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I know that $|\mathbb{N}| = |\mathbb{N}^2|$, and that $|\mathbb{R}| = |\mathbb{R}^2|$. It seems like this might be true for all sets, but I don't know how to go about proving this.

It's easy to prove that $|A| \leq |A \times A|$ (see this question). So to prove the claim it suffices to show that there exists some injective function $f : A \times A \to A$.

For $A = \mathbb{N}$, you can use e.g. $f:p, q \mapsto (p+q)^2 + p$. For $A = \mathbb{R}$ you can construct various space-filling curves. But I'm not sure how such constructive solutions can be generalized, if at all.

Any suggestions on how to approach this?

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augurar
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    The claim is true in general but certainly not easy to prove. And it relies on some aspect of choice. In general, cardinal arithmetic is well-behaved if you accept choice. You can find a proof of the claim in any (good) text on axiomatic set theory, as well as some good texts on logic and set theory. – Ittay Weiss Apr 09 '14 at 22:12
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    @IttayWeiss This not just relies on some aspect of choice, it is in fact equivalent to choice. – Andrés E. Caicedo Apr 09 '14 at 22:23
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    Thanks @AndresCaicedo. Indeed, what I meant is that for a set of given infinite cardinality a slightly weaker choice principle is needed, but for the general result full choice is required. – Ittay Weiss Apr 09 '14 at 22:26
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    There are many questions on this site addressing this, you may want to spend a few minutes looking for them. In short: This is true, but it is equivalent to the axiom of choice. One way of proving the equality is to use choice to well-order $A$, and then use that for infinite cardinals $\alpha$, we have that $\alpha$ and $\alpha\times\alpha$ are in bijection as shown, for instance, using Gödel's pairing. – Andrés E. Caicedo Apr 09 '14 at 22:26
  • @AndresCaicedo Thanks, this is exactly what I was looking for! It seems like Gödel's pairing is a natural way to generalize the usual bijection offered from $\mathbb{N}^2$ to $\mathbb{N}$. Care to upgrade your comment to an answer? – augurar Apr 09 '14 at 23:43
  • @augurar: This shouldn't be answered it should be closed as a duplicate. – Asaf Karagila Apr 09 '14 at 23:50
  • @AsafKaragila If you can substantiate that claim with a link, please do so and I will delete the question. – augurar Apr 09 '14 at 23:53
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    Here we go: http://math.stackexchange.com/q/608538/622 and http://math.stackexchange.com/q/180671/622 and http://math.stackexchange.com/q/54892/622 and there might be a few more. Search through the linked and related questions (appearing on the right of the thread) to find those. – Asaf Karagila Apr 09 '14 at 23:59
  • @AsafKaragila Thanks, I have voted to close. Somehow I didn't see these in my search. The second link you mention is the closest duplicate. – augurar Apr 10 '14 at 00:12

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