Timeline for Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member?
Current License: CC BY-SA 3.0
19 events
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| Jun 23, 2017 at 1:12 | comment | added | Joel David Hamkins | In our paper, we address the role of real parameters and indeed any $\Sigma_2$-definable class of parameters. | |
| Jun 23, 2017 at 1:10 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Sep 15, 2014 at 1:01 | comment | added | Joel David Hamkins | That question is now answered, and led to an improvement here, which I have now edited to add. So $V=\text{HOD}$ is equivalent to the assertion that every $\Pi_2$-definable set has an ordinal-definable element. | |
| Sep 15, 2014 at 1:00 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Sep 14, 2014 at 3:42 | comment | added | Joel David Hamkins | I asked a question whether we can not weaken statement 5 to $\Sigma_2$ or $\Pi_2$ definable sets and maintain the equivalence with $V=\text{HOD}$. mathoverflow.net/q/180810/1946 | |
| Sep 14, 2014 at 0:53 | comment | added | Joel David Hamkins | I just noticed that the question also asks about allowing reals as parameters, which I didn't really consider in my answer. | |
| Sep 13, 2014 at 23:42 | comment | added | Joel David Hamkins | @FrançoisG.Dorais I got the definition of that set down to $\Sigma_2\wedge\Pi_2$, and have edited my answer. It would be interesting to consider the weakening of statement 5 to say only that every $\Sigma_2$-definable set has an OD member. I don't think this implies $V=\text{HOD}$, but I don't have a model yet. | |
| Sep 13, 2014 at 23:40 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Sep 13, 2014 at 14:32 | comment | added | Joel David Hamkins | Great! I'm glad we've got it nailed down now. I think I'll edit this into my answer later, since I think it brings the phenomenon to an even sharper point. | |
| Sep 13, 2014 at 14:27 | comment | added | François G. Dorais | In more detail... There is a (parameter-free) $\Sigma_2$ formula $OD(x)$ which says '$x$ is OD'. The smallest rank $\alpha$ of a non-OD set is $\Sigma_3$-definable by $$\exists U,x(Ord(\alpha) \land U = V_\alpha \land (\forall z \in U)OD(z) \land x \subseteq U \land \lnot OD(x)).$$ So the set of all minimal-rank non-OD sets is $\Sigma_3$-definable. It's actually $\Delta_3$ since the ordinal $\alpha$ is unique. | |
| Sep 13, 2014 at 13:53 | comment | added | François G. Dorais | Sorry, $\Pi/\Sigma$ issue. The $\Sigma_2$ statement says that $x$ is OD, so being non-OD would be $\Pi_2$. I convinced myself that I could gobble up the minimal-rank requirement inside but I might have been too hasty doing that. Worst case $\Sigma_3$? | |
| Sep 13, 2014 at 13:37 | comment | added | Joel David Hamkins | @FrançoisG.Dorais But in that argument, you have an ordinal parameter, whereas here, we want to have no parameters. The question is, how complicated is the definition of the set of minimal-rank non-OD sets, if there are such sets? This cannot be $\Sigma_2$, since we could make those elements definable by coding into the GCH high up, above the witness of the $\Sigma_2$ statement, and this would preserve the $\Sigma_2$ assertion, while destroying that set as consisting of non-OD sets. | |
| Sep 13, 2014 at 13:28 | comment | added | François G. Dorais | I think you can replace 5 by 2: mathoverflow.net/questions/72807/sigma-n-version-of-hod/… | |
| Sep 13, 2014 at 6:30 | vote | accept | user38200 | ||
| Sep 13, 2014 at 2:10 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Sep 12, 2014 at 20:36 | comment | added | Joel David Hamkins | See also this ancient MO post with essentially (eventually) the same idea: mathoverflow.net/a/10415/1946 | |
| Sep 12, 2014 at 20:23 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Sep 12, 2014 at 19:27 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |