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We are given 2 ordinals: $\alpha$ and $\beta$ where $\beta$ does not have a maximal number (So it's transfinite, right?)

We are asked to find $\alpha,\beta$ such that:

$\alpha+\beta > \beta+\alpha$

My problem is, that I don't think such a solution exists when alpha is a finite ordinal (lets say that $\beta=\omega$). if it is, then $\alpha+\beta = \{0,1,2,...,\alpha-1,0^*,1^*,2^*,...\} = \omega$

and $\beta+\alpha = \{0,1,2,...,0^*,1^*,2^*,...(\alpha-1)^*\} > \omega$

So I think $\alpha$ also has to be a transfinite ordinals. But I can't for the life of me think of such an ordinal that this will be true.

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  • $\begingroup$ There are so many ordinals without maximum, for example $0,\omega,2\omega,\omega^2,\omega^3,\omega^\omega,\ldots$ $\endgroup$ – Hagen von Eitzen Jan 8 '14 at 22:31
  • $\begingroup$ I know. I just took $\omega$ as an example. $\endgroup$ – Oria Gruber Jan 8 '14 at 22:32
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HINT: Think of a case where $\beta+\alpha=\alpha$. For example when $\alpha$ is much bigger than $\beta$.

(Also, ordinal is without a maximum if and only if it is not a successor ordinal, which means either zero or a limit ordinal - which has to be infinite.)

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  • $\begingroup$ if $\beta$ is the zero ordinal that's no good to us, because addition with zero commutes. so $\beta$ has to be transfinite. let's look at an example where alpha is much bigger. $\beta=\omega$, $\alpha=\omega^3$. why is $\beta+\alpha=\alpha$? why isn't the other way around true? $\endgroup$ – Oria Gruber Jan 8 '14 at 22:35
  • $\begingroup$ Try $\omega^2$ instead, it will be simpler. It's just $\omega$ copies of $\omega$ stacked one on top of another, so it's easy to understand why $\omega+\omega^2=\omega^2$. On the other hand, $\omega^2$ has an end-segment of $\omega$, whereas $\omega^2$ doesn't. Therefore $\omega^2\neq\omega^2+\omega$, and so it is clearly smaller. $\endgroup$ – Asaf Karagila Jan 8 '14 at 23:11

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