if $\omega^{\alpha} =A \cup B$ then $A$ or $B$ has order type $\omega^{\alpha}$

Question

Show that if $\omega^{\alpha} =A \cup B$ then $A$ or $B$ has order type $\omega^{\alpha}$ where $\alpha \geq 1$ is a ordinal number.

Hint: Use induction on $\alpha$

I don't know how to start with the base case, can someone give me some help

Answer 1

Let us consider the case $\alpha=3$ and assume that our theorem holds for $\alpha=1,2$. You can generalize my argument to any successor cases.

Let $\omega^3=A\cup B$. For $X\subseteq \omega^3$ define $$X[n] = \{\xi<\omega^2\mid \omega^2\cdot n+\xi\in X\}.$$ Then $\omega^2=A[n]\cup B[n]$ for each $n<\omega$. This shows one of $A[n]$ or $B[n]$ has the ordertype $\omega^2$. If there are at most finitely many $n$ such that the ordertype of $A[n]$ is $\omega^2$, then the ordertype of $B$ is $\omega^3$. Otherwise, the ordertype of $A$ is $\omega^3$.

Now suppose that $\alpha$ is a limit ordinal and $\omega^\alpha=A\cup B$. Then we can repeat a similar argument I gave before with $A\cap \omega^\beta$ and $B\cap \omega^\beta$ for $\beta<\alpha$.