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Given an infinite set $A$ - does the cardinality of $A$ equal to the cardinality of $A^2$?

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Yes, assuming the axiom of choice. –  Brian M. Scott Aug 9 '12 at 13:38
    
Do you mean group, which is a set with a binary operation; or just a set? –  Asaf Karagila Aug 9 '12 at 14:17
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See here. –  Martin Sleziak Aug 9 '12 at 14:19
    
My bad - A is a set - not group. –  poli Aug 9 '12 at 14:42
    
As a simple plausibility argument, consider the case where $A=\mathbb{R}$. Then clearly we can take the decimal expansions of two real numbers and combine them into one decimal expansion, like shuffling a pack of cards. It's also pretty easy to do an explicit construction for $\mathbb{Z}^S$, where $S$ is any set. Since we think of big sets as being built from power-sets of smaller sets, this makes it pretty plausible that it holds in general. –  Ben Crowell Aug 9 '12 at 19:03

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up vote 7 down vote accepted

Zermelo proved that every well-ordered infinite set has this property, so if we assume the axiom of choice then the answer is yes. In fact the axiom of choice is equivalent to the assertion that for every infinite set $A$, $A$ and $A^2$ are equinumerous.


However if the axiom of choice is negated there are sets whose cardinality is strictly less of that of their square. We even know how to construct such set:

Suppose that $X$ cannot be well-ordered (such set exists, since we assume the axiom of choice is false, and therefore the well-ordering principle is false). Let $\kappa$ be an ordinal such that there is no $f\colon\kappa\to X$ which is injective.

In this case $A=X\cup\kappa$ has the property that $A^2$ is strictly larger than $A$.

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You proved that $|S|=|S^2|$ implies AC, but the person asking the question is interested in proving $|S|=|S^2|$, presumably in ZFC. –  Ben Crowell Aug 9 '12 at 18:48
    
@Ben: You're right, I was going to edit in the refernce to my answer from Martin's comment, but I forgot. I'll do that now. –  Asaf Karagila Aug 9 '12 at 18:49
    
The story about why Comptes Rendu rejected Tarski's paper is hilarious. I find the theorem surprising in that it implies full AC rather than some restricted version of choice. But I suppose the sets that violate $|A^2|=|A|$ are not explicitly constructible...? –  Ben Crowell Aug 9 '12 at 19:26
    
You mean Lebesgue. That depends on what you mean constructible. I gave you a way to construct such set given a set which is not well-orderable. If the real numbers fails to be well-orderable then you can use them as the starting set. –  Asaf Karagila Aug 9 '12 at 19:54

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