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I have been thinking for a while whether its possible to have bijection between $\mathbb{R}$ and $\mathbb{R}^2$, but I cant think of a solution. So my question is: is there a bijection between $\mathbb{R}$ and $\mathbb{R}^2$ (with proof)?

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Since you tagged this real-analysis, do you want the bijection to be continuous? If yes, then no such bijection exists. If no, then one exists. –  Tobias Kildetoft Feb 20 '13 at 17:34
    
@Tobias Kildetoft it doesnt have to be continuous. –  Badshah Feb 20 '13 at 17:36
    
Also: math.stackexchange.com/questions/247696/… and the relevant links appearing there. –  Asaf Karagila Feb 20 '13 at 17:40
    
Also related –  JavaMan Feb 20 '13 at 17:41
    
Amongst the zillion duplicates, I chose this one. I encourage others to add other duplicates when voting to close. –  Asaf Karagila Feb 20 '13 at 17:43
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marked as duplicate by Asaf Karagila, David Mitra, Clayton, 5pm, tomasz Feb 20 '13 at 18:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Yes there is. I think it is one of the results of Cantor. Take two real numbers and combine them by interposing their digit in the decimal expansion.

example: $$ (0.1415\dots,0.7172\dots) \mapsto (0.17411752\dots) $$

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This doesn't quite work. See the first answer to this post. –  David Mitra Feb 20 '13 at 17:44
    
well, this is only the idea... you can easily make it rigorous using the Cantor-Bernstein theorem. –  Emanuele Paolini Feb 20 '13 at 17:49
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