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Misha comments on Harry Potter and the Methods of Rationality discussion thread, part 8 - Less Wrong
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Is it often the case that you need to assume the negation of AC for a proof to hold? AC comes up in seemingly-unrelated areas when you need some infinitely-hard-to-construct object to exist; I can't imagine a similar case where you'd assume not(AC) in, e.g., ring theory.
As usual, the negation of a useful statement ends up not being a useful statement. I don't think anyone works with not(AC), they work with various stronger things that imply not(AC) but actually have interesting consequences.
That's intriguing. Do you have any examples of what people actually work with?
Sniffnoy may have more examples, but here are some that I know:
Adding the first two to ZF + DC (dependent choice) is consistent (assuming that ZFC + Con(ZFC) is consistent, as just about everybody believes), and this gives a "dream universe" for analysis in which, for example, any everywhere-defined linear operator between Hilbert spaces is bounded.
This isn't quite right. The consistency of ZF + DC + "every subset of R is Lebesgue measurable" is equivalent to the consistency of an inaccessible cardinal, which is a much stronger assumption then the consistency of ZFC + Con(ZFC).
Sorry, my mistake. Still, set theorists usually believe this.