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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
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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