OK, Cohen has constructed a model in which both ZFC and ~CH are true. Isn't this model an answer to the continuum problem? Hasn't he showed that it is indeed possible to construct a set with cardinality between that of the integers and that of the reals? Why is it still not considered sufficient to settle CH? Why is one model not enough? Why for all models? In other words, why do we have to answer whether "ZFC |- CH" instead of just "CH" itself?

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    $\begingroup$ Cohen showed that (given that there is a model of ZFC) a model can be built in which ZFC holds and CH does not hold. For some, this (along with work of Goedel) settles the issue: ZFC, if it is consistent, does not prove CH. For others, CH is a natural enough condition that, like Euclid's parallel postulate, one wonders if there is some other basic "truth" that should be assumed for set theory to hold. Gerhard "Ask Me About System Design" Paseman, 2010.06.19 $\endgroup$ – Gerhard Paseman Jun 20 '10 at 5:32
  • $\begingroup$ I mean, set theory plus this "truth" (and so CH) to hold. Gerhard "Ask Me About System Design" Paseman, 2010.06.19 $\endgroup$ – Gerhard Paseman Jun 20 '10 at 5:34
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    $\begingroup$ There seem to be several questions on this precise topic already, e.g., mathoverflow.net/questions/10227/… and they have some very well-written answers. Is there something specifically unsatisfying about these answers? $\endgroup$ – S. Carnahan Jun 20 '10 at 7:05
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    $\begingroup$ The question makes some sense to me, if phrased this way: why doesn't the construction of a model of set theory in which CH let us show that CH fails in the standard model? I don't think that is answered very well (if at all) in the answers to the question Scott Carnahan linked to. I'm not saying it's an extremely deep question, though. (Aside: would someone with the technical ability please edit the question to correct the spelling of "Cohen"? ) $\endgroup$ – Carl Mummert Jun 20 '10 at 11:54
  • $\begingroup$ As I read Cohen's result now, I notice it has an assumption, that is ZFC is consistent. Without this assumption, Cohen could have resolved CH. With this assumption, he only showed how to construct a set from an assumed set. Without the assumed set, his method does not work. And Godel showed that no one can construct the assumed set. $\endgroup$ – Zirui Wang Dec 31 '10 at 8:01

I am sympathetic to this question, which often arises for those first learning of Cohen's theorem, and I don't think it is an idle question. I recall my sophomore undergradatue self being confused about it when I first studied the set-theoretic independence phenomenon. And I think that Carl is right, that this particular issue is not addressed on the other CH questions.

I view the question as arising from the following line of thought: Cohen proved that it is possible that CH fails. Thus, it is possible that there is a set of reals whose cardinality is strictly between the integers and the continuum. But if we can decribe how such a set can exist, then haven't we actually described a set of such intermediate cardinality? That is, doesn't this mean that CH is simply false?

This line of thinking may be alluring, but it is wrong. The reason it is wrong, as Gerhard explains in his comment, is that it doesn't appreciate the role of models, or what might be called the set-theoretic background. What Cohen did was to show that if ZFC is consistent, then so is ZFC + ¬CH. (In contrast, Goedel proved that if ZFC is consistent, then so is ZFC + CH.) Thus, Cohen's intermediate-cardinality set has the property that it is intermediate in cardinality in the model that Cohen describes, with respect to that set-theoretic background, but it will not be intermediate-in-cardinality with respect to other set-theoretic backgrounds. The property of being intermediate in cardinality is dependent on the set-theoretic background in which this property is considered. For example, a set $X$ is uncountable if there is no function from the natural numbers onto it. But perhaps there is no such function mapping onto $X$ in a model of set theory $M$, but there is a larger model of set theory $N$, still having $X$, but for which now there IS a function mapping the natural numbers onto $X$. In fact, this very situation follows from Cohen's forcing method: any set can be made countable in a forcing extension.

Thus, whether a set forms a counterexample to CH cannot be observed looking only at that set---one must consider the set-theoretic universe in which the set is considered, and the possible bijective functions that might witness its countability or not. The very same set of reals can be countable in some models of set theory and bijective with the continuum in others.

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    $\begingroup$ Yes, that is precisely what happens. The way forcing works is that you set up a partial order consisting of small pieces of the new set to be added, and then argue that there is a generic manner of assembling these pieces together so as to form a new generic object. For example, if you force with finite partial functions from the natural numbers to $X$, then the generic function will be a map from $\mathbb{N}$ to $X$, making $X$ countable, even if it was originally uncountable. $\endgroup$ – Joel David Hamkins Jun 20 '10 at 13:41
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    $\begingroup$ Yes, that's exactly the reason why a larger model of set theory can have more bijections. But there is a subtle point that the larger model might not only have more functions, it might also have more real numbers. So even if two sets from the smaller model have the same cardinality in the larger model, maybe neither of these sets is "the set real numbers" in the larger model. For example, because ZFC proves the set of real numbers is uncountable, if I collapse the set of real numbers in a particular model to be countable in some larger model, the larger model must contain new real numbers. $\endgroup$ – Carl Mummert Jun 20 '10 at 13:48
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    $\begingroup$ This is probably a misguided question, but I think it may be on the minds of other non-logicians: why is that the naturals/rationals have a certain objective "reality" - that allows us to say that certain propositions not provable in ZFC, are nonetheless true - but the reals (defined as classes of converging sequences of rationals) don't seem to have that same "reality"? Is it the notion of "subset" that becomes somewhat more arbitrary for uncountable sets? Or is it the notion of "sequence of rationals" that has already lost the above "objectivity"? $\endgroup$ – Yaakov Baruch Jun 20 '10 at 15:11
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    $\begingroup$ @Boyarsky: although the actual set Z[[x]] depends on the model of set theory, the "non-set-theoretic" properties that can be directly proved in ZFC do not. So algebraists do not ordinarily need to worry about independence results. However, Shelah's work on the Whitehead problem shows that there are interesting algebraic questions that cannot be settled in ZFC ( en.wikipedia.org/wiki/Whitehead_problem ). $\endgroup$ – Carl Mummert Jun 20 '10 at 17:18
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    $\begingroup$ Yaakov, it's worse than you think. Even the natural numbers $\mathbb{N}$ themselves are dependent on the set-theoretic background. The usual uniqueness proof is, after all, a set-theoretic argument. Different models of set theory can have different, non-isomorphic sets of natural numbers, with different arithemtic truths. (Forcing does not affect arithmetic, but the models are built by other means.) One can sensibly hold that seemingly absolute nature of the natural numbers is an illusion that has not yet been shattered by the rise of a technique comparable to forcing, but for arithmetic. $\endgroup$ – Joel David Hamkins Jun 21 '10 at 1:06

There are several ways of describing "what you're doing" when you construct a model by forcing. One of these ways is to recast the whole proof as a syntactical consistency proof. That's not going to help here, because your question is about actual properties of models. In particular, if you want to talk about settling CH then you have to commit to the existence of some standard model of ZFC, and then ask whether CH hold in this model.

The most common description in elementary books is that you start with a countable transitive model of ZFC and you use it to construct a second countable transitive model of ZFC. The reason that you start with a countable model is to make it easy to prove that the generic filter (the primary thing you "add" by forcing) actually exists. If you start with an uncountable model, it's not apparent that the necessary generic filters exist. Worse, if you are committed to the existence of a "standard model" containing "all sets" then the generic filters you want often cannot exist over that model. Because, trivially, it's impossible to find any "new" sets if you already have "all" sets. Since your question is directly about the standard model, this is a severe limitation.

The reason that the countable transitive model method does not tell you anything about the standard model is that it only works with countable models. So, basically, it takes one nonstandard model and produces a second nonstandard model. This method can be used to show statements are independent of ZFC, but it gives no information at all about the standard model.

Another way to view forcing is that you start with an arbitrary transitive model of set theory and construct a Boolean-valued model from it. But the Boolean-valued model you construct is not even a classical two-valued "model", so it again tells you nothing about the standard model. To turn an arbitrary Boolean-valued model into a two-valued transitive model, you have to find a suitable generic ultrafilter to mod out the truth values, and constructing this ultrafilter is essentially the same problem as constructing a generic filter over the original poset.

In the end, the only way to show that the standard model has some property is to prove that property from axioms that hold in the standard model. The independence of CH from ZFC means that you would need to assume some additional axioms beyond ZFC to prove or disprove CH in the standard model. There is more information about that in the following MO question: Solutions to the Continuum Hypothesis

  • $\begingroup$ Carl, it seems we were writing at the same time! $\endgroup$ – Joel David Hamkins Jun 20 '10 at 12:56
  • $\begingroup$ You can use any ultrafilter on the Boolean algebra when taking the quotient of the Boolean-valued model, to get a 2-valued model. Nothing special is required. The problem is only that you may end up with a non-well-founded model. $\endgroup$ – Joel David Hamkins Jun 20 '10 at 12:58

I think the answer to the question doesn't really have anything to do with how Cohen actually proved the independence of CH, it is more about what a model of set theory is.

