I've been having trouble with the following:

Assume that a, b are ordinals with 1< min(a, b) and omega <= max(a, b). Then |a^b| =max(a, b).

Anything would help, thanks.

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- Mar 12th 2010, 01:58 PMculturalreferenceInfinite ordinal exponentiation doesn't raise cardinality
I've been having trouble with the following:

Assume that a, b are ordinals with 1< min(a, b) and omega <= max(a, b). Then |a^b| =max(a, b).

Anything would help, thanks. - Mar 12th 2010, 05:37 PMwgunther
- Mar 13th 2010, 09:02 AMculturalreference
your counter example is not true if b = anything greater than or equal to omega, like it has to be, by the hypothesis.

The hypothesis says that one or both of the ordinals is greater than or equal to omega, and if one is not greater than omega then it can't be less than two.

thanks. - Mar 13th 2010, 09:15 AMwgunther
$\displaystyle 2^\kappa=|\wp(\kappa)|>\kappa$ for all kappa, even omega. 2^omega has cardinality continuum which can be very large, def larger than omega. The thing that is similar to what you are saying has extra hypothesizes about cofinality. So I'm still not sure what your question means.

- Mar 13th 2010, 09:45 AMclic-clac
@wgunther: $\displaystyle 2^\omega=\omega$ ;) this is ordinal exponentiation, not cardinal exponentiation.

@culturalreference: Have you tried by ordinal induction? - Mar 13th 2010, 09:56 AMwgunther
- Mar 13th 2010, 10:10 AMculturalreference
clic-clac: thanks for the hint, but i'm struggling with what the mapping would go to. In other words,

so i show by induction that a^n = a (for a > omega), for n < b < omega. Then i feel like the next step has to be to show a map from U a^n ---> something, but i feel like this map can't just be to omega. does it have to be to omega x omega, or a x omega, or...?

Thanks, and sorry if this is muddled.