Why is Cohen's result insufficient to settle CH? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:29:58Z http://mathoverflow.net/feeds/question/28806 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28806/why-is-cohens-result-insufficient-to-settle-ch Why is Cohen's result insufficient to settle CH? Zirui Wang 2010-06-20T05:06:46Z 2010-06-20T19:02:38Z <p>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?</p> http://mathoverflow.net/questions/28806/why-is-cohens-result-insufficient-to-settle-ch/28836#28836 Answer by Carl Mummert for Why is Cohen's result insufficient to settle CH? Carl Mummert 2010-06-20T12:50:31Z 2010-06-20T13:23:49Z <p>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. </p> <p>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 <em>cannot</em> 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. </p> <p>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. </p> <p>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. </p> <p>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: <a href="http://mathoverflow.net/questions/23829/solutions-to-the-continuum-hypothesis" rel="nofollow">http://mathoverflow.net/questions/23829/solutions-to-the-continuum-hypothesis</a></p> http://mathoverflow.net/questions/28806/why-is-cohens-result-insufficient-to-settle-ch/28837#28837 Answer by Joel David Hamkins for Why is Cohen's result insufficient to settle CH? Joel David Hamkins 2010-06-20T12:54:40Z 2010-06-20T12:54:40Z <p>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.</p> <p>I view the question as arising from the following line of thought: <em>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?</em></p> <p>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 + &not;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. </p> <p>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. </p> http://mathoverflow.net/questions/28806/why-is-cohens-result-insufficient-to-settle-ch/28844#28844 Answer by Michael Greinecker for Why is Cohen's result insufficient to settle CH? Michael Greinecker 2010-06-20T14:08:17Z 2010-06-20T14:43:12Z <p>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.</p> <p>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.</p> <p>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.*</p> <p>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). </p> <hr> <p>*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. </p> http://mathoverflow.net/questions/28806/why-is-cohens-result-insufficient-to-settle-ch/28873#28873 Answer by T. for Why is Cohen's result insufficient to settle CH? T. 2010-06-20T18:32:58Z 2010-06-20T19:02:38Z <p>CH has not been "settled" (and there are obstacles to settling it) in any of the following senses:</p> <p>$\quad 1.$ 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.</p> <p>Here the main approach is blocked, because large cardinal axioms don't directly decide CH.</p> <p>$\quad 2.$ Finding compelling arguments for replacing set theory, wherever it is used (e.g., as a foundation or formalization scheme), with set-theory-plus-CH.</p> <p>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.</p> <p>$\quad 3.$ 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. </p> <p>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: <a href="http://mathoverflow.net/questions/27755/knuths-intuition-that-goldbach-might-be-unprovable/27914#27914" rel="nofollow">http://mathoverflow.net/questions/27755/knuths-intuition-that-goldbach-might-be-unprovable/27914#27914</a></p> <p>(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.) </p> <p>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.</p>