User blog comment:Rgetar/Simple way to create lists of ordinals/@comment-35392788-20181008074754/@comment-32213734-20181009055542

Syst3ms, I think yes.

For example, let α is countable ordinal. Take pair 0 - α. Enumerate


 * ordinals of its single expansion in increasing order (0, ...),
 * ordinals of its double expansion, absent in its single expansion, in increasing order
 * ordinals of its triple expansion, absent in its double expansion, in increasing order

And there are no pairs of countable ordinals with infinite single (double, triple, ...) expansion.

Proof: let for a given fundamental sequence system b0 is least ordinal such as there is pair a - b0, a < b0, with infinite single expansion. Its small expansion is a, b1. Pair a - b1 also has infinite single expansion. But b1 < b0. Contradiction.

Note: this is also true for uncountable ordinals (that is single expansion of any pair of ordinals is finite), but we cannot enumerate all ordinals up to an uncountable ordinal this way, since for uncountable ordinals we need uncountable number of expansion levels.

But we can "enumerate" all ordinals up to any ordinal of cardinality Ω with ordinals < Ω this way using Ω levels of expansions (Generally, all ordinals up to any ordinal of cardinality Ωα, with ordinals < Ωα using Ωα levels of expansions).

For example, let Ω[n] = n (here n is countable ordinal). For pair 0 - Ω


 * single expansion: 0, 1
 * double expansion: 0, 1, 2
 * triple expansion: 0, 1, 2, 3
 * ω-th expansion: 0, 1, 2, 3, ..., ω
 * ω + 1-th expansion: 0, 1, 2, 3, ..., ω, ω + 1
 * ω + 1-th expansion: 0, 1, 2, 3, ..., ω, ω + 1

So, "enumeration" of all ordinals up to Ω is 0, 1, 2, 3, ... ω, ω + 1, ..., that is they are just placed in increasing order.