Timeline for Set-theoretic geology: controlled erosion?
Current License: CC BY-SA 4.0
39 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2020 at 13:38 | vote | accept | Mirco A. Mannucci | ||
| Sep 2, 2020 at 17:40 | comment | added | Mirco A. Mannucci | You like cheap shots, don't you? No I do not claim that. What I claim is this: assume you start with $M =L[0^\#, x]$. You want to erode x from M, presumably reducing it to $M =L[0^\#]$ . Now, let us call $ $M_0 =L[0^\#]$. Now I want somehow reach M_0 from below. I believe that , one can express M_0 as a union of forcing extension , NOT as any one of them . PS This is my last reply to you as a comment: either you post an answer, I promise that if it is well written I will give you my vote, or light sabers in private chat | |
| Sep 2, 2020 at 16:43 | comment | added | Asaf Karagila♦ | Wait, are you claiming that $0^\#$ lies in some set generic extension of $L$? | |
| Sep 2, 2020 at 13:22 | comment | added | Mirco A. Mannucci | emendation: the set of all extensions of L which do not contain A AND such that their are NOT A-grounds (in the sense specified by The answer) | |
| Sep 2, 2020 at 13:21 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 24 characters in body
|
| Sep 2, 2020 at 13:16 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 24 characters in body
|
| Sep 2, 2020 at 12:14 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 51 characters in body
|
| Sep 2, 2020 at 12:09 | comment | added | Mirco A. Mannucci | now, your counterexample: let us say that we work in $M =L[0^\#, x]$ without knowing that it is a forcing extension of $M =L[0^\#]$ and the devil chooses x. I consider the set of all submodels of M which are generic extensions of L and which omit x, ordered by inclusion. This set is not empty (trivial). Assume I take the UNION of all of them. The union is exactly your friend, namely $M =L[0^\#]$. | |
| Sep 2, 2020 at 12:01 | comment | added | Mirco A. Mannucci | Asaf, to begin with, even if your counterexample worked, it would simply imply that in SOME cases my bottom up method does not work, not that it "can't work". Basic logi. Secondly, if you have an answer to this question which goes beyond what Jonas has already put forward, I strongly invite you to post it. MO has already complained that our back and forth goes beyond the standards of acceptable comments, and suggested continuing the conversation in chat. If you wish to do it, bring the laser saber... :) | |
| Sep 2, 2020 at 8:42 | comment | added | Asaf Karagila♦ | Your bottom up really can't work. Suppose that $x$ is a generic real over $L[0^\#]$, make it even minimal, just for kicks. Now you want to remove $x$ from $L[0^\#,x]$. By minimality we even know what the result must be, $L[0^\#]$. But if you start with $L$, no matter what you do, you will never, ever, ever, reach $L[0^\#]$ by adding more and more generics. | |
| Sep 1, 2020 at 20:53 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 1476 characters in body
|
| Aug 31, 2020 at 19:42 | comment | added | Mirco A. Mannucci | Two cases: either this fat submodel is M itself, in which case zero sharp is NOT erodable, or it is a proper submodel of M. In that case, I call it M_0 and add zero sharp. PS I would bet that the envelope of the lattice is M itself | |
| Aug 31, 2020 at 19:42 | comment | added | Mirco A. Mannucci | Answer: I would not erode it :) Rather, assuming there is a CORE (not a mantle, something much more solid, such a L itself restricted to M ) I would begin growing the core in all possible forcing directions (think of a directed lattice of extensions). Now, if I understand you (remember you are the set theorist, not I) , zero sharp is in none of the extensions in the lattice. What about then taking the minimal submodel of M which contains ALL the forcing extensions? | |
| Aug 31, 2020 at 19:24 | comment | added | Asaf Karagila♦ | So how would you erode $0^\#$ from $L[0^\#]$? | |
| Aug 31, 2020 at 15:52 | answer | added | jonasreitz | timeline score: 9 | |
| Aug 31, 2020 at 10:54 | comment | added | Mirco A. Mannucci | I see your point now. Well done. But, rather than deter me, this makes me even more confident that if you want to erode without taking out "too much", the best strategy would be to start from some basic ground, grow it in all possible ways as long as it does stay inside M, and just try to detect where and when G finally pops up. | |
| Aug 31, 2020 at 10:08 | comment | added | Asaf Karagila♦ | But you removed too much. You can keep the subset of the Cohen real with the even coordinates. You can even save a large number of subsets, provided you're willing to forego the axiom of choice. | |
| Aug 31, 2020 at 0:55 | comment | added | Mirco A. Mannucci | If by that you mean M = L[c] then obviously the case G= c is trivial: Just define M_0 = L (relativized to M) and we are done. If G is not c, then it is not so trivial, at least to me. Intuitively, the simple minded approach would work is G has , so to speak, the same definability strength of c.: c is in L[G] and conversely. | |
| Aug 31, 2020 at 0:04 | comment | added | Asaf Karagila♦ | Well, then let's go to the case of $L[c]$ where $c$ is a Cohen real. | |
| Aug 30, 2020 at 23:01 | comment | added | Mirco A. Mannucci | @Asaf, sorry I was not clear. What I meant is not to erode till I get to L, rather, I want to have a ground INSIDE M to start my campaign, and if this ground exists, grow it (add forcing layers) in the hope that at some point I stumble upon my selected G. Notice that assuming that there is a ground, there are many ascending chains of forcing extensions, and so I will have to be able to see if G gets intercepted in any of them. Makes sense? | |
| Aug 30, 2020 at 22:53 | comment | added | Asaf Karagila♦ | @Mirco: I don't really understand your question. My point is that there is no maximal model without $0^\#$. You said that you want to avoid removing "too many sets". Going all the way down to $L$ is going to remove a lot of sets that you could have kept. | |
