if $\alpha<\beta$ then $\omega^\alpha+\omega^{\beta}=\omega^{\beta}$

Question

Prove that if $\alpha<\beta$ then $\omega^{\alpha}+\omega^{\beta}=\omega^{\beta}$

Proof: I'm not too sure what to start with to attempt this, I was thinking that I would need to perform transfinite induction either $\alpha$ or $\beta$ but i'm not sure, if someone could give me some direction on what to do that would be great.

Answer 1

Hint : There is $\delta >0$ such that $\alpha + \delta = \beta$, so that $\omega^\alpha + \omega^\beta = \omega^\alpha + \omega^\alpha \omega^\delta = ...$

Answer 2

Consider ordinals as well-ordered sets in the usual way. As $\omega^\beta\ge\omega^{\alpha+1}$ then $\omega^\beta$ has an initial segment order-isomorphic to $\omega^{\alpha+1}$. Removing this gives a well-ordered set, order-isomorphic to an ordinal $\gamma$, so $$\omega^\beta=\omega^{\alpha+1}+\gamma.$$ Thus it suffices to prove that $\omega^{\alpha+1}=\omega^\alpha +\omega^{\alpha+1}$. But $\omega^{\alpha+1}$ is order-isomorphic to a $\omega^\alpha$ concatenated with itself "$\omega$" times (the lexicographic product of $\omega^\alpha$ with $\omega$). Adding an extra $\omega^\alpha$ at the front won't change this order type, so indeed $\omega^\alpha+\omega^{\alpha+1} =\omega^{\alpha+1}$.