 Solution

First let $\delta ={\omega }^{\alpha }$. If $\beta ,\gamma <\delta$, every exponent appearing in their Cantor normal forms is strictly below $\alpha$. The rules for ordinal addition can delete lower terms but cannot introduce an exponent at least $\alpha$. Thereforeso $\delta$ is additively closed.

Conversely, suppose nonzero $\delta$ is additively closed. Part (a) givesIf $\rho >0$, then both ${\omega }^{\alpha }n$ and $\rho$ are smaller than $\delta$, but their sum is $\delta$, contradicting closure. Hence $\rho =0$. If now $n>1$, thenwhileanother contradiction. Thus $n=1$, andThis proves the classification of additively closed ordinals.