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I know that we cannot prove the consistency of $ZFC$ from within $ZFC$, but for today I will assume that it is consistent, so that it has models.

Now, we know that the Continuum Hypothesis ($CH$) is independent from $ZFC$, meaning that we cannot prove $CH$ using the axioms of $ZFC$. Consequently, $ZFC$ must have models where $CH$ holds and models where the negation of $CH$ holds. What it is clear (I think) is that, given an specific model $V$ of $ZFC$, either $CH$ holds in $V$ or its negation does.

On the other hand, when we do "usual math", we can talk about the natural numbers $\mathbb{N}$ and the real numbers $\mathbb{R}$, whose construction is carried on in $ZFC$ without any considerations on $CH$. So, does it make sense to ask if there is a set $A$ such that $$|\mathbb{N}|<|A|<|\mathbb{R}|?$$ I mean, this must be either true or false in the specific model we are working and that contains our "version" of $\mathbb{N}$ and $\mathbb{R}$.

Thanks in advance and sorry for any inaccuracies.

Earnur
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    You misunderstood the independence of CH: What it precisely states is that if $\mathsf{ZFC}$ is consistent then $\mathsf{CH}$ is independent of $\mathsf{ZFC}$. – Hanul Jeon Mar 25 '25 at 14:46
  • It is true that for every model $M$ of $\mathsf{ZFC}$, $M$ thinks either $\mathsf{CH}$ holds or its negation holds. We do not know what holds unless we are given more information about $M$. Also, I do not see what you want to ask about in the last paragraph: It just asks again about the validity of $\mathsf{CH}$, but you know it is independent. Could you clarify what you want to ask? – Hanul Jeon Mar 25 '25 at 14:49
  • What I am asking is that, assuming that $ZFC$ is consistent and fixing a model of it, then in that model either $CH$ holds in the sense that such a set $A$ exist, or not. What I don't understand then is if it is possible that our "version" of $\mathbb{R}$ contains such a set or not. – Earnur Mar 25 '25 at 14:49
  • You may overthink the question: If $M$ thinks $\mathsf{CH}$ is false, then $M$ thinks $\mathbb{R}^M$ (the set of reals computed in $M$) contains a subset whose cardinality is intermediate between $|\mathbb{N}|$ and $|\mathbb{R}|$. If $M$ thinks $\mathsf{CH}$ is true, $M$ thinks there is no such set. – Hanul Jeon Mar 25 '25 at 14:52
  • You specifically just mentioned that in a model of ZFC, it is either true or false. That doesn't mean that in all models, the answer is the same. The whole point of independence is that it means we can certainly ask if $A$ exists, but we can't answer without some additional axiom. – Thomas Andrews Mar 25 '25 at 14:52
  • So, if somehow, one day we discover a set $A$ like that, it will just mean that we have built the real numbers in a model of $ZFC$ that "thinks" $CH$ fails. Yes? – Earnur Mar 25 '25 at 14:53
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    How do you "discover" a set, outside of set theory? @Earnur If we can find such a set in set theory, then set theory is inconsistent. – Thomas Andrews Mar 25 '25 at 14:55
  • But in which model? We cannot find any 'definable' counterexample of $\mathsf{CH}$ just from $\mathsf{ZFC}$ due to the independence result. – Hanul Jeon Mar 25 '25 at 14:55
  • That is precisely the point that I don't understand. If we are working with $\mathbb{R}$ in a model of $ZFC$ where $CH$ fails, then $\mathbb{R}$ must contain a set like the $A$ I described. But the point is that, although that set could exist, we will never be able to find it? I thought that we could find it, although it would not be "definable" (because if it was definable, it would also exist in all the other models, and that is not the case). Does it make sense? – Earnur Mar 25 '25 at 14:58
  • Well, we might be able to define such a set, but not prove its cardinality. Usually, we construct a model which is specifically designed to have CH or have counterexamples. Note, models are set theory objects, they aren't really concrete. Roughly speaking, we show that in any model of ZFC, we can construct a different model where CH is false. @HanulJeon – Thomas Andrews Mar 25 '25 at 15:01
  • I think that now I see the point. If we could find such a set, we would prove its existence and cardinality using $ZFC$ axioms, because they are "all" we have. And that is precisely what we cannot do. Am I right now? – Earnur Mar 25 '25 at 15:07
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    There are a lot of sets we can't define. The axiom of choice gives a lot of ways to how a set exists without defining them. That is why intuitionists and constructivists don't like AoC. @Earnur – Thomas Andrews Mar 25 '25 at 15:07
  • Indeed, axiom of choice can be used to show that there is a minimal cardinality greater than any other cardinality. We just can't answer in ZFC, in the case of $\aleph_1>\aleph_0,$ if that cardinality is equal to the continuum. – Thomas Andrews Mar 25 '25 at 15:13
  • @ThomasAndrews I know the points you mentioned. I should have said some set theorists try to determine $\mathsf{CH}$ from extra 'reasonable' axioms, but I thought this POV would confuse the OP more. – Hanul Jeon Mar 25 '25 at 15:26
  • @ThomasAndrews Also, your explanation, 'models are set-theoretic objects and they are not concrete,' I do not think it is a precise explanation: They are set-theoretic objects, but set theory proves they exist. Under reasonable assumptions like inaccessible cardinals, we can construct a concrete one. It is true that the ways we used to prove the independence of $\mathsf{CH}$ (say, inner model and forcing) a relative constructions, so it gives a new model from a given model. (Alternatively, from a proof-theoretic view, they provide an interpretation of extensions of $\mathsf{ZFC}$.) – Hanul Jeon Mar 25 '25 at 15:30
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    @ThomasAndrews It is inaccruate to say $\mathsf{AC}$ is the source of undefinable sets. You can find a pointwise definable model of $\mathsf{ZFC}$ (if there is a transitive model of $\mathsf{ZFC}$), in which every element is definable. The reason why constructivists do not like $\mathsf{AC}$ is not definability-related issues (Even Brouwer mentioned the notion of 'lawless sequences' in his intuitionistic mathematics) but more about BHK-interpretation-related issues. – Hanul Jeon Mar 25 '25 at 15:33

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When we do "usual math" we don't do it by selecting and working inside a specific model of ZFC at all; it isn't possible to exhibit such a model using ZFC alone, so doing this would not count as using ZFC as a foundation. ZFC is (supposedly) acting as a foundation syntactically; that is, it tells us (supposedly) which proofs are valid.

When we talk about $\mathbb{N}$ and $\mathbb{R}$ that is (supposedly) because we produced syntactically valid constructions of those sets from the ZFC axioms; as a result those sets exist in any model of ZFC, if you want to think about what's happening semantically. Furthermore any results we prove about them from the ZFC axioms are true in any model of ZFC.

The independence of the continuum hypothesis from ZFC means exactly that we can't produce either a syntactic proof or disproof, from the ZFC axioms, of the existence of a set $A$ of cardinality intermediate between $\mathbb{N}$ and $\mathbb{R}$. Note that this does not require or involve thinking about models of ZFC, it is just a limitation on what kinds of syntactically valid proofs we can write down when we use ZFC as a syntactic foundation.

This may make you concerned about what the "ontology of usual math" is supposed to be, i.e. which sets are supposed to "really exist" and so forth. That's a big question but you can see, for example, Hamkins' The set-theoretic multiverse for a discussion.

Qiaochu Yuan
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  • Very nice answer. There is some more discussion here about why, foundationally speaking, models play a secondary role to syntactic proofs. – Joe Jun 21 '25 at 20:11