Timeline for Can only the constructible sets be proven to exist in $ZF$ without benefit of extra assumptions?
Current License: CC BY-SA 3.0
27 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2017 at 9:08 | comment | added | Asaf Karagila♦ | @Stefan: I don't think it has no value, it has some value. But I understand the deletion votes. I don't object to them, but I won't partake in the process. If any other set theorist feels that this should be deleted, so be it. | |
| Oct 2, 2017 at 9:06 | comment | added | Stefan Kohl♦ | The question is closed and has negative score, and there remains only one vote for deletion of the thread. At least superficially, the thread and the discussion in the comments don't seem to be of lasting value for readers, though I cannot really judge the contents. As your answer has positive score, would you object deletion? -- If yes, you might consider voting to reopen the question. | |
| Aug 27, 2017 at 9:22 | vote | accept | Thomas Benjamin | ||
| Aug 25, 2017 at 21:50 | comment | added | Noah Schweber | @ThomasBenjamin What we should call $\mathcal{P}_{def}$, or what the phrase "predicative powerset" means, is irrelevant to the mathematics; do you understand why $\mathcal{P}_{def}$ cannot be the same as $\mathcal{P}$ in any model of ZF (including one satisfying "V=L")? And even this doesn't really have anything to do with your question, which Asaf has (I also have, but more confusingly) thoroughly addressed. You are trying to form a big web of ideas, but you still don't understand the basics and wind up mixing several confusions. You should stick to your question, which has been answered. | |
| Aug 25, 2017 at 18:53 | comment | added | Thomas Benjamin | @AsafKaragila: Please reread pp. 85-87 of Paul Cohen's _Set Theory and the Continuum Hypothesis _ for (at least his) motivation as to the validity of defining $\mathscr P_{Def}$ as a 'predicative power-set' (although Cohen actually writes, "...it was G$\ddot o$del's idea that if we iterate predicative constructions up to any ordinal the resulting sets could furnish an adequate model for $ZF$.") If $\mathscr P_{Def}$ is not a 'power-set operation', what would you would define it as ? | |
| Aug 25, 2017 at 18:31 | comment | added | Asaf Karagila♦ | @Thomas: Until you actually suggest a predicative power set operation, this is all moot. Yes, any operation which is a reasonable "power set" one would not be $\mathcal P_{\rm def}$. At the very least, it would have to capture all the subsets in one shot, otherwise it is not quite a power set operation. You seem to continuously ignore this, despite this being pointed out by several people. So I am going to reserve my energy and do something else, and I encourage the others to do so as well. | |
| Aug 25, 2017 at 18:28 | comment | added | Thomas Benjamin | @NoahSchweber: And yet it is defined as such--it is called the "predicative" power-set operation. If one was able to define a predicative version of $ZF(C)$, would this version of the power-set operation be considerably different? Also, would $\mathscr P_{Def}$ = $\mathscr P$ for $ZF$ $-$ Infinity? | |
| Aug 25, 2017 at 16:24 | comment | added | Noah Schweber | @ThomasBenjamin What constitutes a powerset-like notion? There is certainly no model of ZF in which $\mathcal{P}_{def}$ is the actual powerset operation - indeed, ZF proves that as long as $A$ is infinite and $S$ is any set containing $A$, there will be subsets of $A$ which are not definable in $(S, \in)$ with parameters from $A\cup\{A\}$. And as usual ZF is massive overkill for this. The powerset operation in $L$ does indeedcome from a transfinite iteration of $\mathcal{P}_{def}$, but this is quite different from the idea that $\mathcal{P}_{def}=\mathcal{P}$, which is ZF-provably false. | |
| Aug 25, 2017 at 14:38 | comment | added | Thomas Benjamin | @AsafKaragila: Do you believe that $\mathscr P_{Def}$ should be considered a 'power-set' at all, due to the observation given in your answer? | |
| Aug 17, 2017 at 5:21 | comment | added | Asaf Karagila♦ | @Monroe: Ah, okay. Then there is no clash between what we said... (Which unfortunately means that we cannot publish a paper about ZFC being inconsistent, well, at least not because of this argument... ;-)) | |
| Aug 17, 2017 at 5:18 | comment | added | Monroe Eskew | I was just guessing that the idea that we always construct new reals might be based on the internal undefinability of truth, and saying this doesn't do it. My example was meant to say, certainly the theory of $L_\alpha$ is in $L_\beta$, but surprisingly it can be in $L_\alpha$ too. | |
| Aug 17, 2017 at 5:15 | comment | added | Asaf Karagila♦ | @Monroe: I must have misunderstood you, then. Because if $L_\alpha\prec L_{\omega_1}$, then there is some $\beta$ large enough such that the theory of $L_\alpha$ lies in $L_\beta$. That was my point. I guess I just didn't fully understand yours. | |
