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Feb 18, 2011 at 19:21 vote accept Gabriel Furstenheim
Feb 5, 2011 at 22:38 answer added Andreas Blass timeline score: 5
Feb 5, 2011 at 19:44 answer added Jim Belk timeline score: 7
Feb 5, 2011 at 19:33 answer added Not Mike timeline score: 2
Feb 5, 2011 at 18:00 answer added Gerald Edgar timeline score: 7
Feb 5, 2011 at 15:42 comment added Gabriel Furstenheim @Qiaochu The examples I'm looking for are like the Cantor-Bernstein theorem.
Feb 5, 2011 at 15:34 comment added Qiaochu Yuan I don't think the question is well-defined as stated. You can trivially insert "by the Axiom of Choice, ..." into the proof of any statement and then remove it. I think you meant to say "if a result apparently depends on AC, does it necessarily?" and then there are a million counterexamples.
Feb 5, 2011 at 15:01 comment added Asaf Karagila The Cantor-Bernstein theorem was first proved using AC but is provable without it as well.
Feb 5, 2011 at 14:39 comment added darij grinberg Yes, but few of their applications do. If you read some French, try out this: hlombardi.free.fr/liens/constr.html (particularly "Algèbre Commutative"). There are lots of results of the kind "any sufficiently elementary theorem provable in ZFC is provable in ZF or even constructively", where "sufficiently elementary" means things like "1st order formula", "geometric formula" and the likes. Some of these results, ironically, require ZFC themselves, while others don't. Alas, I am not an expert in this field ("program extraction from classical proofs").
Feb 5, 2011 at 14:23 comment added Gabriel Furstenheim @darij At least the basic results in algebra need AC: the different definitions of noetherian aren't equivalent without DC. ( (mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma).
Feb 5, 2011 at 14:06 comment added darij grinberg I don't understand your question. 98% of theorems in advanced algebra that are usually proven with the AC don't depend on the AC.
Feb 5, 2011 at 13:52 history asked Gabriel Furstenheim CC BY-SA 2.5