Are you referring to the line with “to arbitrary precision” on the bottom of page 17?
Although they don’t express themselves as clearly as they could, I don’t think that they mean anything like, “and hence we arrive at the exact regrading Θ in the limit by sending the number of atoms with the same valuation to infinity.” Rather, I think that they mean that a larger number of atoms with the same valuations puts stronger constraints on the regrading Θ, but it is never so constrained that it can’t exist.
In other words, their proof accommodates arbitrarily many atoms with the same valuation, but it doesn’t require it.
The more closely I’ve read their proof, the more confident I’ve become that they prove the following:
Let L be a finite lattice satisfying equations (0)–(3), and let a valuation m: L → R and a binary operation ⊕ on R satisfying axioms (0)–(3) be given. (Here, I take the equations and axioms to be corrected as described here).
Then there exists a strictly monotonically increasing function Θ: R → R such that Θ(a ⊕ b) = Θ(a) + Θ(b) for all a, b ∈ R such that a, b, and a ⊕ b are in the range of m.
Well, they claim that the interleaving of a and b is linear. When they prove that if a:b>5/3 then a/b>3/2 (this implication is necessary to be consistent once we have 5 copies of a), they use 9 copies of a. It is easy to prove this particular case without appealing to more than 5 copies of a, but you need to do the things that the authors seem to specifically avoid.
The worst part is that they seem to use “to arbitrary precision” argument to prove how adding a and b would work.
Let me look up the paper once more..
OK. I give up.
Do they require commutativity or not? They say they don’t need to preassume it.
a=1;
b=1.01;
a+b=2.02;
b+a=2.03;
That’s all. Two atoms. We have no other additions to check. There is not enough atoms to check associativity. Axiom 2 coincides with Axiom 1. It is impossible to regrade it with a strictly increasing Θ into commutative addition.
Ha! Nice. I’d forgotten that they didn’t require join to be commutative. But they very clearly and intentionally do not. I don’t see any way to wriggle out of your counterexample if join isn’t commutative. (If join is commutative, then, even if a+b and b+a were distinct, they wouldn’t both be in the image of the lattice. My formulation of their theorem might still hold.)
I still have no idea whether your statement is true. It requires checking. But I hope now it is clear that no part of their proof can be trusted without some editing.
If you have enough interest to try to write a claim and a proof without references to the paper, I guess it would be nice to post it as a direct comment to the post.
Sufficiently large to be larger than inverse of our precision requirements.
Are you referring to the line with “to arbitrary precision” on the bottom of page 17?
Although they don’t express themselves as clearly as they could, I don’t think that they mean anything like, “and hence we arrive at the exact regrading Θ in the limit by sending the number of atoms with the same valuation to infinity.” Rather, I think that they mean that a larger number of atoms with the same valuations puts stronger constraints on the regrading Θ, but it is never so constrained that it can’t exist.
In other words, their proof accommodates arbitrarily many atoms with the same valuation, but it doesn’t require it.
The more closely I’ve read their proof, the more confident I’ve become that they prove the following:
Let L be a finite lattice satisfying equations (0)–(3), and let a valuation m: L → R and a binary operation ⊕ on R satisfying axioms (0)–(3) be given. (Here, I take the equations and axioms to be corrected as described here).
Then there exists a strictly monotonically increasing function Θ: R → R such that Θ(a ⊕ b) = Θ(a) + Θ(b) for all a, b ∈ R such that a, b, and a ⊕ b are in the range of m.
Well, they claim that the interleaving of a and b is linear. When they prove that if a:b>5/3 then a/b>3/2 (this implication is necessary to be consistent once we have 5 copies of a), they use 9 copies of a. It is easy to prove this particular case without appealing to more than 5 copies of a, but you need to do the things that the authors seem to specifically avoid.
The worst part is that they seem to use “to arbitrary precision” argument to prove how adding a and b would work.
Let me look up the paper once more..
OK. I give up.
Do they require commutativity or not? They say they don’t need to preassume it.
a=1; b=1.01; a+b=2.02; b+a=2.03;
That’s all. Two atoms. We have no other additions to check. There is not enough atoms to check associativity. Axiom 2 coincides with Axiom 1. It is impossible to regrade it with a strictly increasing Θ into commutative addition.
Ha! Nice. I’d forgotten that they didn’t require join to be commutative. But they very clearly and intentionally do not. I don’t see any way to wriggle out of your counterexample if join isn’t commutative. (If join is commutative, then, even if a+b and b+a were distinct, they wouldn’t both be in the image of the lattice. My formulation of their theorem might still hold.)
I still have no idea whether your statement is true. It requires checking. But I hope now it is clear that no part of their proof can be trusted without some editing.
If you have enough interest to try to write a claim and a proof without references to the paper, I guess it would be nice to post it as a direct comment to the post.
btw, I mentioned the work you two have been doing here to the author and tried to get him to respond here but unfortunately he hasn’t agreed to.