Timeline for Why is Cohen's result insufficient to settle CH?
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12 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2010 at 2:31 | comment | added | Carl Mummert | Indeed; this is one reason set theory is not a universal fix-it for worries about the foundations of mathematics. | |
| Jun 21, 2010 at 1:06 | comment | added | Joel David Hamkins | 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. | |
| Jun 20, 2010 at 20:38 | comment | added | Yaakov Baruch | @Carl: I think this last comment of yours also answers my question, i.e. that the notion of what constitutes a sequence of rationals (or an element of Z[[x]], or a real) is already model dependent, not "objective" in the way N is. I'm hesitant to post my question above as a separate one because I can't tell to what extent it duplicates other questions/answers on CH. Anyone who would like to post it: feel free to go ahead. | |
| Jun 20, 2010 at 17:18 | comment | added | Carl Mummert | @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 ). | |
| Jun 20, 2010 at 17:12 | comment | added | Carl Mummert | @Yaakov Baruch: That is a good question, but there is not enough room in these comments to answer it. I think it would work better as a separate question, particularly because I think there is more than one possible answer. | |
| Jun 20, 2010 at 16:56 | comment | added | Jason DeVito - on hiatus | @Yaakov, This is probably a misguided answer, but here is my understanding (I'm a grad student geometer who occasionally dabbles in logic): What usually happens is that "unprovable" statements about natural numbers are unprovable assuming the axioms of Peano Arithmetic, but ARE provable instead assuming ZFC (which also proves all of the Peano Arithmetic axioms). I know that this is the case concerning, e.g., the Goodstein Sequence. | |
| Jun 20, 2010 at 16:00 | comment | added | Boyarsky | Carl, is your point about the "set of real numbers" changing simply a consequence of the definition in terms of "equivalence classes of Cauchy sequences of rationals" (so may get new Cauchy sequences, etc.)? It is akin to saying that the ring $\mathbb{Z}[[x]]$ (or equivalently, the group $\prod_{n=0}^{\infty} \mathbb{Z}$) depends on the model of set theory since it is the "set of maps" from $\mathbb{Z}_ {\ge 0}$ to $\mathbb{Z}$. This is distressing, to say the least. | |
| Jun 20, 2010 at 15:11 | comment | added | Yaakov Baruch | 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"? | |
| Jun 20, 2010 at 13:48 | comment | added | Carl Mummert | 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. | |
| Jun 20, 2010 at 13:41 | comment | added | Joel David Hamkins | 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. | |
| Jun 20, 2010 at 13:14 | comment | added | Boyarsky | Would it be accurate to say that the way functions between sets (e.g., bijections) seem to pop in and out of existence depending on the model of set theory is because functions from $A$ to $B$ are (by definition...) certain kinds of subsets of $A \times B$? So if you "enlarge" your model of set theory (whatever that means) then $A \times B$ may acquire more subsets and in particular there may arise a bijection between them which wasn't there before? In this way, some weird subset of $\mathbb{R}$ with intermediate cardinality before can become bijective with $\mathbb{R}$ in the bigger model? | |
| Jun 20, 2010 at 12:54 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |