Let $\alpha$ and $\beta$ two infinite cardinal numbers.
Can we have $\alpha = \beta^\alpha$?
This problem comes from a situation where I am dealing with the cardinal of a set of functions.
2026-04-06 18:15:08.1775499308
Infinite cardinal comparation
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No, we cannot. Cantor's diagonalization argument shows that $\alpha<2^\alpha,$ and it is readily apparent that $2^\alpha\le\beta^\alpha$ for any $\beta\ge 2.$