Timeline for Smallest $\beta$ such that it is provable that $2^{\aleph_\beta} > 2^{\aleph_0}$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2019 at 12:29 | vote | accept | Dominic van der Zypen | ||
| Jan 17, 2019 at 10:38 | comment | added | user57888 | I think one natural reformulation of the question is a s follows: Look at the set of all formulae $\phi$ such that provably in ZFC they define a cardinal and ZFC proves $\forall x \ \ \phi(x) \rightarrow x > 2^{\aleph_0}$. On these formulae you have a partial order: $\phi \leq \psi$ if provably in ZFC the inequality holds between the elements defined with the formulae. Note that this is not a linear order and it isn't well-founded. Does this set have a minimum? Can you characterise it? As I see it, it has nothing to do with Solovay's theorem. | |
| Jan 17, 2019 at 3:30 | review | Close votes | |||
| Jan 18, 2019 at 10:31 | |||||
| Jan 16, 2019 at 18:27 | answer | added | Andrés E. Caicedo | timeline score: 17 | |
| Jan 16, 2019 at 17:33 | comment | added | Wojowu | @YCor Over all models of ZFC it doesn't quite make sense, but we can fix a model of ZFC and restrict attention to models with the same cardinals. Then Easton's theorem implies that among forcing extensions, which I believe (in this case) do not collapse any cardinals, continuum can be arbitrarily large, and all ordinals up to it can have the same cardinality of the power set. | |
| Jan 16, 2019 at 17:07 | comment | added | YCor | @user57888 thanks, you're right. So the question sounds strictly equivalent to (in ZFC): let $\gamma$ be the min of the (nonempty) set of cardinals $\alpha\le 2^{\aleph_0}$ such that $2^\alpha>c$. What is $\gamma$ (or, writing $\gamma=\aleph_\delta$, what is $\delta$)? Under GCH, $\delta=1$, and there are models of ZFC for which $\delta>0$. It sounds like the intended question somewhat is "what is the sup of $\delta$ over all models of ZFC?" and I'm not sure this makes sense. | |
| Jan 16, 2019 at 14:50 | comment | added | user57888 | @YCor "Such that it is provable in ZFC" obviously is a formula of ZFC. Although I do agree that the question could be clarified a bit. | |
| Jan 16, 2019 at 10:56 | comment | added | YCor | Just an analogy: I recently read a paper (published in a serious journal) in which the authors apply Cohen's results on independence of GCH as "for every cardinal $\kappa$ there exists a model of ZFC in which $\kappa<2^{\aleph_0}$". Applying this to $\kappa=2^{\aleph_0}$ results in an obvious paradox, showing that some care is needed is formulating those results. | |
| Jan 16, 2019 at 10:48 | comment | added | YCor | The question is ill-posed. It asks about min of a certain "collection" of cardinals. But this "collection" is not definable in ZFC. I mean, the scheme of specification cannot be applied, because "such that it is provable in ZFC" is not a formula of ZFC. Maybe logicians would interpret the question by allowing some embedding between models, but it's not currently asked this way. | |
| Jan 16, 2019 at 9:52 | comment | added | Wojowu | I guess $\beta=2^{\aleph_0}$ is one such answer. The question is a bit subtle though - ZFC can't directly talk about a statement for arbitrary ordinal, we need the ordinal to be defined by some formula. And the formula can essentially say "$\beta$ is the least such that $2^{\aleph_\beta}>2^{\aleph_0}$"... One statement Asaf (I presume) refers to is Easton's theorem which is a model-theoretic in nature, but should answer some reasonable version of this question. | |
| Jan 16, 2019 at 9:37 | comment | added | Asaf Karagila♦ | There is no such $\beta$. All the many theorems about the size of the continuum say exactly otherwise. | |
| Jan 16, 2019 at 8:40 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |