Indecomposable limit ordinals

Question

A limit ordinal $\gamma>0$ is said to be indecomposable iff $\nexists\alpha,\beta<\gamma$ such that $\alpha+\beta=\gamma$. In view of this definition, I’m trying to prove the equivalence of the following statements for limit ordinals $\gamma$:

(1) $\gamma$ is indecomposable.

(2) $\alpha+\gamma=\gamma$ $\forall\alpha<\gamma$.

(3) $\gamma=\omega^\alpha$ for some $\alpha\in\mathrm{ON}$.

I’ve been able to get $(1)\implies(2)$ and $(3)\implies(1)$, but I have nothing more than scratch work for $(2)\implies(3)$. Any suggestions would be appreciated.

Answer 1

It's easy to show $(1)\implies(3)$ and $(2)\implies(1)$, and deduce the final implication. Because both of these are really obvious.

To see $(1)\implies(3)$ note that if $\gamma$ is indecomposable then its Cantor normal form has to have only one term, and it has to be $\omega^\alpha$ for some $\alpha$.

To see that $(2)\implies(1)$ note that if $\alpha,\beta<\gamma$ then we have that $\alpha+\beta+\gamma=\gamma$, but if $\alpha+\beta=\gamma$ then $\gamma+\gamma=\gamma$, which is impossible.

Answer 2

Suppose that $\gamma$ is not a power of $\omega$. Let $A=\{\alpha\le\gamma:\alpha=\omega^\beta\text{ for some }\beta\in\mathbf{ON}\}$, and let $B=\{\beta\in\mathbf{ON}:\omega^\beta\in A\}$; then $\sup A=\omega^{\sup B}$, so $\sup A<\gamma$. Let $\alpha=\sup A$ and $\beta=\sup B$, and let $\eta$ be the unique ordinal such that $\alpha+\eta=\gamma$. Suppose that $\eta=\gamma$; then $\alpha+\gamma=\gamma$, and by an easy induction $\alpha\cdot n+\gamma=\gamma$ for each $n\in\omega$. But then $\alpha\cdot n\le\gamma$ for each $n\in\omega$, and therefore

$$\omega^{\beta+1}=\omega^\beta\cdot\omega=\alpha\cdot\omega=\sup_{n\in\omega}\alpha\cdot n\le\gamma<\omega^{\beta+1}\;,$$

an obvious contradiction.