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I could not find a question that seemed to answer my specific query, despite lots of material on the Continuum Hypothesis (CH) on MSE and MO. If there is already a question on this, I'd greatly appreciate a link.

My question has to do with the resilience of CH to being resolved. I realize that it is independent of ZFC and "most" large cardinal axioms, and that CH can fail as "badly as you want" (see this question).

My question is: why is this the case, and is there intuition as to why "ZFC + any large cardinal axiom" cannot seem to anchor this problem? I realize the "multiverse" approach is simply to study CH in whatever universe you are in. But my question whether there is intuition as to why CH has such an indeterminate status no matter what axioms you throw at it. Thanks!

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There are axioms that determine the truth of $\mathsf{CH}$. For example, $\diamondsuit$ and $\mathsf{PFA}$. –  William Aug 20 at 0:19
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@William: Sure, as well as CH, GCH, and Woodin's work. But my question is about the context within large cardinal axioms. Since large cardinals don't settle CH, some set theorists are comfortable saying that this unsettled state is the natural order (multiverse). I'm interested in the intuition as to why we should accept the unsettled large cardinal state of affairs. –  trb456 Aug 20 at 0:25
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@trb456: There's a lot about history, tradition, the fact that we are interested in both CH holding and failing since both have interesting consequences, and of course, we aim to have a minimal axiomatic setting. So once the dust settled and $\sf ZFC$ seemed like a good candidate, it's hard to add more axioms to the "hardcore of mathematics" without a good reason. –  Asaf Karagila Aug 20 at 0:57
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Large cardinal axioms are not accepted as "correct" either. They are popular because rich consequences can be derived from them with current methods. Before that forcing axioms were popular, some still are. Before that GCH itself was popular, etc. The reason why they weren't permanently adopted is not specific to CH: none of them are needed in most of mathematical practice. –  Conifold Aug 20 at 1:08
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@zibadaw: No, that's not quite right. The continuum hypothesis and its negations have implications to operator theory and they continue from there. All the way to physics. To say that no one but set theorists care for its value is plain false; in fact I don't think that many set theorists care for the value of the continuum, and perhaps the emerging school of multiverse approach to set theory (rather than formalist or Platonist) will end up as the dominant one in the future. Precisely because there are so many conflicting interests. –  Asaf Karagila Aug 20 at 1:47

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up vote 7 down vote accepted

Let me address the question as stated first.

Why does $\sf CH$ has no determinate provability from any of the axioms we throw at it? This is false. As remarked in the comments. Plenty of axioms prove $\sf CH$ or disprove it. Things like $V=L$ or $\lozenge$ imply $\sf CH$ whereas things like $\sf PFA$ and similar forcing axioms imply its negation (these in particular imply that $2^{\aleph_0}=\aleph_2$ of all values).

Specifically you might be asking about large cardinals. Well, this is because large cardinals are large, and the continuum is small. Historically, I believe, the motivation for the thought that large cardinals might settle the continuum hypothesis came from the fact that we can prove the continuum hypothesis for Borel sets. Namely every uncountable Borel set has size continuum, and we can push this to analytic sets, but not further.

However, if there exists a measurable cardinal then we can prove the continuum hypothesis for co-analytic sets as well. Which is a push forward. So perhaps by having "enough" large cardinals, or a strong enough large cardinal axiom we can push this proof and go through all the possible sets of reals?

Well, no, not really. While these things definitely can occur (in the presence of Woodin cardinals we can push the continuum hypothesis higher up the projective hierarchy), there are more sets of real numbers than that. And at some point the large cardinals exhaust their power.

In particular we have the Levy-Solovay phenomenon:

Suppose that $\kappa$ is an inaccessible cardinal, and $\Bbb P$ is a forcing such that $|\Bbb P|<\kappa$. Then $\Vdash_\Bbb P\check\kappa\text{ is inaccessible}$.

Namely, small forcings cannot change the largeness of an inaccessible cardinal. The same is true for many other types of large cardinals, e.g. weakly compact cardinals, measurable cardinals, Woodin cardinals, etc.

On the other hand we always have a forcing of size $\aleph_2$ which violates the continuum hypothesis: adding $\aleph_2$ Cohen reals; and a forcing of size continuum which restores the continuum hypothesis: Levy collapse of $2^{\aleph_0}$ to $\aleph_1$ (and it doesn't add real numbers too!)

So with particularly small forcings we can switch the truth value of the continuum hypothesis on and off as we like. Therefore large cardinals, which are unaffected by small forcings, cannot possibly decide the truth value of $\sf CH$.

(That been said, there are large cardinals properties which can be destroyed by small forcings, but so far these were shown consistent with $\sf CH$ and with its negation by various means, e.g. by showing that we can perhaps destroy the large property, but then we can have another forcing which restores it and doesn't change the value of the continuum.)

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Therefore large cardinals, which are unaffected by small forcings, cannot possibly decide the truth value of CH. I think this is the key statement. The fact that we cannot decide which combination of large cardinal and small forcing is "correct" is a major factor in not being able to resolve CH. –  trb456 Aug 20 at 1:44
    
It's like you don't even read what I wrote in all those comments and this long answer. :-) We're not even trying to decide what is "correct". You asked why large cardinals didn't decide the continuum hypothesis, this is why. Moreover saying that a "combination of large cardinal and small forcing is "correct" is a major factor ..." shows complete lack of understanding what forcing is used for and how it is used for. –  Asaf Karagila Aug 20 at 1:50
    
We use forcing to introduce new objects to a model of set theory, and thus show that a certain statement is consistent. We don't use it (often) to decide a particular model of set theory is "the one true model of set theory". Philosophically this is hardly makes any sense. If you insist that $\sf CH$ is necessary true, then you can't really use any type of forcing which changes the value of the continuum. In fact, just adding real numbers seems morally objectionable once you decide that $\sf CH$ is true. Forcing is a force majeure in how little we work towards settling the truth value of CH. –  Asaf Karagila Aug 20 at 1:53
    
Look, I'm not a professional mathematician, and I did read what you wrote, so please don't be insulting. I have an advanced, but not professional, background in mathematics. And I highly regard your work! I'm trying to narrow down to why we cannot resolve CH beyond that it eliminates lots of interesting set theory. As you noted, we could decide all groups are abelian, but we don't, for good reasons. So given that there is lots of interesting set theoretical exploration, why does none of this settle the problem? And this is where I acknowledge that there may be no good intuition as to why. –  trb456 Aug 20 at 2:04
    
I'm sorry if it came off as insulting. I even put a little smiley there to make sure it reads with a smile, as a joke. As for the rest of the comment, "can't prove it from the axioms we regard as standard + either truth value has interesting results + the technique which revolutionized modern set theory will be handicapped" are three excellent reasons not to want to settle the problem "once and for all". –  Asaf Karagila Aug 20 at 2:20

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