# A limit ordinal $\gamma$ is indecomposable if and only if $\alpha + \gamma = \gamma$ if and only if $\gamma = \omega^{\alpha}$ for some $\alpha$. [duplicate]

I am trying to prove ex 2.13 from Jech's book on set theory:

Ex 2.13: A limit ordinal $\gamma$ is indecomposable if and only if $\alpha + \gamma = \gamma$ if and only if $\gamma = \omega^{\alpha}$ for some $\alpha$.

I think this is equivalent to proving that the 3 following statement are equivalent:

1. $\gamma$ is indecomposable (i.e. there are no $\alpha,\beta < \gamma$ s.t. $\gamma = \alpha + \beta$).

2. For every $\alpha < \gamma$, $\alpha + \gamma = \gamma$.

3. There exists $\alpha$ s.t. $\gamma = \omega^{\alpha}$.

I think that I can prove $(1) \rightarrow (3)$ since it seems almost straight forward from Cantor's normal form theorem. I can also prove that $(1) \Rightarrow (2)$ since, for every $\alpha < \gamma$, there exists a unigue $\delta$ s.t. $\alpha + \delta = \gamma$. Given, (1), $\delta \geq \gamma$. But $\delta > \gamma \Rightarrow \alpha + \delta > \gamma$ and we are left with $\delta = \gamma$.

My problem is with the other directions. Any help?

For $(2) \rightarrow (3)$ assume that $\gamma$ can't be written as $\omega^\alpha$. Then by Cantor's normal form theorem $\gamma = \omega^{\beta_1}\cdot k_1 + \dots + \omega^{\beta_n}\cdot k_n$ with $k_1 > 1$ or $n>1$. But then $\omega^{\beta_1} < \gamma$ and also \begin{align*} \omega^{\beta_1} + \gamma = \omega^{\beta_1} + (\omega^{\beta_1}\cdot k_1 + \dots + \omega^{\beta_n}\cdot k_n) &= (\omega^{\beta_1} + \omega^{\beta_1}\cdot k_1) + \dots + \omega^{\beta_n}\cdot k_n\\ &= \omega^{\beta_1}\cdot (k_1+1) + \dots + \omega^{\beta_n}\cdot k_n\\ &> \omega^{\beta_1}\cdot k_1 + \dots + \omega^{\beta_n}\cdot k_n = \gamma \end{align*} which together contradicts $(2)$.
For $(3) \rightarrow (1)$ let $\beta_1,\beta_2<\gamma=\omega^\alpha$. By Cantor's normal form theorem you'll easily find $\alpha'<\alpha$ and $k\in\omega$ with $\beta_1,\beta_2<\omega^{\alpha'}\cdot k$. This implies $\beta_1 + \beta_2 < \omega^{\alpha'}\cdot (k+k) < \omega^\alpha=\gamma$.
• Thank you for your answer. Most of it I understood. Exapt one thing: In $(2) \rightarrow (3)$, at the end of the proof, how do we get that $\omega^{\beta_1} + \gamma > \gamma$? Sep 16 '14 at 14:30
• In $\omega^{\beta_1}+\gamma$ write $\gamma$ as $\omega^{\beta_1}\cdot k_1 + \dots + \omega^{\beta_n}\cdot k_n$, then the sum is $\omega^{\beta_1}\cdot (k_1+1) + \dots + \omega^{\beta_n}\cdot k_n$. Sep 16 '14 at 14:58