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Why is $\Omega$ the first uncountable ordinal number and not $\omega^\omega$? Isnt the latter a countably infinite product of countably infinite sets and hence uncountable?

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    $\begingroup$ and isn't a countable product of countable sets countable? ;) $\endgroup$ – W_D Aug 26 '13 at 6:34
  • $\begingroup$ I meant a countably inf product of countably inf sets. That is not countable, unlike the case of unions. For example, use some form of cantor diagonal argument $\endgroup$ – renrenthehamster Aug 26 '13 at 6:36
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The ordinal exponentiation $\omega^\omega$ is a countable set. But is is not a countable product of countable sets. Instead it is the limit of $\omega^n$, all of which are countable.

As cardinal exponentiation $\omega^\omega=2^\omega$, and the question whether or not there is a smaller uncountable cardinal is known as the continuum hypothesis, and is unprovable from the usual axioms of set theory, that is from $\sf ZFC$.


Related:

  1. Cardinal Arithmetic versus Ordinal Arithmetic
  2. Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
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