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Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$

However without the axiom of choice this doesn't have to be true anymore. For example if $A$ is an amorphous set (infinite set that cannot be written as a disjoint union of two infinite sets), then it is actually true that $2^A<3^A<4^A<\ldots< A^A$. The reason these inequalities hold is that $A^A$ is actually Dedekind-finite, so whenever we remove elements we strictly decrease in cardinality.

Of course there are still sets that obey the equation $A^A=2^A$, even if $A$ cannot be well-ordered. For example given any set $A$ it is not hard to verify that $A^\omega$ has the property $A^\omega\times A^\omega=A^\omega$. From this follows:

$$2\lt A^\omega\leq 2^{A^\omega}\implies 2^{A^\omega}\leq\left(A^\omega\right)^{A^\omega}\leq \left(2^{A^\omega}\right)^{A^\omega}=2^{A^\omega}$$ (In fact we can replace $\omega$ by any set $\tau$ such that $\tau+\tau=\tau$)

But I have a hard time to believe that these two things are equivalent: $$A^A=2^A\iff A\times A=A.$$

Question I. Is there anything known on the properties of sets for which $A^A=2^A$?

Question II. If $2^A=A^A$ does not characterize the sets for which $A\times A=A$, does the axiom "For every infinite $A$, $2^A=A^A$", imply the axiom of choice?

If the answer is unknown, does this question (or variants, or closely related questions) appeared in the literature?

It seems like a plausible question by Tarski or SierpiƄski. I found several other questions I have asked before to be questions that have been asked in one a paper or another.

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Given that $A\times A=A$ for all $A$ is equivalent to AC, you could also ask whether the same is true of $A^A=2^A$. –  Eric Wofsey Feb 24 '13 at 6:21
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Because of the question "When does $A^A=2^A$ without the axiom of choice?", I could't help but comment - when A=2... –  Daniel Spector Feb 24 '13 at 9:10
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Which for some reason is excluded from the initial statement :) –  Adam Epstein Feb 24 '13 at 9:54
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@Daniel Spector, also when $A=0$ (as some people define $0^0=1$). :) –  Joel Reyes Noche Feb 25 '13 at 0:47
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Does anyone have an idea why this was downvoted? –  Asaf Karagila Feb 26 '13 at 6:09
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