Timeline for Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable element?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2017 at 17:16 | vote | accept | Joel David Hamkins | ||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 14, 2014 at 19:24 | comment | added | Joel David Hamkins | OK, I'll post it later. | |
| Sep 14, 2014 at 19:18 | comment | added | Emil Jeřábek | That’s great! If you are in the mood to write up question 2 as well, that would be awesome, as you have a much better grasp of what’s going on here than me. | |
| Sep 14, 2014 at 17:19 | comment | added | Joel David Hamkins | I was able to push that observation through to a positive answer for question 1, by iterating. | |
| Sep 14, 2014 at 17:18 | answer | added | Joel David Hamkins | timeline score: 9 | |
| Sep 14, 2014 at 15:42 | comment | added | Joel David Hamkins | I've realized that my observation on $\Sigma_2$ shows that there can be no $\Sigma_2$ formula that provably defines a nonempty set with no ordinal-definable members, under $V\neq\text{HOD}$, since one can always do coding high up so as to preserve that set being defined by that formula, but making one of its elements definable. This means that there can be no uniform counterexample to question 1, in the way that Emil's idea has now seemed to make a uniform counterexample to question 2. | |
| Sep 14, 2014 at 14:57 | comment | added | Joel David Hamkins | We have also to add that $\theta$ is minimal, which amounts to the claim that $V_\theta$ thinks that there is no smaller $\theta'<\theta$ such that $V_{\theta'}$ is able to ordinal-define all the sets not in $A$ of that rank or lower. Will you write it up as an answer? | |
| Sep 14, 2014 at 14:44 | comment | added | Joel David Hamkins | I think it works! You will define $A\times V_\theta$ as the unique set of pairs, whose second coordinate is a $V_\theta$, such that $A$ consists of the minimal-rank non-OD sets inside $V_\theta$, and such that every set in $A$ is really not in OD (in $V$). This seems altogether to be $\Pi_2$, but it has no ordinal-definable element. This answers question 2 negatively, and improves the other theorem from $\Sigma_2\wedge\Pi_2$ to just $\Pi_2$. | |
| Sep 14, 2014 at 14:41 | comment | added | Joel David Hamkins | Oh, but maybe that doesn't preclude your idea with $A\times V_\theta$. Does it work? | |
| Sep 14, 2014 at 14:34 | comment | added | Joel David Hamkins | I don't think that the set $A$ of all minimal-rank non-OD sets can be provably $\Pi_2$ definable, since then sets that are not $A$ are recognized by having a certain property in some large $V_\theta$, and we can change whether some $B\subset A$ is all the non-OD sets of that rank or not by forcing above $\theta$. That is, I can build models where $A$ gets smaller in a forcing extension that preserves $V_\theta$. | |
| Sep 14, 2014 at 14:15 | comment | added | Emil Jeřábek | That should read $x\notin A$. Anyway, it doesn’t work, as the two elements of $\langle A,V_\theta\rangle$ are in fact definable. What about $A\times V_\theta$? | |
| Sep 14, 2014 at 14:04 | comment | added | Emil Jeřábek | I guess I rather want to define $\langle A,V_\theta\rangle$, where $A$ is the set of minimal-rank non-OD sets, and $\theta$ is minimal such that every $x\in A$ of rank less than the rank of $A$ is definable in $V_{\theta'}$ for some $\theta'<\theta$. Or maybe with the sequence $\langle V_{\theta'}:\theta'<\theta\rangle$ in place of $V_\theta$ if it makes the formula simpler. | |
| Sep 14, 2014 at 13:52 | comment | added | Joel David Hamkins | Every OD set $x$ does indeed have a minimal ordinal $\theta$ such that $x$ is definable in $V_\theta$. So you are proposing to define the set of such pairs $\langle x,\theta\rangle$? | |
| Sep 14, 2014 at 13:49 | comment | added | Emil Jeřábek | The relevant $\Sigma_2$ properties in your answer (i.e., the last two conditions) say that a bunch of sets are in OD. As far as I understand it, a witness to the outer existential quantifier here is an ordinal and a defining formula (or something to that effect), and these are explicitly well ordered. Maybe you could use this to $\Pi_2$-define the pair $\langle x,y\rangle$, where $x$ is the original set, and $y$ is the least witness to the existential quantifier? | |
| Sep 14, 2014 at 12:59 | comment | added | Joel David Hamkins | The former. If $V\neq\text{HOD}$, then the answer at the other question shows that the set of minimal-rank non-OD sets is characterized as the unique set with a certain $\Sigma_2\wedge\Pi_2$ property. | |
| Sep 14, 2014 at 12:40 | comment | added | Emil Jeřábek | By a $\Sigma_2$-definable set, you mean a unique set satisfying a $\Sigma_2$-formula $\phi(x)$, or a set of the form $\{x:\phi(x)\}$ for a $\Sigma_2$-formula $\phi(x)$? | |
| Sep 14, 2014 at 2:21 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
More polite
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| Sep 14, 2014 at 0:59 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
small fix
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| Sep 14, 2014 at 0:46 | history | asked | Joel David Hamkins | CC BY-SA 3.0 |