| Aug 16, 2017 at 22:53 | comment | added | Monroe Eskew | @Asaf what did I say that's wrong? | |
| Aug 16, 2017 at 22:52 | comment | added | Asaf Karagila♦ | @Monroe: Wait, so how can there be a club of elementary submodels of $L_{\omega_1}$? Since obviously $L_{\omega_1}$ knows its own theory (it's just not aware that this is its own theory). I mean, it's a real, and at some point it will go into the $L$ hierarchy, and there will be a larger model which has that theory. So what you're saying can't be fully right, or I might be misunderstanding it. | |
| Aug 16, 2017 at 22:49 | comment | added | Monroe Eskew | @Asaf: It is somewhat counterintuitive. You'd think that if $\alpha < \beta$ are countable and each satisfies a strong enough theory, then the set of Godel numbers of true sentences in $L_\alpha$ would be in $L_\beta \setminus L_\alpha$, but apparently not. Of course by Tarski, this real cannot be definable in $L_\alpha$. | |
| Aug 16, 2017 at 22:42 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
added 19 characters in body
|
| Aug 16, 2017 at 22:41 | comment | added | Asaf Karagila♦ | @Monroe: I knew that this is probably false the moment I re-read the answer. But I was waiting for someone to come along and correct me, because I wasn't 100% sure if it is in fact false or not. My point, of course, is that no countable step captures the entire power set of $\omega$. | |
| Aug 16, 2017 at 22:34 | comment | added | Monroe Eskew | @Andrés: Thanks for pointing that out. | |
| Aug 16, 2017 at 22:15 | comment | added | Andrés E. Caicedo | @Monroe Or simply take a countable elementary substructure of a large fragment of the universe and collapse. The behavior of gaps is actually quite interesting. | |
| Aug 16, 2017 at 22:08 | comment | added | Monroe Eskew | @AsafKaragila: Actually, new subsets of $\omega$ are not added at every successor step. Force with $\mathcal P(\omega_1)/\mathrm{NS}$, so that we get a generic elementary embedding $j : L \to M$ with critical point $\omega_1^L$. $M$ is well-founded up to $\omega_2^L$. Therefore, by absoluteness, it sees that there is a large gap after $\omega_1^L$ where no new subsets of $\omega$ are added. By reflection this happens a lot below $\omega_1^L$. | |
| Aug 16, 2017 at 17:57 | comment | added | Asaf Karagila♦ | @Thomas: They are definable class-functions. They can be seen as syntactic objects (as they should be), since they are classes (and not sets), and their properties are provable from ZF. The point is that "being constructible" is a property which, while syntactic, does not tell you much about whether certain sets satisfy this property of not (otherwise V=L would be provable or disprovable from ZF itself). | |
| Aug 16, 2017 at 17:54 | comment | added | Thomas Benjamin | @AsafKaragila: Thanks for the clarification. Are the G$\ddot o$del operations (out of which the constructible sets can be 'constructed') part of syntax or semantics (or both)? | |
| Aug 16, 2017 at 17:42 | comment | added | Asaf Karagila♦ | @Thomas: If ZF proves some set exists, then it proves that its power set exists; that its Hartogs and Lindenbaum numbers exist; that a bunch of other sets which have canonical definitions exist. The question has no "concrete" answer. Unlike PA, which has some nontrivial terms and can prove the existence of every natural number but not much more, ZF has only one relation symbol in the language of set theory so there are no non-trivial terms, and in particular there is no easy way to explain "which sets exist". Even more so, since "existence" is a semantic property, and provability is syntactic. | |
| Aug 16, 2017 at 17:38 | comment | added | Thomas Benjamin | @AsafKaragila: What sets can $ZF$ prove exist? And can $ZF$ prove some sets are constructible? "Some sets exist" is obviously not a question. I attempted to ask these questions, but in a different way. Apparently (somehow) I didn't manage to delete all of it and this must have been the remnant. Thanks for deleting this remnant for me--greatly appreciated. By the way, is that all $ZF$ can prove--that some sets exist? | |
| Aug 16, 2017 at 14:18 | comment | added | Asaf Karagila♦ | @Thomas: That some sets exist. I don't understand your question. And frankly, I think that you don't understand your question either. | |
| Aug 14, 2017 at 16:46 | comment | added | Noah Schweber | +1. To clarify your last paragraph for the OP with an example: note that every real in $L_{\omega+1}$ is arithmetic; however, $0^{(\omega)}$ is a non-arithmetic real which is in $L_{\omega+2}$. | |
| Aug 14, 2017 at 15:36 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |