15 '18 at 19:50
  • @RothX "So are you saying, even given some model of ZFC, there's no way to know whether CH is true in that model or not?" How precisely would we be "given" a model of ZFC? Furthermore, going back to the basic assumptions, on what basis do you think mathematicians "have access to" a model of ZFC? – Noah Schweber Sep 15 '18 at 19:55
  • @NoahSchweber I see what you're saying. Also, you keep accusing me of claiming that this "standard universe" exists, which I never did. In fact, my question was asking whether it did. If this seems unclear, is there a way I could edit my question to make it more clear? Also, the reason I didn't consider NF is because I'd never heard of it, but I'll look into it. – RothX Sep 15 '18 at 19:55
  • @RothX "the model of set theory used almost universally" implies that there is such a model. This might not have been what you intended, but it's what you wrote. (As to NF, the details don't matter - the point is just that there are alternatives to ZFC, even very weird ones!) – Noah Schweber Sep 15 '18 at 19:56
  • @NoahSchweber Right, I guess I meant that it seems like there's a model of set theory used universally but I'm not sure. I will edit my question and try to clarify this. – RothX Sep 15 '18 at 19:57
  • @RothX Ah, yes, that makes things clearer. My apologies for the misunderstanding! – Noah Schweber Sep 15 '18 at 19:58
  • @NoahSchweber Also, as for your other comment: "on what basis do you think mathematicians "have access to" a model of ZFC?" Wow, you're really digging into me here, but I like it. I guess I really don't have a basis. I'm far from a professional mathematician, and I really have no idea how much rigor mathematicians use. I kind of assumed that to work with sets, you had to know a certain amount about them, but I guess I was wrong. – RothX Sep 15 '18 at 20:00
  • @RothX Sorry if I'm coming off as aggressive - text isn't the best way of communicating tone of voice (and I'm not the best at text, myself). I think the issue is what "certain amount" means, and also what "work with sets" means. Most mathematicians' attitude towards ZFC (in my experience) is that it's an "ultimate bulwark" - we'll make Platonist-sounding mathematical claims ("there are infinitely many prime numbers") but can (almost) always use ZFC to walk them back to formalist ones ("ZFC proves that there are infinitely many prime numbers"). (cont'd) – Noah Schweber Sep 15 '18 at 20:02
  • So the common role of ZFC is paradoxically to make people not have to care about foundational matters; and in particular, the claim that ZFC suffices for the vast majority of mathematics is one which most mathematicians tend to just trust those of us with specific ZFC-experience on. – Noah Schweber Sep 15 '18 at 20:03
  • I think one of the reasons this is hard to untangle is that mathematicians ourselves are not particularly philosophically consistent, and are happy to jump between wildly contradictory philosophies depending on the day. The fact that this doesn't get in the way of "doing mathematics" is ironically something which can be used to argue for almost any philosophical perspective: the Platonist could say "of course it doesn't matter, you're all looking at the same universe even if you don't know it, your philosophy is just window dressing"; (cont'd) – Noah Schweber Sep 15 '18 at 20:06
  • the formalist could say "of course it doesn't matter, all you're doing is pushing symbols around; we've more-or-less agreed to restrict ourselves to implicitly work within a single formal framework (ZFC), so there's no possible meaningful disagreement;" the multiversalist could say "Of course it doesn't matter; in fact, this philosophical pluralism just reflects, deep down, the fact that you all know that "the mathematical universe" is a fluid concept, with some features constant (the ZFC axioms) but others flexible (CH);" and so forth. – Noah Schweber Sep 15 '18 at 20:09
  • @NoahSchweber Ok, that makes a lot of sense now. So I guess the summary is that there is currently no universal model, and we don't know if there could be one. In addition, most mathematicians don't know exactly what model they're using, and probably can't know. – RothX Sep 15 '18 at 20:11
  • @RothX ... or even if they are using a model in any meaningful sense (e.g. the formalist would say no). – Noah Schweber Sep 15 '18 at 20:11
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    @RothX I've edited my answer; I think this addresses (and represents) your question better, but if you prefer the old version feel free to rollback (or let me now how I could improve this one). – Noah Schweber Sep 15 '18 at 20:25
  • In connection with the claim that most mathematicians don't know what the axioms of ZFC are, may I ask whether you think most mathematicians have never read Halmos's book Naive Set Theory. If that is the case, it's a great shame. – Rob Arthan Sep 15 '18 at 20:34
  • @NoahSchweber Yes, I think this answers my question better. I think the only thing that it's missing is a straightforward answer to "is there a maximal model for ZFC?" The answer seems to be "we don't know" or maybe "we can't know." You sort of touch on it at the beginning with "there is actually no universal agreement on what the 'mathematical universe' looks like, or whether it even exists," but I think this is more talking about whether these is a "true" mathematical universe, as opposed to just a maximal universe if we assume ZFC. Other than that, it's great, and thanks for all your help! – RothX Sep 15 '18 at 20:37
  • @RobArthan I would say yes, but I would also say that they have seen all the concepts, at least enough to be comfortable (e.g. I don't know any mathematicians who would find it hard to think of indexed families of sets as functions, even if they don't tend to think of them this way). Incidentally, I think this is a huge positive re: Halmos: it shows the extent to which the book correctly captured informal set as used by the majority of the mathematical community. – Noah Schweber Sep 15 '18 at 20:43
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  1. It's true that there is a ZFC model that satisfies CH and one that doesn't.

