Timeline for Is definability of a basis for $\mathbb{R^N}$ independent of ZFC?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2013 at 23:28 | vote | accept | Colin McLarty | ||
| Nov 20, 2013 at 13:54 | comment | added | Asaf Karagila♦ | Colin, yes. Since the structure of the vector space is certainly in $\sf HOD(\Bbb R)$, we have that being a basis is something which depends on the finite subsets which are certainly absolute to that inner model. | |
| Nov 20, 2013 at 13:23 | comment | added | Colin McLarty | You say "to some extent absolute," but I believe you mean it is rigorously true that $B$ is a basis in $V$ iff it is in $V$. Right? I think the point is that $B$ being a basis is a matter relating individual elements of $V$ to finite subsets of $B$ and the field. Is that right? Or at least reasonably well stated? | |
| Nov 19, 2013 at 16:54 | comment | added | Asaf Karagila♦ | Colin, do note that under "reasonable" assumptions (e.g. that $V$ is the collapse of an inaccessible cardinal) this model does not satisfy the axiom of choice. However my argument is that essentially being a basis of a definable vector space is to some extent absolute. So $B$ is a basis in $\sf HOD(\Bbb R)$ if and only if it is a basis in $V$. | |
| Nov 19, 2013 at 16:51 | comment | added | Colin McLarty | I am learning about this from reading the comments. But it seems to me that being in $\sf HOD(\Bbb R)$ is at least as broad a sense of definable as I would want to know about. | |
| Nov 18, 2013 at 21:13 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |