Timeline for If a vector space has a basis then its dual vector space has a basis
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2021 at 17:22 | vote | accept | Michal R. Przybylek | ||
| Jul 22, 2021 at 19:56 | answer | added | Harry West | timeline score: 1 | |
| Jul 4, 2021 at 16:03 | history | edited | Michal R. Przybylek | CC BY-SA 4.0 |
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| Jul 4, 2021 at 14:43 | comment | added | Michal R. Przybylek | @HarryWest, I am interested in any variation of the statement "if a vector space has a basis then its dual vector space has a basis". If you can find a counterexample to this statement in the second Fraenkel-Mostowski model, then I will be happy to award you with a bounty. | |
| Jun 30, 2021 at 17:54 | comment | added | Harry West | It is not clear to me what would count as an answer. What is the strength of a choice axiom? A comparison to all the axioms in Howard-Rubin? Separately, I'm pretty sure this doesn't hold in the second Fraenkel-Mostowski model (at least for char K $\neq$ 2). | |
| Jun 24, 2021 at 16:03 | comment | added | Michal R. Przybylek | @JeremyRickard, It is not that obvious. It is true in general set theories by the above two examples (notice that in any non-trivial ZFA the Axiom of Choice does not hold). I think it is also true over ZF (I have some ideas, but I am still looking for a cheap argument). | |
| Jun 23, 2021 at 15:42 | comment | added | Asaf Karagila♦ | @Jeremy: Over ZF? Not that obvious. (IAAST) | |
| Jun 23, 2021 at 13:10 | comment | added | Jeremy Rickard | Is it obvious that the statement is strictly weaker than AC? (IANAST) | |
| Jun 23, 2021 at 12:36 | comment | added | Michal R. Przybylek | @AsafKaragila, yes, this is a good point :-) In fact it is a part of my question --- I suspect that the statement holds in every sufficiently well-behaved set theory with atoms (assuming AC in the metatheory). Understanding what it means to be "sufficiently well-behaved" can help us with the general statement. E.g. does it hold in every set theory with atoms over an $\omega$-categorical structure? | |
| Jun 23, 2021 at 12:13 | comment | added | Asaf Karagila♦ | Yes, but how do you plan on moving the statement to ZF from ZFA? In ZFA the power set of an ordinal can be well-ordered (at least in the standard "ZFC holds in the pure part" situation). In ZF the power set of an ordinal will fail to be well-orderable if AC failed. Especially when asking about power sets, this seems to be at least somewhat relevant. | |
| Jun 23, 2021 at 12:10 | comment | added | Michal R. Przybylek | @AsafKaragila, in both the first and the ordered Fraenkel-Mostowski model MC does not hold. So, I do not think it should be a big issue. | |
| Jun 23, 2021 at 12:05 | comment | added | Michal R. Przybylek | @JohannesHahn, in the first Fraenkel-Mostowski model UF does not hold, whereas in the ordered Fraenkel-Mostowski model UF principle holds. On the other hand, in both of these models the statement from the question holds. | |
| Jun 23, 2021 at 11:43 | comment | added | Asaf Karagila♦ | If you are interested in bases of vector spaces, working with permutation models is the wrong way to go, since "Every vector space has a basis" might be weaker than AC in ZFA. | |
| Jun 23, 2021 at 11:35 | comment | added | Johannes Hahn | Since you mentioned power sets as $\mathbb{F}_2$-vector spaces: Have you investigated how this compares to the ultrafilter principle? | |
| Jun 23, 2021 at 11:01 | history | asked | Michal R. Przybylek | CC BY-SA 4.0 |