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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?
  • 0
    1.$|\alpha|^{|\beta|}$ (Ordinal exponentiation).2012-06-06
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    2.$|\alpha|^{|\beta|}$ (Cardinal exponentiation)2012-06-06
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    Are they equal?2012-06-06
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    No, they aren't ... $|\omega^\omega| = \omega$ (ordinal exponentiation) ...2012-06-06
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    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.2012-06-06

1 Answers 1

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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.