= 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 $\alpha$. Therefore
$$
\beta+\gamma<\omega^\alpha=\delta,
$$
so $\delta$ is <additively closed ordinal>[additively closed].
Conversely, suppose nonzero $\delta$ is additively closed. Part (a) gives
$$
\delta=\omega^\alpha n+\rho,
\qquad
0<n<\omega,\quad \rho<\omega^\alpha.
$$
If $\rho>0$, then both $\omega^\alpha n$ and $\rho$ are smaller than $\delta$, but their sum is $\delta$, contradicting closure. Hence $\rho=0$. If now $n>1$, then
$$
\omega^\alpha(n-1)<\delta,
\qquad
\omega^\alpha<\delta,
$$
while
$$
\omega^\alpha(n-1)+\omega^\alpha
=\omega^\alpha n=\delta,
$$
another contradiction. Thus $n=1$, and
$$
\boxed{\delta=\omega^\alpha}.
$$
This proves the classification of <additively closed ordinals>.
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