@Joel: math.stackexchange.com/questions/184816

@Joel: I'll read your simplified argument now. So that you don't feel ignored in the meantime: I thought your original argument was very nice because (i) I was mildly perturbed that I hadn't used the non-increasing property of these sequences yet and (ii) I think that handling the largest classes first makes it clearest that you avoid problems. So you gave me those two bonuses also. I am currently thinking about how to say more about the assignments within each constant segment. Since $\delta_{k_0} = \delta$ is the number of classes in the segment, it at least gives me a limit to work with.

@Gerhard: If you are curious, I think I know what happens with bijective reductions when everything is Borel, and it doesn't look like the generalization breaks anything. $E \cong F$ ($E$ bijectively reduces to $F$) iff $s_f(E)_i = s_f(F)_i$ for all $i$, where I am using slightly different notation from Hamkins. $s_f(E)$ is the sequence indexed by cardinals $I$ where $s_f(E)_i$ is the number of $E$-classes of size exactly $i$. [cont above...] (These are out of order because I mistyped your name the first time and had to delete the comment. It appears comments can't be edited here?)

I have been calling this the fine shape of $E$ and the analogous sequence where $s_c(E)_i$ is the number of $E$-classes of size at least $i$ the coarse shape of $E$. Comparing the shapes of two relations componentwise (using the relations on cardinals) tells you exactly when reductions, bireductions, embeddings, biembeddings, and isomorphisms exist. The case above was by far the most difficult to prove in the Borel case.

Perhaps I should mention another probably familiar trick: 1,1,2,1,2,3,1,2,3,4,.... You can use the pattern in this sequence to interleave ctbly many sequences by letting the 1st occurrence of 1 be the 1st term of sequence 1 and generally the ith occurrence of n be n_i. You can do this repeatedly, interleaving many sequences into one, then interleaving this with many others, etc. So if you can chunk up the problem into countable pieces, you can possibly weave together an order. Thanks for the welcome, @David. I have heard of JDH, which is why I am scared I won't finish this proof soon.

@Trevor: I edited some further explanation into my question.

@Trevor: It sounds plausible to me (?) that sending a class to the smallest available one that works will avoid fatally wasteful assignments regardless of the order in which you make assignments. My eventual restrictions make the size of a class $\leq\omega$ and the number of classes $\leq 2^\omega$, changing the array in the link only by adding columns for $\omega$ and $2^\omega$. Snake through the finite part as shown, and separately snake through the two additional columns the same way. Then interleave your snakes, giving $(1,1),(\omega,1),(1,2),(\omega,2),(2,1),(2^\omega,1),\ldots$. Works?

@Joel: Uh-oh. Does that mean it's not getting an answer soon? :^) This is just for an expository paper where I am trying to establish what happens with reductions of equivalence relations generally before I confine myself to Borel reductions of countable Borel equivalence relations. I want to be as constructive as possible now to inform later proofs. I might have made it too difficult by removing the cardinality constraints, esp. because I don't think I appreciate the vastness of the ordinals or even cardinals, which leaves me unsure of whether I am considering all that can happen.