What if "ZFC +CH is false" has no set universe which interprets it as true?

Question

We have proved that "CH is false" is consistent with ZFC. So assuming ZFC is consistent itself, "ZFC+CH is false" is good as an axiomatic system.

But is it also possible that this axiomatic sytem is merely a string manipulation game, and that there exists no universe of collections of objects which interprets these axioms as true?

If yes, then shouldn't "CH is true" be the preferable axiom?

If no, then have we found that universe and its collections of objects?

Answer 1

Cohen's proved the consistency of ZFC+(not CH) by giving a model. That's a "universe of collections of objects which interprets these axioms as true", I would say.

It is true that he had to assume axiom SM to construct his model. He showed how his argument could be converted into a purely syntactic argument ("a string manipulation game"), avoiding SM. However, most set theorists are willing to accept SM, at least if they believe that ZFC is true for an actual universe of objects.

Later on people came up with Boolean-valued models. These might be regarded as "collections of objects" satisfying ZFC+(not CH), though that depends on how far you're willing to stretch the meaning of "collection".

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More details.

First, axiom SM: this says that there is a "standard" model of set theory whose universe is a set. That is, there's a set $M$ such that the pair $(M,\in)$ satisfies all the ZFC axioms, with $\in$ being the element-of relation of the "actual universe".

You can give a fairly convincing argument that SM "ought" to be true, even with $M$ countable, by mimicking the proof of the Löwenheim-Skolem theorem. You can prove formally ("inside ZFC") that a model $(M,\in)$ exists for any finite subset of the ZFC axioms. That's enough for the consistency proof. Cohen outlines all this in his book Set Theory and the Continuum Hypothesis.

Next, the construction of Cohen's model of ZFC+(not CH). Cohen showed how to "add elements" to the model $(M,\in)$ given by SM. Rough idea: letting $M$ be countable, there are "really" only countably many subsets of $\omega$ in $M$. That is, $P(\omega)\cap M$ is countable in the "actual universe". $M$ thinks that $2^{\aleph_0}$ is uncountable because the bijection between $\omega$ and $P(\omega)\cap M$ is also missing from $M$. So there are many subsets of $\omega$ that you could add to $M$. Of course, once you add them, you have to add many other things, but eventually you can get a model $N\supseteq M$ of ZFC. There are many obstacles, and Cohen's method of forcing was devised specifically to surmount these. The method is flexible enough to achieve many different effects. With one particular variant, you can make $N$ satisfy ZFC plus $2^{\aleph_0}=\aleph_2$.

Bottom line: the model $(N,\in)$ is a universe of objects, and ZFC+(not CH) holds in it.

The Boolean-valued model approach messes around with the notion of "set". Let $B$ be a boolean algebra. Of course, the usual two-valued boolean algebra {T,F} (call it $B_2$) is the easiest example. Now if $S$ is some set, then a function $f:S\to B_2$ is basically the same as a subset of $S$. Generalize: a function $f:S\to B$ is a "set", except the truth-value of the assertion $s\in S$ can be any element of $B$, not just T or F.

The actual construction of a Boolean-valued model involves lots of technicalities, way too many for me to describe here. But the bottom line: one can construct such a model where the truth-value of all the ZFC axioms is T (the top element of $B$), but the truth-value of $2^{\aleph_0}=\aleph_1$ is not T. That's enough to show that ZFC can't prove $2^{\aleph_0}=\aleph_1$.

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Now just a smidgen of philosophy. You ask whether CH or something else should be the "preferable axiom". Of course, one could prefer one axiom or another purely on pragmatic grounds. What allows you to prove the most theorems? Or the prettiest theorems? Or something like that.

Since no "value" for $2^{\aleph_0}$ has turned out to be clearly preferable by this criterion, people handle it by stating explicit assumptions. For example, take a look at Wetzel's problem.

If one is a strong Platonist, one would want to know the "true value" of $2^{\aleph_0}$. Both Gödel and Cohen were Platonists. Gödel spent the last few years of life trying to prove that $2^{\aleph_0}=\aleph_2$ (of course, adding new "true" axioms to ZFC for this purpose). Cohen said this in his book:

A point of view which the author feels may eventually come to be accepted is that CH is obviously false. The main reason one accepts the Axiom of Infinity is probably that we feel it is absurd to think that the process of adding one set at a time can exhaust the entire universe. Similarly with the higher axioms of infinity. Now $\aleph_1$ is the set of countable ordinals and this is merely a special and the simplest way of generating a higher cardinal. The set $\mathfrak{c}$ is, in contrast, generated by a totally new and more powerful principle, namely the Power Set Axiom. It is unreasonable to expect that any description of a larger cardinal which attempts to build up that cardinal from ideas deriving from the Replacement Axiom can ever reach $\mathfrak{c}$. Thus $\mathfrak{c}$ is greater than $\aleph_n$, $\aleph_\omega$, $\aleph_\alpha$ where $\alpha = \aleph_\omega$ etc. This point of view regards $\mathfrak{c}$ as an incredibly rich set given to us by one bold new axiom, which can never be approached by any piecemeal process of construction. Perhaps later generations will see the problem more clearly and express themselves more eloquently.

Finally, some links. First I will plug my own set of notes: Notes on Smullyan & Fitting, and on Forcing (skip to section 21). (These are from a set-theory meetup.) Also, you may find this mathoverflow item helpful.