| Aug 30, 2020 at 21:02 | comment | added | Mirco A. Mannucci | @MonroeEskew as I mentioned in the question, I am a newbie in Set Theoretical Geology, but I do know a few fact, like the ones you mentioned (not their proof!) by reading a nice slides intro. Definitely relevant. I hope some expert will chime in (included you and Asaf) to tell me what can be done and what is unknown. Meanwhile, thanks to you both for the useful comments | |
| Aug 30, 2020 at 20:59 | comment | added | Mirco A. Mannucci | Asaf, I apologize for my ignorance, but are you saying that if I start from $M =L[0^\#]$ L truncated to M is not a model of ZF? | |
| Aug 30, 2020 at 20:38 | comment | added | Monroe Eskew | You might find the notion of grounds and mantle relevant here. It’s a nontrivial result that the statement “V is a set-generic extension of some inner model” is actually first-order expressible, and moreover, the intersection of all such inner models, the mantle, is also first-order definable. Usuba proved that the intersection of set-many grounds is itself a ground. | |
| Aug 30, 2020 at 20:33 | comment | added | Asaf Karagila♦ | In general, there's no reason to expect a minimal model. In $L[0^\#]$, for example, I don't believe there is a maximal model without $0^\#$ itself. Or even adding a Cohen real, there's no reason to expect a maximal model where "only the Cohen real is missing", because you will therefore still want the Cohen real restricted to various subsets of $\omega$. | |
| Aug 30, 2020 at 19:44 | comment | added | Mirco A. Mannucci | emendation to what I have said: actually, if G appears at some point of the expansion, it is certainly removable, but unless it is so to speak in the outermost boundaries of the expansion, it will not qualify to describe M as M_0[G] for some M_0 submodel of M. It is, so to speak, "buried inside some layer of M". | |
| Aug 30, 2020 at 19:13 | comment | added | Mirco A. Mannucci | elaborating this idea a bit further: M_minimal , M1, M2,..... etc all forcing extensions of the previous one. What one would like to see is whether our G is reachable by this chain.Even if it is, though, there is an issue: the extension which contains G may not reach the "crust" of M... | |
| Aug 30, 2020 at 19:10 | comment | added | Mirco A. Mannucci | @AsafKaragila Todah Rabbah! Yes, it is related, especially your answer there. Now, problem is, we do not know if there is an inner model inside, although in fact we do (the constructibel minimal model) . Now, M_minimal is definitely inside M, but perhaps it is too "thin". But we may be able to salvage your idea: how about expanding M_minimal by forcing extensions all the while remaining inside M? If we are lucky, at some point we may meet G, in which case we win. But I suspect there are situations in which there is an ascending chain inside M, and none contains G... | |
| Aug 30, 2020 at 18:20 | comment | added | Asaf Karagila♦ | Possibly related: math.stackexchange.com/q/291088/622 | |
| Aug 29, 2020 at 21:42 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 35 characters in body
|
| Aug 29, 2020 at 21:35 | history | edited | Andrés E. Caicedo | CC BY-SA 4.0 |
added 1 character in body; edited title
|
| Aug 29, 2020 at 21:26 | comment | added | Mirco A. Mannucci | Post Scriptum I am sure you already see where I am trying to go. The inverse of forcing, or "selective erosion"< assuming of course that is feasible and not trivial, would play the reversal role of forcing: rather than judiciously "fattening" a model a bit, and add new facts unknown to the ground model (but "true" outside) here you would like to have a method for removing facts from a model. Some kind of inner model construction to obliterate some facts from your initial model | |
| Aug 29, 2020 at 21:11 | comment | added | Mirco A. Mannucci | take out also a minimal set of sets in M which are related to G, in such as way that after that I end up with a transitive submodel of M of the same height and moreover if I throw in G back, close with respect to definable operations, again BANG! I get M. Now, it is clear to me that tons of Gs are not yankable, right? But some possibly are. So, again, think of this as putting a MINUS sign to forcing: M+ G = M[G], now the equatin would be : M -G is a M_0 such that M_0[G] = M. Kind of an inverse operation | |
| Aug 29, 2020 at 21:06 | comment | added | Mirco A. Mannucci | If you weren't here already to help me either tuning my often sloppy questions, or "yank them out" altogether, I should pay you to do so. So, let me be a bit more precise: YES, the idea is to get M_0 as close as possible to M (hence the name, I could have as well called judicious surgery) Secondly, I do not just want to "peel off" anything, rather suppose taht someone (say the devil) comes up to you and say: Andreas, you have this wonderful M. Let me pick a set in it, say G. Now, you have to tell me if this set is yankable or not, meaning that I can yank it out, | |
| Aug 29, 2020 at 21:01 | comment | added | Andreas Blass | Your question and the terminology "selective erosion" suggest that you want the submodel $M_0$ to be in some sense close to $M$. Otherwise, you could simply say that $G$ is yankable iff it's not in the minimal model, and the result of yanking is the minimal model. But in what sense should $M_0$ be close to $M$? The same ordinals? Then you can yank any $G$ that isn't in $L^{(M)}$. Maybe $M$ should be a forcing-extension of $M_0$? More specifically, a forcing extension by re-adjoining $G$? Or maybe $M$ should satisfy some form of the covering lemma over $M_0$? Or maybe $\dots$? | |
| Aug 29, 2020 at 19:44 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 126 characters in body
|
| Aug 29, 2020 at 19:40 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals
|
| Aug 29, 2020 at 19:25 | history | edited | Mirco A. Mannucci | CC BY-SA 4.0 |
added 2 characters in body
|
| Aug 29, 2020 at 19:14 | history | asked | Mirco A. Mannucci | CC BY-SA 4.0 |