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  1. $|A|^{|B|} = |A^B|$ ? (cardinal exponentiation)
  2. Let $\alpha$ and $\beta$ be ordinals and $\gamma$ = $|\alpha|^{|\beta|}$ (Ordinal exponentiation)
    Then is $\gamma$ an initial ordinal(thus cardinal) and can the ordinal exponentiation in this case be understood as a cardinal exponentiation?
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1.$|\alpha|^{|\beta|}$ (Ordinal exponentiation). – Katlus Jun 6 '12 at 9:08
2.$|\alpha|^{|\beta|}$ (Cardinal exponentiation) – Katlus Jun 6 '12 at 9:09
Are they equal? – Katlus Jun 6 '12 at 9:10
No, they aren't ... $|\omega^\omega| = \omega$ (ordinal exponentiation) ... – martini Jun 6 '12 at 9:40
There is one countably infinite cardinal, and uncountably many countablby infinite ordinals. Some of them defined as exponents of others. Hence exponentiation works differently between the two. – Arthur Jun 6 '12 at 10:03

With regards to 1: Cardinal exponentiation is not equal to ordinal exponentiation.

Take for example, $2^\omega$. If we are doing cardinal exponentiation, then this is the cardinality of the continuum, whereas if we are doing ordinal exponentiation, this is the limit of the sequence:

$\{2,4,8,16,32,64...\}$, which equals $\omega$.

With regards to 2:

$\alpha^\beta$ is not necessarily a cardinal. Take for example $\omega^2$. With ordinal exponentiation, this is equal to the limit of the sequence:

$\{ \omega,\omega\cdot2,\omega\cdot3,\omega\cdot4\dots\}$

Being a countable union of countable sets, $\omega^2$ is countable and strictly greater than $\omega$. Thus, it is not a cardinal.

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