  2. ZFC is not maximal in your sense. In fact, there can't be a theory of sets that is maximal in your sense, because of Gödel's incompleteness theorem.

  3. In ZFC, there is a set with the cardinality of the reals, albeit it may not be obvious from the axioms, you can "construct" it from the axioms. That is, you can prove it's existence by actually building it. The reals exist in every ZFC model you can think of. The same can't be done with CH, hence, the big deal of the unprovability of it.

  4. The model we use is the "standard interpretation" of ZFC, for most things. This (kinda) means that if you can build it in ZFC, then it "exists". I am not aware on any consensus of incorporating CH or it's negation in the standard model.

Fernando Chu
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    To point 2, I never asked anything about ZFC being maximal. I only asked about whether, give ZFC, there was a maximal model of it. – RothX Sep 15 '18 at 19:51
  • As for point 3, could there be a model of ZFC without any set that has the cardinality of the reals? Or does such a set exist in every model? – RothX Sep 15 '18 at 19:51
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    @RothX It depends what you mean - by "the reals," do you mean the "true" reals? In that case, consider countable models of ZFC. However, each model has a set it thinks is the reals, and so in the "internal" sense every model has a set of size $\mathbb{R}$ - namely, (the thing it thinks is) $\mathbb{R}$. – Noah Schweber Sep 15 '18 at 19:59
  • @NoahSchweber I guess it's hard to say exactly what I mean. What I want to mean is sets with cardinality equal to the set of reals that we usually use (Dededekind cuts or some equivalent construction), but we can't really create bijections across models anyway, so it's a hard question to ask I suppose. But I suppose if there are models with only countable sets, that would answer my question satisfactorily. – RothX Sep 15 '18 at 20:03
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    @RothX Incidentally, the particular case of LS for set theory has a special name: Skolem's paradox. Ironically, within logic the LS theorem has become viewed as a generally positive feature of ZFC, even though (IIRC) Skolem originally used it as an argument against FOL! – Noah Schweber Sep 15 '18 at 20:04
  • @RothX My bad, there is a maximal model of ZFC (it must only be one, by the maximality definition). However working with models in general is hard, maybe even impossible, hence we use theories for that. Another related question is, shouldn't we use a theory that gets closer to the maximal model? This can't be the case, as it could distort some of our usual constructions. Something a long the lines of having a natural number between $1$ and $2$. (this example in specific is false, but should convey the idea) – Fernando Chu Sep 15 '18 at 21:34
  • @Ryunaq "there is a maximal model of ZFC" Citation needed. It's true if we adopt a Platonist attitude, although in that case we need to stretch the definition of "model" to "class model," but it's false under e.g. the multiverse perspective. I also don't understand what "Another related question is, shouldn't we use a theory that gets closer to the maximal model? This can't be the case, as it could distort some of our usual constructions." means. – Noah Schweber Sep 15 '18 at 22:09
  • @Noah I think maximal in his sense means $ZFC\not\vdash \lnot \exists xP(x)$ then there is a model $M$ where $M\vDash \exists xP(x)$. You could create a maximal $M$ by well ordering such propositions, and adding them to the basic model by recursion, I think (just realized it won't be unique). I think I worded it wrongly, my bad. I think RothX is suggesting to have this following (meta) schema: If $ZFC$ is consistent with $\exists x P(x)$ then $\exists x P(x)$ is an axiom of $ZFC^*$. I think adding this metas chema would create more problems than solving them. – Fernando Chu Sep 15 '18 at 22:56
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    @Ryunaq I disagree - I think that "maximal" is meant in the sense of "containing all the sets" (which of course begs the question). But this could indeed be clarified. And your analysis of that (and ZFC*) is right. – Noah Schweber Sep 15 '18 at 22:56
  • @NoahSchweber What I meant by "a maximal model" is that $U$ is a maximal model of ZFC if: $\forall M$, where $M$ is a model for ZFC, $A \in M \implies A \in U$. Or equivalently, all models of ZFC are subsets of $U$. – RothX Sep 16 '18 at 02:26
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I don't have anything to contribute on the philosophical questions here; instead, I'll look at this:

