Timeline for Can only the constructible sets be proven to exist in $ZF$ without benefit of extra assumptions? [closed]
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Sep 3, 2017 at 5:06 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 3, 2017 at 5:00 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 3, 2017 at 4:50 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 2, 2017 at 5:55 | review | Reopen votes | |||
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| Sep 1, 2017 at 13:47 | history | closed |
Emil Jeřábek R.P. Jan-Christoph Schlage-Puchta Henry.L Yoav Kallus |
Needs details or clarity | |
| Sep 1, 2017 at 12:23 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 1, 2017 at 12:13 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 1, 2017 at 12:02 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Sep 1, 2017 at 3:03 | review | Close votes | |||
| Sep 1, 2017 at 13:47 | |||||
| Aug 29, 2017 at 13:26 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Aug 28, 2017 at 10:26 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Aug 27, 2017 at 9:22 | vote | accept | Thomas Benjamin | ||
| Aug 25, 2017 at 12:02 | history | edited | Todd Trimble | CC BY-SA 3.0 |
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| Aug 24, 2017 at 19:25 | comment | added | Thomas Benjamin | @Burak: That is indeed helpful. Thank you. | |
| Aug 24, 2017 at 19:18 | comment | added | Burak | @ThomasBenjamin: I don't understand what you mean by "introducing operations as supplementary axioms". That does not really make sense. If you are talking about adding each sentence that specifies a property of a Gödel operation, ZF can already prove these, so you don't really have to add anything. I don't really understand what you are after but my impression is that, once you are able to precisely state what you are thinking, you will end up getting the answers "No, ZFC cannot prove the existence of non-constructible sets. Yes, the constructible sets are the smallest inner model". | |
| Aug 24, 2017 at 16:14 | comment | added | Thomas Benjamin | @Burak: So it would seem that to make that formula non-trivial, one might need to use the G$\ddot o$del operations to construct constructible sets from the empty set (or use $\mathscr P_{Def}$, of course). For my part, I like the G$\ddot o$del operations, but to make things syntactically correct should I introduce them as supplementary axioms (or can $\mathscr F_1$,..., $\mathscr F_{10}$ be derived as theorems from $ZF$)? What to do, what to do.... | |
| Aug 24, 2017 at 15:51 | comment | added | Burak | That $\forall x \exists \alpha \in Ord \ x \in L_{\alpha}$ is literally the statement V=L and is not provable from ZFC, provided that it is consistent. That $\exists x \exists \alpha \in Ord \ x \in L_{\alpha}$ is trivially true as the empty set belongs to L. | |
| Aug 24, 2017 at 15:29 | comment | added | Thomas Benjamin | @Burak: Here is the formula I wish to have $ZF$ prove: ($\exists$$x$)($\exists$$\alpha$$\in$ $Ord$)($x$$\in$$L_{\alpha}$). Also, please answer my question regarding the G$\ddot o$del operations. Thanks. | |
| Aug 23, 2017 at 3:03 | review | Close votes | |||
| Aug 23, 2017 at 9:07 | |||||
| Aug 22, 2017 at 6:30 | comment | added | Thomas Benjamin | (Cont.) or (better put) would the proper way to add the G$\ddot o$del operations into $ZF$ would be to add them to $ZF$ as supplementary axioms? | |
| Aug 22, 2017 at 2:38 | comment | added | Thomas Benjamin | @Goldstern: I will try again to clarify. I'm sorry that I have been 'clear as mud' thus far. Question: Since the properties of the G$\ddot o$del operations are provable from $ZF$, is the proper way to introduce them into $ZF$ is to add them as supplementary axioms to $ZF$? | |
| Aug 18, 2017 at 19:08 | comment | added | Goldstern | I am sorry. To me it is "unclear what you are asking". | |
| Aug 18, 2017 at 14:50 | comment | added | Thomas Benjamin | (cont.) $\varphi$. | |
| Aug 18, 2017 at 14:49 | comment | added | Thomas Benjamin | (cont) models of $ZF$ ($L$, for example) in which $0^{\sharp}$ does not exist and models of $ZF$ (where measurable cardinals also 'exist', say) in which $0^{\sharp}$ 'exists'. I find it interesting that in every transitive model $\mathfrak M_{ZF}$ of $ZF$, $L$ is the smallest submodel of $\mathfrak M_{ZF}$. If one restricts the class of models of $ZF$ to only the transitive models, one could possibly use G$\ddot o$del's completeness theorem (since every trasitive model of $ZF$ has $L$ as a submodel) to relate provability with the existence of constructible sets. Am working on a suitable | |
| Aug 18, 2017 at 14:33 | comment | added | Thomas Benjamin | @Goldstern and Burak: Thank you for your comments. Perhaps in order to express the notion of 'exists' syntactically I would need to treat 'existence' as a predicate (there are logics in which 'exists' is a predicate, of course). But as it stands, if 'exists' is a semantic notion, one could define 'exists' as 'being an element of a universe $M$ of a model $\mathfrak M$ of some theory $T$ ($ZF$ in this case). So to say (for example) that '$0^{\sharp}$ exists' is to say that '$0^{\sharp}$ is an element of the universe $M$ of some model $\mathfrak M$ of $ZF$'. There are, of course, | |
| Aug 17, 2017 at 21:35 | comment | added | Goldstern | @ThomasBenjamin Regarding your comment that ZF is a "very incomplete description of sets": That is true. (And, after 50 years of experience with forcing, rather obvious.) | |
| Aug 17, 2017 at 21:33 | comment | added | Goldstern | @ThomasBenjamin 1. If you are asking whether "ZF proves 'there are constructible sets' ", then the question should be closed. 2. I cannot see a "syntactic notion of 'provably existent'" in your question. If you are talking about a specific syntactic notion, please write it down explicitly. | |
