- $|A|^{|B|} = |A^B|$ ? (cardinal exponentiation)
- 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?
Ordinal and cardinal exponentiation
3
$\begingroup$
elementary-set-theory
-
1There 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
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.