No something sillier. You can prove the axiom of choice from the assumption that every set can be well-ordered. (Proof: use the well-ordering to construct a choice function by taking the least element in every part of your partition.)
If one doesn’t wish to assume that every set has a well-ordering, but only a single set such as the real numbers, then one gets a choice-style consequence that’s limited in the same way: you can construct choice functions from partitions of the real numbers.
I’d hardly call a well-ordering on one particular cardinality “almost the full strength of AC”! I guess it probably is enough for a lot of practical cases, but there must be ones where one on 2^c is necessary, and even so that’s still a long way from the full strength...
Hm, agreed. I guess not so much “the full strength” but “the full counterintuitiveness”? Where DC uses hardly any of the counterintuitiveness, and ultrafilter lemma uses nearly all of it?
No something sillier. You can prove the axiom of choice from the assumption that every set can be well-ordered. (Proof: use the well-ordering to construct a choice function by taking the least element in every part of your partition.)
If one doesn’t wish to assume that every set has a well-ordering, but only a single set such as the real numbers, then one gets a choice-style consequence that’s limited in the same way: you can construct choice functions from partitions of the real numbers.
I’d hardly call a well-ordering on one particular cardinality “almost the full strength of AC”! I guess it probably is enough for a lot of practical cases, but there must be ones where one on 2^c is necessary, and even so that’s still a long way from the full strength...
I just have a hard time imagining someone who was happy with “c is well-ordered” but for whom “2^c is well-ordered” is a bridge too far.
Hm, agreed. I guess not so much “the full strength” but “the full counterintuitiveness”? Where DC uses hardly any of the counterintuitiveness, and ultrafilter lemma uses nearly all of it?