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How attacking this question?

Show that if $A$ and $B$ are sets such that $A$ is infinite, $|A|+|A|=|A|$, and $|B|\geq 2$, then $|B^A|+|B^A|=|B^A|$

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    @ougao: Furthermore there are plenty of classical results in elementary set theory which hold "immediately" when assuming AC; but are proved in "the hard way" regardless. For example, why bother with $|A\times\{0,1\}|=|A|+|A|$? Just lump it all into $|A|+|A|=\max\{|A|,|A|\}=|A|=\max\{|A|,2\}=|A\times\{0,1\}|$. But there is a very good reason to let students meddle within these sort of results by hands because they help to develop intuition which may be essential for later on (depending on the course, and the student).2012-12-13

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The $\geq$ part is obvious; and to show $\leq$ simply show that the following holds: $|B^A|+|B^A|\leq|B^A|\times|B^A|=|B^{A+A}|=|B^A|$

Now use Cantor-Bernstein to finish the job.

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    @Omar: Well, I suppose you're right. It's slightly overelaborated for a hint. :-)2012-12-13