A retraction
Mar. 31st, 2018 05:44 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo, OK, before I get to other things, I want to post a retraction here. (Not sure that I'm going to go back to relevant earlier entries and edit them.)
Previously here I've claimed to have proven the inequality that, for any ordinal α and any WPO X, one has o(Rev(α,X))≤o(Rev(α,o(X))). (And if X has an initial element, one has o(Rev0(α,X))≤o(Rev0(α;o(X))) -- it's really this latter claim I'm going to focus on.)
My claimed proof of this was wrong. I have not thus far been able to fix it. I still think it is true.
Note that none of this affects my formula for Rev0(α;β) for α and β ordinals and multidimesional generalizations (and other variants) of such; these proofs (which I finally recently got around to writing up properly) are unaffected.
I have been able to recover the inequality in a number of special cases. It's easily seen to be true when α and X are both finite. My original proof basically still works if α is finite; and if α=ω, I can prove it as a corollary of Andreas Weiermann's work on the WPO of finite multisets of a given WPO. Meanwhile, a different proof works if instead X is finite, and I think I can extend it also to the case where o(X)=ω. In fact I think I can possibly go further and get it to work when o(X)<ω2. There are other special cases I've managed to recover too -- generally requiring that we take α<ω² and then putting conditions on the form o(X) takes -- but I'm not going to go listing them all here. Still, in general, I have not been able to recover it thus far.
One other special case I wanted to mention that I can recover is -- well, remember, the original reason I wanted to prove this inequality was to prove the following two inequalites, where ((α multichoose β)) is to be interpreted as o(Rev(β,α)):
((α⊗β multichoose γ)) ≥ ((α multichoose γ)) ⊗ ((β multichoose γ))
((α multichoose β⊗γ)) ≤ (( ((α multichoose β)) multichoose γ))
(These are companion to a third inequality, namely,
((α multichoose β⊕γ)) ≤ ((α multichoose β)) ⊗ ((α multichoose γ)),
but this one can be proven pretty easily without my conjectured inequality.)
Anyway, I can report at least that the two inequalities above are definitely true -- I proved it the stupid way, by computing both sides via my formula and then comparing. Still, this is definitely an unsatisfying proof, compared to the proof we'd obtain with a proof of my conjectural inequality above.
Well, at least the formula in the total case (especially in multiple dimensions) is quite noteworthy and will make a good paper on its own, assuming I can't recover the more general partial case!
Previously here I've claimed to have proven the inequality that, for any ordinal α and any WPO X, one has o(Rev(α,X))≤o(Rev(α,o(X))). (And if X has an initial element, one has o(Rev0(α,X))≤o(Rev0(α;o(X))) -- it's really this latter claim I'm going to focus on.)
My claimed proof of this was wrong. I have not thus far been able to fix it. I still think it is true.
Note that none of this affects my formula for Rev0(α;β) for α and β ordinals and multidimesional generalizations (and other variants) of such; these proofs (which I finally recently got around to writing up properly) are unaffected.
I have been able to recover the inequality in a number of special cases. It's easily seen to be true when α and X are both finite. My original proof basically still works if α is finite; and if α=ω, I can prove it as a corollary of Andreas Weiermann's work on the WPO of finite multisets of a given WPO. Meanwhile, a different proof works if instead X is finite, and I think I can extend it also to the case where o(X)=ω. In fact I think I can possibly go further and get it to work when o(X)<ω2. There are other special cases I've managed to recover too -- generally requiring that we take α<ω² and then putting conditions on the form o(X) takes -- but I'm not going to go listing them all here. Still, in general, I have not been able to recover it thus far.
One other special case I wanted to mention that I can recover is -- well, remember, the original reason I wanted to prove this inequality was to prove the following two inequalites, where ((α multichoose β)) is to be interpreted as o(Rev(β,α)):
((α⊗β multichoose γ)) ≥ ((α multichoose γ)) ⊗ ((β multichoose γ))
((α multichoose β⊗γ)) ≤ (( ((α multichoose β)) multichoose γ))
(These are companion to a third inequality, namely,
((α multichoose β⊕γ)) ≤ ((α multichoose β)) ⊗ ((α multichoose γ)),
but this one can be proven pretty easily without my conjectured inequality.)
Anyway, I can report at least that the two inequalities above are definitely true -- I proved it the stupid way, by computing both sides via my formula and then comparing. Still, this is definitely an unsatisfying proof, compared to the proof we'd obtain with a proof of my conjectural inequality above.
Well, at least the formula in the total case (especially in multiple dimensions) is quite noteworthy and will make a good paper on its own, assuming I can't recover the more general partial case!
