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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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    Hint, $|A|+|A|=|A|=\infty$ is trivial, only need to show $B^A$ is infinite set, this can be done by write $B=\{0, 1\}$ then $B^A$ is the same as the subsets of $A$, which is infite.2012-12-13
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    What? @ougao: That makes no sense. To the OP, I assume this is to be done without the axiom of choice?2012-12-13
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    @Asaf, no need to use the axiom of choice, just note that you can use the subsets which are all single point in A to show all the subsets of A is infinite.2012-12-13
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    @ouago: I still don't understand what you are trying to suggest. I also didn't ask *you* about the axiom of choice, I asked the OP whether or not he is allowed to use the axiom of choice in this argument. Cardinal exponentiation can be dealt with quite differently under the assumption of choice.2012-12-13
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    @ougao: Further reading indicates that you may have misunderstood the question. The question does not require to prove that $A$ is infinite, or that $B^A$ is infinite. Furthermore the question does not allow "setting $B$". Lastly to prove that every infinite set has the property $|A|+|A|=|A|$ requires some of the axiom of choice to hold. I admit that I still can't understand your second comment.2012-12-13
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    @Asaf, maybe I should explain more clearly. I suppose $B=\{0, 1\}$ because this would suffice to show that $B^A$ is an infinite set, I think $B^A$ as all the maps $f: A\to B$, then note that any map $f: A\to B$ is determined uniquely by $f^{-1}(1)$, which is a subset of $A$, write $A=\{a_i\}_{i\in I}$so if you consider all the maps $f_i: A\to B$ such that $f_i(a_j)=\delta_{i,j}$, then $f_i, i\in I$ has the same cardinality with $A$, hence already infinite. Then use the corollary of the first equality to show that second one.2012-12-13
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    @ougao: But the question does not ask to show it is an infinite set. The question asks to show an equality between the cardinality of two sets. Not all infinite sets have the same size, and with the axiom of choice it is certainly consistent that some sets have the property that $|A|+|A|\neq|A|$.2012-12-13
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    @Asaf, I think for such a homework in an elementary set-theory, no need to consider so further issues as you mentioned.2012-12-13
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    @ougao: I happen to know the guy who wrote the homework sheet, and I have seen it myself (and had things gone differently, it may have been me). Regardless to all things concerned, there are many courses in which the axiom of choice is not assumed **implicitly**, it is either expressed and assumed, or consequences depending on its validity are withheld from the class.2012-12-13
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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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    I'd remove the word "Hint" at the beginning.2012-12-13
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    @Omar: Well, I suppose you're right. It's slightly overelaborated for a hint. :-)2012-12-13