Timeline for Is there a "hereditary" construction for $L$?
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2014 at 4:44 | comment | added | Asaf Karagila♦ | @Mohammad: Sure! Thanks for thinking this question over! :-) | |
| Aug 26, 2014 at 4:43 | comment | added | Mohammad Golshani | My argument seems to be wrong at some point so I deleted it. Let me think on it a little more. | |
| Aug 22, 2014 at 18:04 | comment | added | Joel David Hamkins | Thanks! One may hope to learn something new every day... | |
| Aug 22, 2014 at 10:45 | comment | added | Emil Jeřábek | @Joel: Yes, I meant definability in $V$. The argument, which works for arbitrary first-order structures, is that if $a$ is definable by a formula $\phi(x)$, it is also definable by the formula $\forall y\,(\phi(y)\to y=x)$. | |
| Aug 22, 2014 at 7:18 | comment | added | Joel David Hamkins | I understood @EmilJeřábek to be talking about definability in $V$, not just $L$. | |
| Aug 22, 2014 at 2:09 | comment | added | Asaf Karagila♦ | @Joel, the "usual argument" as I know it is for $L$-like construction, where if a set is definable (by a particular type of formula like $\Sigma_n$ or $\Sigma^1_2$) then its complement is definable the next step. I don't know if that is what Emil meant, or if that works here. | |
| Aug 21, 2014 at 22:41 | comment | added | Joel David Hamkins | Is that true? What is the usual argument? | |
| Aug 21, 2014 at 21:18 | comment | added | Emil Jeřábek | Well, $\Sigma_n$-definable elements are $\Delta_n$-definable by the usual argument, aren’t they? | |
| Aug 21, 2014 at 20:59 | comment | added | Joel David Hamkins | I guess the $L[b]$ example also shows that the hereditarily ordinal $\Delta_1$-definable sets can exceed $L$, since $b$ is actually $\Delta_1$-definable from ordinal parameters. | |
| Aug 21, 2014 at 20:45 | comment | added | Asaf Karagila♦ | @Emil: While we're on the subject, let me say that what I said before was nonsense and a step-by-step construction is most definitely not $V_\alpha\cap\sf HOD$. Sorry 'bout that. :-) | |
| Aug 21, 2014 at 20:44 | comment | added | Emil Jeřábek | @Joel: Thanks for the explanation, that’s interesting. | |
| Aug 21, 2014 at 20:41 | comment | added | Joel David Hamkins | @EmilJeřábek Meanwhile, on countable sets, your idea works, since $L_{\omega_1^L}\prec_{\Sigma_1} V$ by Levy-absoluteness. So any $\Sigma_1$-definable set with countable ordinal parameters (countable in $L$) will be in $L$. I suppose this could be seen as a partial answer to Asaf's question, up to $\omega_1^L$. | |
| Aug 21, 2014 at 20:36 | comment | added | Asaf Karagila♦ | @Joel, I do recall this class. Unfortunately I'm looking for something whose "hereditary subclass" is always $L$. | |
| Aug 21, 2014 at 20:34 | comment | added | Joel David Hamkins | Asaf, you might like the class Imp (for "Implicitly definable") in my paper with Cole Leahy: jdh.hamkins.org/algebraicity-and-implicit-definability. The class Imp is obtained by iterating implicit definability, rather than merely definability, and we prove that it can be larger than L. | |
| Aug 21, 2014 at 20:27 | comment | added | Asaf Karagila♦ | @Goldstern: I mean that in the sense that $\sf HOD$ can be defined as first defining some class $X$ which is not transitive, then taking the class of "hereditarily $X$". So in the case of $\sf HOD$ we have that $X=\sf OD$. But is there some $X$ for which $hX=L$ (in $\sf ZFC$, of course, not a consistency result!) | |
| Aug 21, 2014 at 20:27 | comment | added | Joel David Hamkins | @EmilJeřábek The class of hereditarily ordinal $\Sigma_1$-definable sets can be more than $L$. For example, start in $L$ and let $T$ be a Suslin tree with the unique branch property (see jdh.hamkins.org/degreesofrigidity), which means that if we force to add an $L$-generic branch $b$, then $L[b]$ has only one branch though $T$. In $L[b]$, the branch $b$ is hereditarily ordinal $\Sigma_1$-definable, since $T$ is ordinal $\Sigma_1$-definable (being in $L$) and $b$ is the unique branch through it. So in $L[b]$, we get $\text{HOD}_{\Sigma_1}=L[b]$. | |
| Aug 21, 2014 at 20:26 | comment | added | Goldstern | What do you mean by "similar"? I think that $L$ is "similar" to OD, because both are defined using quite explicitly definability relations. (And clearly "hereditarily in L" is the same as "in $L$".) | |
| Aug 21, 2014 at 19:19 | comment | added | Asaf Karagila♦ | Well, the second one is really the same (at least when considering all the formulas), since you really just construct $V_\alpha\cap\sf HOD$ one step at a time. In either case, I should have opened with it in my first comment: I have no idea. :-) | |
| Aug 21, 2014 at 19:18 | comment | added | Emil Jeřábek | The first one. The second one doesn’t seem to make much sense to me. | |
| Aug 21, 2014 at 19:13 | comment | added | Asaf Karagila♦ | @Emil: Do you mean taking $\sf OD$ with only $\Sigma_1$ formulas, and then considering those whose transitive closure is made of such sets, and it is itself $\Sigma_1$-definable, or construct by recursion a continuous hierarchy where at each step you add the $\Sigma_1$-OD subsets of the previous collection? (The point is that I'm not 100% clear that these two approaches are equivalent when talking only about $\Sigma_1$ formulas.) | |
| Aug 21, 2014 at 19:11 | comment | added | Emil Jeřábek | Take the definition of HOD just like above, but require $\varphi(x,\alpha_1,\dots,\alpha_n)$ to be a $\Sigma_1$-formula. | |
| Aug 21, 2014 at 19:07 | comment | added | Asaf Karagila♦ | Emil, what precisely do you mean by that? | |
| Aug 21, 2014 at 19:05 | comment | added | Emil Jeřábek | I’m shooting in the dark, but how do hereditary ordinal $\Sigma_1$-definable sets look like? | |
| Aug 21, 2014 at 18:33 | history | asked | Asaf Karagila♦ | CC BY-SA 3.0 |