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.

  • $\begingroup$ "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 ... $\endgroup$ Sep 15, 2018 at 19:33
  • $\begingroup$ @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. $\endgroup$
    – RothX
    Sep 15, 2018 at 19:38
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    $\begingroup$ 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 ... $\endgroup$ Sep 15, 2018 at 19:39

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$
    – RothX
    Sep 15, 2018 at 19:41
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    $\begingroup$ @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?" $\endgroup$ Sep 15, 2018 at 19:42
  • $\begingroup$ 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. $\endgroup$
    – RothX
    Sep 15, 2018 at 19:45
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    $\begingroup$ @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? $\endgroup$ Sep 15, 2018 at 19:50
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    $\begingroup$ @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). $\endgroup$ Sep 15, 2018 at 20:25
  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.

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    $\begingroup$ 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. $\endgroup$
    – RothX
    Sep 15, 2018 at 19:51
  • $\begingroup$ 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? $\endgroup$
    – RothX
    Sep 15, 2018 at 19:51
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    $\begingroup$ @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) $\mathbb{R}$. $\endgroup$ Sep 15, 2018 at 19:59
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    $\begingroup$ @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! $\endgroup$ Sep 15, 2018 at 20:04
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    $\begingroup$ @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. $\endgroup$ Sep 15, 2018 at 22:56

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.


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