Timeline for Automorphisms of vector spaces and the complex numbers without choice
Current License: CC BY-SA 4.0
14 events
when toggle format | what | by | license | comment | |
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S Mar 8, 2023 at 14:16 | history | bounty ended | THC | ||
S Mar 8, 2023 at 14:16 | history | notice removed | THC | ||
Mar 8, 2023 at 14:16 | vote | accept | THC | ||
Mar 4, 2023 at 0:20 | comment | added | Hanul Jeon | @KevinCasto Blass' proof constructs a field that will be associated with a choice function over a given set (more precisely, a witness for Multiple choice,) so we cannot say the existence of a basis over a $\mathbb{C}$-vector space implies choice from Blass' proof. | |
Mar 1, 2023 at 20:36 | history | edited | Rahman. M | CC BY-SA 4.0 |
deleted 7 characters in body; edited tags
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Mar 1, 2023 at 20:16 | answer | added | Holo | timeline score: 4 | |
Mar 1, 2023 at 17:20 | comment | added | Kevin Casto | Ah fair enough, I didn't read Blass' proof closely enough (it relies on function fields over $\mathbb C$ also having this property). Still, it does in any case seem like a result worth mentioning in the context of the question! | |
Mar 1, 2023 at 16:50 | comment | added | Andrej Bauer | Oh, I see. So how should I understand @KevinCastos comment? | |
Mar 1, 2023 at 16:35 | comment | added | Jeremy Rickard | @AndrejBauer Although I don't think it's known that "every vector space over $\mathbb{C}$ has a basis" implies choice. | |
Mar 1, 2023 at 16:00 | comment | added | Andrej Bauer | To continue @KevinCasto's comment, just in case it's not clear: and once you have choice, you're in ZFC, so you can cook up many automorphisms of $\mathbb{C}$. | |
S Mar 1, 2023 at 14:01 | history | bounty started | THC | ||
S Mar 1, 2023 at 14:01 | history | notice added | THC | Authoritative reference needed | |
Feb 22, 2023 at 20:28 | comment | added | Kevin Casto | "Every vector space has a basis" implies choice; see math.stackexchange.com/questions/207990/vector-spaces-and-ac | |
Feb 22, 2023 at 15:41 | history | asked | THC | CC BY-SA 4.0 |