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
6 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2014 at 22:19 | comment | added | Joel David Hamkins | I don't think you can reason like that, since if it was legitimate, it would work inside a pointwise definable model of $V=L$. The point is that being "definable" is not generally expressible in set theory, and so one really has to talk about models, rather than $V$ like this. | |
| Sep 12, 2014 at 22:10 | comment | added | Bjørn Kjos-Hanssen | @JoelDavidHamkins Right. I guess I was reasoning within $V$. So if $V=L$ then $L$ is uncountable and therefore not pointwise definable. | |
| Sep 12, 2014 at 22:04 | comment | added | Joel David Hamkins | Not every model of $V=L$ or $V=\text{HOD}$ is an example, though, since the OP requested that not every set in the model should be definable. So as Jonas says, one should take a non-pointwise definable model of $V=HOD$, such as any uncountable model of that theory. | |
| Sep 12, 2014 at 18:15 | comment | added | Emil Jeřábek | The same argument applies if V=HOD. There are no other examples, at least for (a): if every definable family has a definable element, then the collection of definable elements forms an elementary submodel in which every element is definable. This implies the submodel (hence the original model) satisfies V=HOD. | |
| Sep 12, 2014 at 18:04 | comment | added | user38200 | Could you give other examples please? | |
| Sep 12, 2014 at 17:54 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |