Ordinal Arithmetic

**whatisthisfor** · November 19th 2009, 06:27 AM

I have a question here on ordinal arithmetic: Suppose $\alpha + \beta = \omega$ ( $\alpha, \beta$ not zero). What are $\alpha \beta , \alpha^\beta$ ?

Am I right in assuming that the sum of two finite ordinals cannot be an infinite ordinal? If so I figure $\beta = \omega$ and $\alpha$ can just be any finite ordinal. So $\alpha \beta = \omega$ and $\alpha^\beta = \omega$ (by two rules in my notes).

Alpha and beta cannot be the other way round as right cancellation does not work.

So I think this proof is sound if the first statement is true. If anyone can comment on what I've done that'd be neat.

**emakarov** · November 19th 2009, 08:03 AM

I think you are right.

Edit: that the sum of two finite ordinals cannot be infinite is obvious. One definition of addition $\alpha+\beta$ is that you make a disjoint union of $\alpha$ and $\beta$ and make every element of $\beta$ greater than any element of $\alpha$ . Well, in taking a union of two finite sets you will never get an infinite set.

**clic-clac** · November 19th 2009, 08:38 AM

Yes as you pointed out, $\alpha$ or $\beta$ must be greater or equal to $\omega$ .

But if one of them is strictly greater than $\omega,$ then, since left and right ordinal addition are increasing "functions", their sum is strictly greater than $\omega.$
Finally, since $\omega+\omega>\omega,$ $\alpha$ or $\beta$ is an integer, and you can end as you did.

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