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

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.

• math.stackexchange.com/questions/2347786/… Nov 18 '17 at 16:26
• I don't see which part of that question is related.
– user395952
Nov 18 '17 at 16:30
• This, on the other hand, is much more relevant, though. Nov 18 '17 at 16:39
• @Gibberish: You are asking for a direction to start. I thought you could find some starting point from that problem. Also there is a link given as a comment which is closely related to your problem. Nov 18 '17 at 16:51
• @Bumblebee oh okay thank you for the contribution. The post that is more relevant uses cantors normal form, which i haven't been taught yet so i was wondering if there is an approach that does not require this?
– user395952
Nov 18 '17 at 18:16

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 $\omega^\alpha+\omega^{\alpha+1} =\omega^{\alpha+1}$.
Hint : There is $\delta >0$ such that $\alpha + \delta = \beta$, so that $\omega^\alpha + \omega^\beta = \omega^\alpha + \omega^\alpha \omega^\delta = ...$