 Solution

For a nonzero ordinal $\alpha$, letThe set is nonempty because $0\in S$, and it is an initial segment because ordinal exponentiation by $\omega$ is strictly increasing. Put $\beta =\sup S$.

If $\beta$ is a successor, the definition of the supremum forces $\beta \in S$. If $\beta$ is a limit ordinal, then continuity of ordinal exponentiation givesso again $\beta \in S$. Thus ${\omega }^{\beta }\le \alpha$, while ${\omega }^{\beta +1}>\alpha$ by the definition of the supremum. Hence $\beta$ is the greatest required exponent.

Apply division by an additively indecomposable ordinal with $\lambda ={\omega }^{\beta }$:Since $\alpha \ge {\omega }^{\beta }$, the quotient $q$ is nonzero. It must be finite: if $q\ge \omega$, thencontradicting the maximality of $\beta$. Writing $q=n$ givesThis is the leading-term decomposition.