User blog comment:Rgetar/Simple way to create lists of ordinals/@comment-35392788-20181008074754/@comment-35470197-20181009072937

@Syst3ms

Yes. The only non-trivial statement is the surjectivity of the expression, which follows from the following argument:

Let \(\beta\) be a countable ordinal with a fixed system of fundamental sequences of all non-zero limit ordinals \(\beta' \leq \beta\). For a successor ordinal \(\beta' \leq \beta\), I denote by \(\beta'[0]\) the predeccessor of \(\beta'\).

Let \(\alpha < \beta\). Repeat the following process:


 * 1) Let \(i = 1\) (unsigned integer) and \(n[] = \) (empty array).
 * 2) If \(\beta\) is a successor ordinal, put \(n[i] = 0\).
 * 3) If \(\beta\) is a non-zero limit ordinal, put \(n[i] = \min \{n < \omega \mid \alpha \leq \beta[n]\}\).
 * 4) Replace \(\beta\) by \(\beta[n[i]]\).
 * 5) If \(\alpha = \beta\), then finish the process.
 * 6) Replace \(i\) by \(i+1\).
 * 7) Go to 2.

By the well-foundedness of ordinals, this programme terminates. Then \(\alpha\) is enumerated in the \(i\)-th level, and the equality \(\alpha = \beta[n[0]][n[1]] \cdots [n[i]]\) holds.