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
18 events
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| Oct 2, 2017 at 9:07 | 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 26, 2017 at 0:01 | comment | added | Todd Trimble | @AsafKaragila and Andrés: Oh sheesh. Yes, got it. Plain as day now. | |
| Aug 25, 2017 at 21:29 | comment | added | Asaf Karagila♦ | [...] for a simpler example one could take $x$ to be $\mathcal P(\alpha)$ for the least ordinal $\alpha$ such that L does not compute its power set correctly; or $0$ otherwise. In other words, either $x$ is empty, or it is the least non-constructible power set of an ordinal. Certainly ZF proves that $x$ exists and that it is a unique object in the universe. But $x$ is empty if and only if V=L is true. (This is a continuation to my previous comment addressed to @Todd...) | |
| Aug 25, 2017 at 21:29 | comment | added | Noah Schweber | Oh jeez, my bad - AndrésE.Caicedo is totally right, I screwed up re: measurable cardinals. Forget I said that. I am tempted to delete this answer given how much confusion it has caused - the right answer is absolutely Asaf's. @ThomasBenjamin Asaf's is the answer you should be reading - although this one is also basically correct (I think my comment re: measurables is the only error - and that can be easily fixed, e.g. $0^\sharp$ implies "there is a nonconstructible set" which V=L clearly contradicts), it will be more confusing than helpful given your current understanding of the issues. | |
| Aug 25, 2017 at 21:24 | comment | added | Asaf Karagila♦ | @Todd: As Andrés points out, taking the power set of $\omega$ works as an example, which is something I have pointed out a ridiculous number of times on this page by now in my answer and in the various comment threads. I agree that the issue with $0^\#$ is probably more confusing than illuminating here without the technical understanding that would make the OP's question moot. The point in Noah's answer, I think, is that there are many examples of sets which have "robust definition" and consistently their existence outright refutes V=L. [...] | |
| Aug 25, 2017 at 20:25 | comment | added | Andrés E. Caicedo | @Todd " what I'm really interested in seeing is an explicit example " That has been addressed explicitly several times: $x=\mathcal P(\omega)$ is an obvious example. It was pointed out at the beginning of the thread and in the answers. The issue about definability of $0^\sharp$ will probably be much more confusing than illuminating in this context. If you are interested, I suggest you ask it as a separate question. | |
| Aug 25, 2017 at 20:06 | comment | added | Todd Trimble | @AndrésE.Caicedo (and Noah and Asaf): Hi guys. The crediting to me in Thomas's first comment comes from a discussion currently at meta about this question. I would be interested though in a discussion of this point about defining $0^\sharp$, which isn't directed just to Thomas but to the community. Or, what I'm really interested in seeing is an explicit example of Asaf's assertion "there are sets which provably exist, but it is not provable that they are constructible" -- because I got the sense that something like this is what Thomas wants. Thanks. | |
| Aug 25, 2017 at 19:08 | comment | added | Andrés E. Caicedo | The second part of the comment is the unreasonable one. I answered the first part. | |
| Aug 25, 2017 at 19:04 | comment | added | Thomas Benjamin | @AndrésE.Caicedo: Is the second part of my previous comment also unreasonable? | |
| Aug 25, 2017 at 18:24 | comment | added | Andrés E. Caicedo | No, Thomas, that is not a reasonable question. The notion of $L$-measurable cardinal is nonstandard. It is a made-up word I used to refer to $L$-indiscernibles that would convey the same feeling that the informal ``there is a measurable cardinal'' in Noah's comment was meant to convey. | |
| Aug 25, 2017 at 18:21 | comment | added | Thomas Benjamin | @AndrésE.Caicedo: Please define the notion of $L$-measurable cardinal. Also, please provide the motivation(s) (mathematical and philisophical) for assuming the existence of $0^{\sharp}$ outright (as in $ZF(C)$ + "$0^{\sharp}$ exists". | |
| Aug 25, 2017 at 17:44 | comment | added | Andrés E. Caicedo | @Noah No, it contains "there is an $L$-measurable cardinal", or some such. But that is too informal. Thomas does not appear too clear yet on how weaker in consistency strength $0^\sharp$ is from a measurable, so that comment would probably increase confusion. | |
| Aug 25, 2017 at 16:17 | comment | added | Noah Schweber | @ThomasBenjamin Yes, it is definable - in fact, it's a $\Pi^1_2$ singleton. This is because it doesn't code truth in $V$, but rather in $L$, and it only exists if $V\not=L$ (and indeed if the theory of $V$ is quite different from that of $L$ - in particular, if $0^\sharp$ exists then the theory of $V$ contains "there is a measurable cardinal" while of course the theory of $L$ can't!). So there is no way that $0^\sharp$ exists and describes truth in $V$. | |
| Aug 25, 2017 at 14:34 | comment | added | Thomas Benjamin | @NoahSchweber: Can $0^{\sharp}$ even be definable in $ZF$ (or $ZFC$) due to Tarski's result on the undefinability of truth (thanks to Todd Trimble for this observation)? | |
| Aug 17, 2017 at 2:29 | comment | added | Andreas Blass | I think the OP would benefit from thinking carefully about what (if anything) is meant by saying, of a set $S$ (as opposed to something like a definition of $S$), that "ZF proves that $S$ exists." | |
| Aug 14, 2017 at 17:35 | comment | added | Noah Schweber | @AsafKaragila Derp. Well, when you have indiscernibles, everything looks like . . . something that needs not to be discerned? I'unno. | |
| Aug 14, 2017 at 17:08 | comment | added | Asaf Karagila♦ | Also, take $\varphi(x)$ to be "$x$ is the power set of $\omega$" works. Which is what I was driving at. And you don't even need to exceed the consistency strength of ZFC for that either. :P | |
| Aug 14, 2017 at 16:53 | history | answered | Noah Schweber | CC BY-SA 3.0 |