How to define the Cantor's normal form, using the base \(\Omega\) (with the purposes of the ordinal collapsing functions)? It seems that writing the ordinal \(\theta(\Omega*\omega^2)\) as \(\theta(\omega^{\Omega+2})\) makes it easier to fit the usual Cantor's normal form, as we avoid multiple multiplications in a single non-additively principal ordinal: \(\theta(\Omega*\omega^2)\) can be expanded to \(\theta(\Omega*\omega*\omega)\), and it's probably escapes the Cantor's normal form. Ikosarakt1 (talk ^ contribs) 10:27, August 11, 2013 (UTC)
I believe Cantor normal form uses only addition and exponentiation, so your ordinal would be \(\theta(\omega^{\Omega+2})\). By the way, this ordinal is additively principal. LittlePeng9 (talk) 12:40, August 11, 2013 (UTC)
Cantor normal form can use multiplications, but only when one number is a finite. This makes more convenient (shortened) situation in the rule \(\omega^{\alpha+1} = \omega^\alpha*n\) instead of \(\omega^\alpha+\omega^\alpha+\cdots+\omega^\alpha+\omega^\alpha\) (with n \(\omega^\alpha\)'s). I wonder how we can create an equivalent system allowing transfinite multiplications. Ikosarakt1 (talk ^ contribs) 17:28, August 11, 2013 (UTC)
- Cantor normal form to a base \(\alpha\) is \(\alpha^{\beta_1} \gamma_1 + \alpha^{\beta_2} \gamma_2 + \ldots + \alpha^{\beta_n} \gamma_n\), where \(\beta_1 > \beta_2 > \ldots > \beta_n\) and \(\gamma_i < \alpha\) for all \(i\). When \(\alpha = \omega\), you can expand things out so that the \(\gamma_i\)'s go away. Deedlit11 (talk) 22:00, October 30, 2013 (UTC)