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It must be posted somewhere, but I can't find it. I've been working on it for a while too without getting anywhere. Does there exist a bijection between $\mathbb{R}\times\mathbb{R}$ and $\mathbb{R}$? Is it possible to give an explicit bijection?

NOTE: This question is not a duplicate of the link suggested. Please see comments for further detail.

Paul S.
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    Yes, there is a isomorphism as vector spaces over $\mathbb{Q}$. – user40276 Jun 21 '13 at 21:24
  • @Paul S. The sentences before the question hint that you're looking for a bijection rather than just wanting to know if a bijection exists. What exactly do you want? And I suggest you clarify this fast, before it gets closed as a duplicate and it actually isn't (if that's the case). – Git Gud Jun 21 '13 at 21:25
  • @GitGud Ok I just read the link that Zev Chonoles posted. I see that they prove a bijection exist, but don't show a bijection. Is it possible to show a bijection? – Paul S. Jun 21 '13 at 21:31
  • Yes, Paul S., they do show a bijection there. – dfeuer Jun 21 '13 at 21:34
  • @dfeuer Yes, but the question itself is not a duplicate. I know you're not saying it is, I'm just pointing it out. – Git Gud Jun 21 '13 at 21:35
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    @dfeuer Don't they show injections in each direction, and then invoke the Schroeder–Bernstein theorem? – Paul S. Jun 21 '13 at 21:36
  • @PaulS. People are still closing this question as a duplicate. I suggest you type in bold that it isn't. – Git Gud Jun 21 '13 at 21:39
  • @PaulS.: yes, and (the standard proof of) Schroeder-Bernstein then gives you an explicit bijection. – Chris Eagle Jun 21 '13 at 22:05
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    I believe it should be possible to get more explicit than that, but it will require a lot of care about sign encoding. – dfeuer Jun 21 '13 at 22:11
  • While it might not be a duplicate of the posts indicated, it can be done using the methods of this thread. – Martin Jun 21 '13 at 22:27

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