0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ math.stackexchange.com/questions/2347786/… $\endgroup$
    – Bumblebee
    Nov 18 '17 at 16:26
  • $\begingroup$ I don't see which part of that question is related. $\endgroup$
    – user395952
    Nov 18 '17 at 16:30
  • $\begingroup$ This, on the other hand, is much more relevant, though. $\endgroup$
    – Asaf Karagila
    Nov 18 '17 at 16:39
  • $\begingroup$ @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. $\endgroup$
    – Bumblebee
    Nov 18 '17 at 16:51
  • $\begingroup$ @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? $\endgroup$
    – user395952
    Nov 18 '17 at 18:16
1
$\begingroup$

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 $\omega^\alpha+\omega^{\alpha+1} =\omega^{\alpha+1}$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ This is a hint, not an answer. $\endgroup$ May 29 '18 at 15:07
  • $\begingroup$ @Alephnull : there is the word "Hint" in boldface indeed.. $\endgroup$ May 29 '18 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy