Ordinal arithmetic and limit ordinals

Question

Suppose $1\leq\xi<\omega_1$ is a countable ordinal and that $1\leq\zeta<\omega^\xi$. Is it always true that $\zeta+\omega^\xi=\omega^\xi$? If so, why?

Answer 1

Suppose first that $\xi$ is a limit ordinal, so that $\omega^\xi=\sup_{\eta<\xi}\omega^\eta$; then

$$\zeta+\omega^\xi=\sup_{\eta<\xi}(\zeta+\omega^\eta)=\zeta+\sup_{\eta<\xi}\omega^\eta=\zeta+\omega^\xi\;.$$

Now suppose that $\xi=\eta+1$. Then $\omega^\xi=\omega^\eta\cdot\omega$, so $\zeta+\omega^\xi=\zeta+\omega^\eta\cdot\omega$. Since $\zeta<\omega^\xi$, there is a unique ordinal $\alpha$ such that $\zeta+\alpha=\omega^\xi$; clearly $\alpha\le\omega^\xi$. Suppose that $\alpha<\omega^\xi$; $\omega^\xi=\sup_{n\in\omega}\omega^\eta\cdot n$, so there is an $n\in\omega$ such that $\zeta<\omega^\eta\cdot n$ and $\alpha<\omega^\eta\cdot n$. But then

$$\zeta+\alpha\le\omega^\eta\cdot n+\omega^\eta\cdot n=\omega^\eta\cdot(2n)<\omega^\xi\;,$$

which is impossible. Thus, $\alpha=\omega^\xi$, and $\zeta+\omega^\xi=\omega^\xi$.