Timeline for Definable collections without definable members (in ZF)
Current License: CC BY-SA 3.0
18 events
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| Jun 23, 2017 at 1:04 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Finally updated answer, give links to my paper with François.
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| Jan 2, 2010 at 22:03 | comment | added | Ashutosh | I guess the "it" in the assertion "it is independent of ZF" refers to the meta-scheme: Every definable nonempty class has a definable member. So that, e.g., "it" holds in (ZF + V=HOD) while "not-it" holds in (ZF + V!=HOD). | |
| Jan 2, 2010 at 21:35 | comment | added | François G. Dorais | It's a minor technical issue which has to do with how the "most direct interpretation of the question" actually reads. Summarized, the question is: does ZF have property (A)? As we now know, this is the same as asking: does ZF imply V=HOD? The answer to that question is no, because V=HOD is independent of ZF. I agree that the answer should be rewritten for clarity. Perhaps writing the new answer above the current one, followed by the historical progression? | |
| Jan 2, 2010 at 21:05 | comment | added | Joel David Hamkins | I don't quite follow you. Surely V=HOD is independent of ZF, if anything is. But I see, you are looking at whether "ZF proves V=HOD" is independent? It seems more natural, however, to omit this double provability aspect from the question. The main issue seems to be whether it is true that every definable set has a definable member, and we've now observed that this is just equivalent to V=HOD. I'm half inclined to re-write the answer completely with this perspective, but I don't know what the recommended MO policy on that is. | |
| Jan 2, 2010 at 20:52 | comment | added | François G. Dorais | Minor edit suggestion: I think the bold phrase in the second paragraph is slightly misleading. The fact that ZF does not imply V=HOD is a definite answer to the question as stated. However, the statement is technically true: if M models ZF and not Con(ZF), then "ZF implies V=HOD" is true in M. | |
| Jan 2, 2010 at 19:56 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Showed property is equivalent to V=HOD
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| Jan 2, 2010 at 19:44 | comment | added | Joel David Hamkins | I see! So the property IS first-order expressible (contrary to what I had said in my first bullet point), and is just equivalent to V=HOD. That is, if a model satisfies V=HOD, then there is a definable well-ordering of the universe, so every nonempty definable set will have a definable element (the least one), and conversely, if the model does not have V=HOD, then the set of minimal rank sets not in OD is definable, but has no definable element. | |
| Jan 2, 2010 at 19:19 | comment | added | Ashutosh | I noticed that your argument for ZF + property (A) -> AC gives a nice characterization of the theories that satisfy property (A): If T is any consistent extension of ZF then, T has property (A) iff T proves V = HOD. | |
| Jan 2, 2010 at 14:45 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Added remark about property implying AC
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| Jan 2, 2010 at 3:28 | comment | added | Joel David Hamkins | Ah, that is also nice. My OD example seems to work the same in ZF as in ZFC. | |
| Jan 2, 2010 at 2:16 | comment | added | Ashutosh | For ZFC, I know a particularly nice formula for which this fails: A = set of all well orderings of reals. A is non empty but ZFC doesn't prove that A has a definable member. This example, however, depends on axiom of choice. | |
| Jan 2, 2010 at 2:10 | comment | added | Joel David Hamkins | By the way, I noticed that you emphasized ZF, as opposed to ZFC, and I went along with this in my answer, but I think the issue of AC does not much affect things here, since the methods for constructing pointwise definable models produce full ZFC models. Although naively one might expect AC to give you non-definable sets, this turns out not to be true, as there are pointwise definable models of full ZFC. | |
| Jan 2, 2010 at 2:08 | comment | added | Ashutosh | That's a nice trick. Reminds me of the proof of relative consistency of "there is no inaccessible". Thanks again. | |
| Jan 2, 2010 at 2:07 | vote | accept | Ashutosh | ||
| Jan 2, 2010 at 1:30 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Added example responding to comment
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| Jan 2, 2010 at 0:31 | comment | added | Joel David Hamkins | Well, V-OD might be empty, and it is in the pointwise definable models. But there is a way to get around this difficulty, and I'll edit my answer above to explain it. Similarly, it is consistent that there are no L-generic Cohen reals. | |
| Jan 2, 2010 at 0:05 | comment | added | Ashutosh | I need a class A such that ZF proves "A is non empty" and for every class b, ZF doesn't prove "b is in A". Does the difference V \ OD qualify as such a class A? Similarly can ZF prove that the set of Cohen reals over L is nonempty - I'd confess that I've never seen forcing extensions over a class? Probably I'm missing something but it would be great if you can explicitly write the class A. Thanks a lot | |
| Jan 1, 2010 at 23:37 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |