# $\alpha = \omega^\beta(\gamma+1)$

This paper mentions that any ordinal $\alpha < \epsilon_0$ can be written uniquely as $\alpha = \omega^\beta(\gamma+1)$, where $\beta<\alpha$.

Does this presentation have a name? Also, how would one get $\omega^\beta(\gamma+1)$ for a given $\alpha$? I played with this for a while and got stuck when setting $\alpha$ to, say, $\omega^\omega+1$. (Or is this just Cantor normal form disguised?)

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$\omega^{\omega}+1 = \omega^0(\omega^{\omega}+1)$ –  Clive Newstead Jan 11 at 13:14
@CliveNewstead Hah, thanks. Seems I somehow implicitly assumed that $\gamma < \alpha$. But is there some general "formula" to break $\alpha$ to the defined form? (I'm trying to prove it but it gets messy...) –  rank Jan 11 at 13:21
Well if $\alpha$ is a successor then you need $\beta=0$. If it's not a successor, factorise it's CNF. –  Clive Newstead Jan 11 at 13:25
Note that given $\alpha < \varepsilon_0$ its Cantor Normal Form is $$\omega^{\beta_1} \cdot k_1 + \cdots + \omega^{\beta_n} \cdot k_n$$ where $\alpha > \beta_1 > \cdots > \beta_n$ and $0 < k_1 , \ldots , k_n < \omega$. Now essentially factor out $\omega^{\beta_n}$ to the left to get the representation (as $0 < k_n < \omega$ whatever is left over after factoring out must be a successor ordinal).