Sewing-Machine comments on Harry Potter and the Methods of Rationality discussion thread, part 8 - Less Wrong

8 Post author: Unnamed 25 August 2011 02:17AM

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Comment author: [deleted] 07 September 2011 05:35:07PM 1 point [-]

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.

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.

Comment author: Eliezer_Yudkowsky 08 September 2011 01:09:30AM 0 points [-]

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.

Comment author: [deleted] 08 September 2011 01:22:11AM 4 points [-]

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.

Comment author: Sniffnoy 08 September 2011 01:42:04AM 2 points [-]

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...

Comment author: [deleted] 08 September 2011 01:49:00AM 1 point [-]

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.

Comment author: Sniffnoy 08 September 2011 01:53:17AM *  3 points [-]

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?

Comment author: Sniffnoy 08 September 2011 01:42:53AM *  3 points [-]

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.

Comment author: TobyBartels 11 September 2011 08:52:22PM *  2 points [-]

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.

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.