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Thread: Ordinal Arithmetic

  1. #1
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    Ordinal Arithmetic

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

    Am I right in assuming that the sum of two finite ordinals cannot be an infinite ordinal? If so I figure $\displaystyle \beta = \omega $ and $\displaystyle \alpha$ can just be any finite ordinal. So $\displaystyle \alpha \beta = \omega$ and $\displaystyle \alpha^\beta = \omega$ (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.
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  2. #2
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    I think you are right.

    Edit: that the sum of two finite ordinals cannot be infinite is obvious. One definition of addition $\displaystyle \alpha+\beta$ is that you make a disjoint union of $\displaystyle \alpha$ and $\displaystyle \beta$ and make every element of $\displaystyle \beta$ greater than any element of $\displaystyle \alpha$. Well, in taking a union of two finite sets you will never get an infinite set.
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  3. #3
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    Yes as you pointed out, $\displaystyle \alpha$ or $\displaystyle \beta$ must be greater or equal to $\displaystyle \omega$.

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