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Let $\omega$ denote the countably infinite cardinality and $A$ a cardinality that is strictly larger (i.e., an uncountable one). Is it true that $A^\omega$ has strictly larger cardinality than $A$?

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up vote 8 down vote accepted

No, it is not necessarily true. For example, let $A$ be the cardinality of the Cantor set, $2^{\aleph_0}$. Then $A^\omega=(2^{\aleph_0})^{\aleph_0}=2^{(\aleph_0\times\aleph_0)}=2^{\aleph_0}=A$.

On the other hand, it is not necessarily false. For example, if $A$ is the supremum of the sequence $\aleph_0,2^{\aleph_0},2^{2^{\aleph_0}},\dots$, then $A^\omega>A$.

This is because of a general fact due to König: The cofinality ${\rm cf}(A)$ of a cardinal $A$ is the smallest $\kappa$ such that $A$ can be written as a union of $\kappa$ sets, each of size smaller than $A$. König showed that $A^{{\rm cf}(A)}>A$ for any $A$ infinite--this is a generalization of Cantor's easier result that $2^\kappa>\kappa$. The $A$ in the previous paragraph explicitly has cofinality $\omega$.

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It is also true when $A\lt 2^{\aleph_0}$. – Jonas Meyer Dec 12 '10 at 0:31
@Jonas: Sure, if $1<A\le 2^\kappa$, then $A^\kappa=2^\kappa$. In particular, if $A<2^{\aleph_0}$, then $A^\w=2^{\aleph_0}$. – Andrés E. Caicedo Dec 12 '10 at 0:36
Thanks, Andres! – Michele Kakusi Dec 12 '10 at 0:50
So, similarly to Jonas's comment, this is true for any cardinality $A$ that is not the form of $\kappa^{\aleph_0}$, since $A^{\aleph_0}$ has at least the cardinality of $A$, but if $A$ is not of that form, then they can't be equal. Right? – Michele Kakusi Dec 12 '10 at 0:59
@user4654 : Yes, exactly. – Andrés E. Caicedo Dec 12 '10 at 2:02

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