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Let $A$ be an infinite set and ${\mathcal C}$ be the collection of all the subsets of $A$ which have the same cardinality with $A$. Does ${\mathcal C}$ have the same cardinality with the power set of $A$?

Yiorgos S. Smyrlis
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    Haha. @Andres, that is so delightfully ironic that you point out the duplicate of this question is something that you answered and I pointed out was a duplicate. :-) – Asaf Karagila Dec 07 '13 at 23:54
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    @AsafKaragila Oh, that's funny! I don't think I knew it had been closed; I remembered I had answered this recently, and looked through my answers for the link. – Andrés E. Caicedo Dec 07 '13 at 23:56

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Assuming the axiom of choice, yes.

The reason is that $A$ can be written as $B\cup C$ where both $B$ and $C$ have the same cardinality as $A$. It is easy to see now that taking every $C'\subseteq C$ we have $B\cup C\in\cal C$, therefore $|\mathcal P(C)|\leq|\cal C|$.

Not assuming the axiom of choice it is consistent that the answer is no. For example if $A$ is an infinite Dedekind-finite set, then the only subset of $A$ which has the same cardinality as $A$ is $A$ itself.

Asaf Karagila
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