### An inequality that doesn't work

Sep. 18th, 2016 06:56 pm**sniffnoy**

So, here's an idea I had the other day: My inequality that o(Rev(α,B))≤o(Rev(α,o(B))) doesn't work if you replace o(B) by B', an extension of B. But what if you replace it by B', an extension of B satisfying o(B')=o(B)?

(We can also more generally replace α by A, a WPO, although then we have to make the additional assumption that everything involved is well-defined, and also in that case I haven't actually proven the original inequality in general)

I mean, it's a fairly natural generalization, right? And like obviously it works if A and B are both finite. And there's also some cases where, based on what I already know, one can see that they have to actually be equal (both being equal to o(Rev(α,o(B)))).

But in fact I'm pretty sure now that this generalization is false! One can take α=ω, B=

(Without that restriction, it's only slightly different; ω

So, so much for that! That said, it can probably still be proven under more restrictive conditions. For instance, maybe if A is finite. And I'd definitely bet on it being true if A=α is a power of ω and B is finite; probably more generally if A=α is total and B is finite. (Probably also if o(A) is initial and B is finite.) Beyond that it's hard to say. One can probably get it to work with other smallness conditions -- I had to take o(B)=ω³ to get a counterexample, can one prove that's the smallest? -- but I don't know to what extent I'll bother...

(I think this case should work also with A=ω, but I haven't done a proper check. I could be wrong there.)

So, basically, this one is unsalvageable. It works in the trivial case when A and B are both finite, and some other trivial or not-so-trivial cases I can prove, but not in any decent generality. I'm calling this one a wrap.

-Harry

(We can also more generally replace α by A, a WPO, although then we have to make the additional assumption that everything involved is well-defined, and also in that case I haven't actually proven the original inequality in general)

I mean, it's a fairly natural generalization, right? And like obviously it works if A and B are both finite. And there's also some cases where, based on what I already know, one can see that they have to actually be equal (both being equal to o(Rev(α,o(B)))).

But in fact I'm pretty sure now that this generalization is false! One can take α=ω, B=

**N**²+**N**³, and B'=**N**+**N**³. Since**N**² can be extended to**N**, B' is an extension of B; and o(B)=o(B')=ω³. And yet things don't work out quite the same way when you take weakly decreasing maps from ω into these partial orders. Because, and I haven't checked this*totally*thoroughly, but I'm pretty sure that (with the additional restriction that the functions must be eventually 0) the left hand side is equal to ω^{ω5}, while the right hand side is ω^{ω4}.(Without that restriction, it's only slightly different; ω

^{ω5+2}+ω^{ω3+3}vs ω^{ω4+1}+ω^{ω3+3}.)So, so much for that! That said, it can probably still be proven under more restrictive conditions. For instance, maybe if A is finite. And I'd definitely bet on it being true if A=α is a power of ω and B is finite; probably more generally if A=α is total and B is finite. (Probably also if o(A) is initial and B is finite.) Beyond that it's hard to say. One can probably get it to work with other smallness conditions -- I had to take o(B)=ω³ to get a counterexample, can one prove that's the smallest? -- but I don't know to what extent I'll bother...

**EDIT**: OK, here's a counterexample with B finite: Take α=ω once again, take B'=2×2, and take B that looks as follows:* | * * \ / *Maybe it can still work if A is finite, though, at least!

**EDIT**: Nope, it fails then too! Take A=2, and take B=(1∪ω)+(1∪ω), where here by "union" I mean "disjoint union"; you can extend the lower copy of 1∪ω to an ω to get B'=ω+(1∪ω). Then o(Rev(2,B))=ω²3+ω+2, but o(Rev(2,B'))=ω²3+ω+1.(I think this case should work also with A=ω, but I haven't done a proper check. I could be wrong there.)

So, basically, this one is unsalvageable. It works in the trivial case when A and B are both finite, and some other trivial or not-so-trivial cases I can prove, but not in any decent generality. I'm calling this one a wrap.

-Harry