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Sewing-Machine comments on Harry Potter and the Methods of Rationality discussion thread, part 8 - Less Wrong
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This makes it sound like believing in an uncountable ordinal is equivalent to AC, which would make things easier - lots of mathematicians reject AC. But you might not need AC to assert the existence of a well-ordering of the reals as opposed to any set, and others have claimed that weaker systems than ZF assert a first uncountable ordinal. My own skepticism wasn't so much the existence of any well-ordering of the reals (though I'm willing to believe that no such exists), my skepticism was about the perfect, canonical well-ordering implied by there being an uncountable ordinal onto whose elements all the countable ordinals are mapped and ordered. Of course that could easily be equivalent to the existence of any well-ordering of the reals.
On the contrary, you need almost the full strength of AC to establish that a well-ordering of the reals exists. Like you say, you don't need it to construct uncountable ordinals, or to show that there is a smallest such. Cantor's argument constructively shows that there are uncountable sets, and you can get from there to uncountable ordinals by following your nose.
Is this because you can't prove aleph-one = beta-one? I'm Platonic enough that to me, "well-order an uncountable set" and "well-order the reals" sound pretty similar.
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?
Uh, that's a lot more than "Platonism"... how was anyone supposed to guess you've been assuming CH?
Edit: To clarify -- apparently you've been thinking of this as "I can accept R, just not a well-ordering on it." Whereas I've been thinking of this as "Somehow Eliezer can accept R, but not a cardinal that's much smaller?!"
Edit again: Though I guess if we don't have choice and R isn't well-orderable than I guess omega_1 could be just incomparable to it for all I know. In any case I feel like the problem is stemming from this CH assumption rather than omega_1! I don't think you can easily get rid of a smallest uncountable ordinal (see other post on this topic -- throwing out replacement will alllow you to get rid of the von Neumann ordinal but not, I don't think, the ordinal in the general sense), but if all you want is for there to be no well-order on the continuum, you don't have to.
That's how I remember it, although I don't know a reference (much less a proof). All we know is that omega_1 is not larger than R.