Question

Many books describe how one can construct "by hand" a table of ordinals $1,\ 2,\ \ldots,\ \omega,\ \omega +1,\ \omega +2,\ \ldots,\ \omega\cdot 2,\ \omega\cdot 2 +1,\ \ldots,\ \omega^{2},\ \ldots,\ \omega^{3},\ \ldots\ \omega^{\omega},\ \ldots,\ \omega^{\omega^{\omega}},\ \ldots, \epsilon_{0},\ \ldots$. But does this span the entire ordinal class? For some reason I can't seem to prove it. Is there an easy way to see that? Thanks!

Answer 1

Since ordinal numbers have a unique division, logarithm and subtraction properties, when given an ordinal $\alpha$ you can write any other ordinal as a finite polynomial in $\alpha$, when $\alpha = \omega$ you get what's known as "Cantor normal form of $\gamma$ for the base $\omega$".

I.e., any ordinal $\gamma$ can be written as a finite sum: $$\gamma = \sum_{i=0}^n \omega^{\alpha_i}\cdot\beta_i$$ Where $\alpha_i$ is a decreasing chain of ordinals, and $\beta_i$ is finite.

(More generally, you can take some base $\zeta$ and then $\beta_i < \zeta$)

Thing is that we only have a finite number of symbols, so at most we can represent (uniquely) a countable number of numbers, since we have a proper class of ordinals, which is a mind boggling concept of infinitude, you obviously can't write them all. But still, any given ordinal can be presented as a finite polynomial in $\omega$, thus spanning the table discussed.