It also seems like this model is "maximal" in the sense that if the existence of a set $A$ doesn't violate any of the axioms of ZFC, then $A$ exists in the universe of sets.

Unfortunately, this idea doesn't work; we can see why with the Continuum Hypothesis.

In a model where CH is true, then by definition we have a bijection $\phi_1 : 2^{\aleph_0} \leftrightarrow \aleph_1$. That's what it means for the two cardinals to be equinumerous. Also, remember that everything—including functions—is a set. In particular, $\phi_1$ is just a set of ordered pairs: it's a subset of $2^{\aleph_0}\times\aleph_1$ demonstrating a 1-to-1 correspondence between the two. In other models, ones where CH doesn't hold, $2^{\aleph_0}$ will be equal to some other aleph number.* E.g., there's a model where we have a bijection $\phi_2 : 2^{\aleph_0} \leftrightarrow \aleph_2$.

See the problem? $\phi_1$ is a set that exists in some models, and $\phi_2$ is a set that exists in others, but they can't both exist in the same model. Otherwise, we'd just compose the two to get a bijection proving $\aleph_1 = \aleph_2$, which is provably false (and therefore false in every model). So there can't be a "maximal model", because despite each set being valid in some model (and so each on their own is consistent with ZFC), no model can contain both.

This is a case of a more general phenomenon: models can really screw with your intuition of what it means for a statement to be "true". My favorite example is the fact that the set of von-Neumann naturals $\omega = \{0,1,2,\ldots\}$ (where each ordinal is the set of those less than it) doesn't model ZFC (e.g., it doesn't satisfy the axiom of pairing), but it DOES satisfy the powerset axiom! This is despite the fact that the "real" powerset of $2 \in \omega$ is $\mathscr{P}(2) = \mathscr{P}(\{0,1\}) = \{\{\},\{0\},\{1\},\{0,1\}\} \notin \omega$. See the linked article for an explanation of how this possibly makes sense. Basically, inside the model, $\mathscr{P}(2) = \{\{\},\{0\},\{0,1\}\} = 3$. While in "reality", $\{1\} \subset \{0,1\}$, inside the model the set $\{1\}$ doesn't even exist (and so in particular asking whether it's a subset of anything isn't even a well-formed question).

The bottom line is that the perspective inside a model doesn't necessarily match the perspective inside another model or from the "outside". This is because statements like "$X$ and $Y$ are equinumerous" or "$X$ is the powerset of $Y$" are actually statements not just about $X$ and $Y$, but about other sets in the universe (namely, a bijection $\phi$ between $X$ and $Y$, and subsets $S$ of $X$), and those other sets may or may not exist in the model alongside $X$ and $Y$.

* Actually, showing that every cardinal number is an aleph number requires the axiom of choice. But that doesn't affect the argument, because every model of ZFC is also a model of ZF; as such, the same two contradictory models still exist without Choice.

greatBigDot
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