There are some axioms defining what a group is. These axioms can be written in ordinary first-order propositional logic. A model of the theory is a set with a binary operation and a distinguished element such that it satisfies the group axioms. That is, a model of the theory is simply a group. Similarly, a model of set theory is simply a set with a relation (representing) such that it satisfies the axioms of set theory. Now there are theorems that follow directly from the group axioms. And these are just the things that hold for every model of the theory. They hold for every group. Other statements, such as "For all a and b, ab=ba" hold for some models but not for others. They are independent of the axioms.

Now, things are really the same in set theory. You can have models (S,e) and (S'.e') such that the continuum hypothesis holds in one (S,e), but not in (S',e). Now these are simply sets with a binary relation defined on them. Such sets exists since, loosely speaking, every consistent theory has a model.*

But this doesn't really tell us anything about the world of sets we are working with, apart from the fact that both could be true. All we know is that the theory we are working with has, provided it is consistent, a model. But there could be many models with different properties, and, since they are consistent with the axioms, we cannot use the axioms to identify a "true" model. In particular we cannot learn from them that something holding in a model is true, we can only use them to show something is not provable false (in our case, the CH and its negation).

*Usually, to make things simpler, one works with models which are closer to the real sets in that for $a,b\in S$ one has $a e b$ just in case $a\in b$ and includes sufficiently many elements such that one can actually treat them in some respects like usual sets. That is where the countable, transitive, standard models come from.

  • $\begingroup$ If something is provable from the axioms, and we believe the axioms are true, then it follows that the thing that was proved is also true. The difficulty with CH is that there is no known extension of ZFC that both decides CH and is widely believed to be true. (There are several extensions that decide CH, though.) $\endgroup$ – Carl Mummert Jun 20 '10 at 14:30
  • $\begingroup$ "If something is provable from the axioms, and we believe the axioms are true, then it follows that the thing that was proved is also true." Did I write something contradicting this? $\endgroup$ – Michael Greinecker Jun 20 '10 at 14:35
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    $\begingroup$ I was responding to the sentence, "... we cannot learn from them that something is true, we can only use them to show something is not provable ...". $\endgroup$ – Carl Mummert Jun 20 '10 at 14:37

CH has not been "settled" (and there are obstacles to settling it) in any of the following senses:

$\quad 1.$ Finding a compellingly natural extension of standard set theory (more natural than ZF+CH) that decides CH, i.e., proves CH or proves its negation.

Here the main approach is blocked, because large cardinal axioms don't directly decide CH.

$\quad 2.$ Finding compelling arguments for replacing set theory, wherever it is used (e.g., as a foundation or formalization scheme), with set-theory-plus-CH.

This approach is blocked by the lack of "material consequences" of CH. For example, the set of true first-order sentences of arithmetic is not affected by assuming CH, so there would be no concrete statement such as the Twin Prime Conjecture that could be proved only with the use of CH. For similar reasons, it is unlikely that there exists a proof of any concrete statement that is much shorter or easier with CH than without it.

$\quad 3.$ Finding a compellingly natural alternative to standard axiomatic set theory (one whose theorems are not a subset or superset of the theorems in ZFC, and which comes to be preferred over ZFC) that can formulate and decide CH.

This development would be a lot more significant than deciding CH, and would presumably affect a large number of other questions. So to the extent that this possibility is relevant it should be discussed directly, and CH itself is immaterial. More on this argument in the earlier thread: Knuth's intuition that Goldbach might be unprovable

(The same comments also apply to the negation of the Continuum Hypothesis; everything above is phrased in terms of CH only to avoid clunky qualifiers in the sentences.)

EDIT: I am not counting another possibility, where partial answers to CH are accepted as the best that can be done, or the original problem comes to be seen as the wrong formulation (but better formulations are decidable in ZFC). For example, there are theorems to the effect that "any reasonably defined set of real numbers satisfies CH", and PCF theory that tries to capture the ZFC content of set theoretic cardinality questions while avoiding independence phenomena. For purposes of this answer I refer only to approaches that would "settle" CH by exhibiting a formal system that is strong enough to derive CH or its negation, and is also adequate in other respects, such as being extremely psychologically or pragmatically compelling compared to systems (such as ZFC) that don't decide CH.

  • $\begingroup$ I should also mention that the lack of first-order arithmetic consequences of CH (and the similar lack for not(CH)) means that there is no way of settling CH by deducing it from concrete statements such as the Goldbach conjecture or Twin Primes. $\endgroup$ – T.. Jun 20 '10 at 20:08

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