Exception in the statement $(\alpha ^ \beta) ^ \gamma = \alpha ^ {(\beta \gamma)}$ for arbitrary cardinals.

Question

$(\alpha ^ \beta) ^ \gamma = \alpha ^ {(\beta \gamma)}$ for arbitrary cardinals.

I found many proofs of this statement on this site. However, none of them considers an exception in this statement.

When $\alpha > 0$ and $\beta = \gamma = 0$, $(\alpha ^ \beta) ^ \gamma = 0^0 = 1 > 0 = \alpha ^ 0 = \alpha ^ {(\beta \gamma)}$.

Am I missing something here or are all the answers from this site proving this statement missing this exception?

@SassatelliGiulio Okay, I see that I've made a mistake in assuming $1^0 = 0$ for cardinals. It should be $1^0 = 1$ because there is an empty map from an empty set to a set with cardinality 1. — Chanhyuk Park, Apr 11 '25 at 22:42
I advise against using Greek letters for cardinals. You usually do that for ordinals, and there the arithmetic is different. — Jakobian, Apr 11 '25 at 23:50
@Jakobian Unless the Greek letter in question is $\kappa$, in which case it's standard. :P — Sassatelli Giulio, Apr 12 '25 at 08:08

score 3 · Accepted Answer · answered Apr 11 '25 at 23:23

As Sassatelli Giulio has noted in the comments, I've made a mistake in assuming $1^0 = 0$ for cardinals, but rather it should be $1^0 = 1$ because there is exactly one map from the empty set to any set. The statement holds true for all arbitrary cardinals.

For the reason why there is exactly one map from the the empty set to any set, please check out this answer (https://math.stackexchange.com/a/475620/1426819) from this site.

Exception in the statement $(\alpha ^ \beta) ^ \gamma = \alpha ^ {(\beta \gamma)}$ for arbitrary cardinals.

1 Answers1