| Aug 17, 2017 at 21:10 | comment | added | Burak | @ThomasBenjamin: If you provide us with a formula $\varphi(x)$ in the language of set theory which you think captures your notion of "the set $x$ being proven to exist", one can easily answer your question. | |
| Aug 17, 2017 at 21:01 | comment | added | Burak | @ThomasBenjamin: Your axiom "all sets definable in ZF exist" does not make sense grammatically. Notice that you are trying to quantify over the same object with two different quantifiers, namely, "all" and "exist". How are you going to state this with a sentence in the language of set theory? An answer to this may help others understand what you mean by "sets proven to exist". (There are also issues regarding the notion "definable" because of Tarski's theorem on truth but let's ignore those for a second since there are notions of definability you can work with such as ordinal definable etc.) | |
| Aug 17, 2017 at 15:02 | comment | added | Thomas Benjamin | (cont) conflation at its worst? But if so, what sets exist? If $ZF$ only proves that some sets exist, then $ZF$ would seem to me to be a very incomplete description of sets. | |
| Aug 17, 2017 at 14:52 | comment | added | Thomas Benjamin | @Goldstern: $\varphi$ would be (in this case), "there are constructible sets". What I am attempting to do is, by analogy to provably recursive sets in $PA$, define a syntactic notion of 'provably existent' definable in $ZF$ and show that the 'provably existent' sets are precisely the constructible sets. Of course, one could assume the following axiom: "All sets definable in $ZF$ exist", but since $0^{\sharp}$ is definable in $ZF$(?), one has as an easy consequence of $ZF$ + "All sets definable in $ZF$ exist", $V$ $\neq$ $L$. But is equating existence with definability | |
| Aug 17, 2017 at 0:09 | comment | added | Goldstern | If you are asking about the ZF-provability of a specific statement $\varphi$ (such as "there are nonconstructible sets" or "all sets are constructible"), then please tell us what $\varphi$ is. (For the two examples I gave, the answers are "no" and "no", assuming that ZF is consistent.) | |
| Aug 16, 2017 at 14:19 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Aug 16, 2017 at 14:04 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Aug 16, 2017 at 13:51 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Aug 15, 2017 at 14:16 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
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| Aug 15, 2017 at 14:09 | comment | added | Thomas Benjamin | @AndrésE.Caicedo: I guess what I really want to ask is, Are the constructible sets the only sets that the axioms of $ZF$ alone can prove to exist? Also, $\mathscr P_{Def}$($x$) is just the predicative powerset operation used to form the constructible hierarchy discussed in Koepke's mini-course preprint. | |
| Aug 15, 2017 at 14:04 | comment | added | Thomas Benjamin | (cont.) alone, without benefit of extra assumptions. Does this help at all? | |
| Aug 15, 2017 at 13:58 | comment | added | Thomas Benjamin | @AndrésE.Caicedo: I have corrected the false basic distinction (hopefully). Yes, it is something more subtle (I think). Perhaps I should start with the question, "What sets can be proven to exist in $ZF$". As is well known, the existence of a non-constructible set of integers ($0^{\sharp}$) can only be proven by assuming the existence of a measurable cardinal. Though a model of $ZF$ + "There exists a measurable cardinal" is certainly a model of $ZF$, it assumes something extra--the existence of a measurable cardinal. What I want to know is, what sets can be proved to exist in $ZF$ | |
| Aug 15, 2017 at 13:50 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
added phrase for clarification
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| Aug 15, 2017 at 13:41 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
corrected mistake
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| Aug 14, 2017 at 17:00 | review | Close votes | |||
| Aug 14, 2017 at 22:27 | |||||
| Aug 14, 2017 at 16:53 | answer | added | Noah Schweber | timeline score: 10 | |
| Aug 14, 2017 at 16:48 | comment | added | Andrés E. Caicedo | Also, what exactly do you mean by your question? Are you asking whether, if $\mathsf{ZF}$ proves that a formula defines a set, then the set is in $L$? (This is false.) Is it something more subtle? | |
| Aug 14, 2017 at 16:46 | comment | added | Andrés E. Caicedo | Also, I have the feeling your use of $\mathcal P_{\mathrm{Def}}(x)$ is somewhat non-standard. What precisely do you mean by this: $\mathcal P(x)\cap L[x]$, $\mathcal P(x)\cap L(x)$? $L_{\alpha+1}\cap \mathcal P(x)$ for $x\in L$ and $\alpha$ least such that $x\in L_\alpha$? Something else? | |
| Aug 14, 2017 at 16:43 | history | edited | Andrés E. Caicedo |
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| Aug 14, 2017 at 16:43 | comment | added | Andrés E. Caicedo | No! This was already explained to you in a previous version of this question: $\mathsf{ZF}+V=L$ is not a model, minimal or otherwise. It is a theory. You need to be more careful with these basic distinctions. | |
| Aug 14, 2017 at 15:36 | answer | added | Asaf Karagila♦ | timeline score: 11 | |
| Aug 14, 2017 at 15:29 | comment | added | Mohammad Golshani | If one can prove the existence of non constructive sets in ZF then one can prove in ZF that $V \neq L$ but this contradicts the consistency of ZF+V=L. | |
| Aug 14, 2017 at 15:10 | history | asked | Thomas Benjamin | CC BY-SA 3.0 |