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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2Since 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
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@Tobias Kildetoft it doesnt have to be continuous. – Badshah Feb 20 '13 at 17:36
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Also: http://math.stackexchange.com/questions/247696/does-mathbb-r2-contain-more-numbers-than-mathbb-r1/ and the relevant links appearing there. – Asaf Karagila Feb 20 '13 at 17:40
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Also related – JavaMan Feb 20 '13 at 17:41
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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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One way is to come up with a bijection between $\mathbb R$ and $P(\mathbb N)$, the set of all subsets of the natural numbers. It is easy to find a bijection between $P(\mathbb N)$ and $P(\mathbb N)\times P(\mathbb(N))$ by splitting a subset of $\mathbb N$ into the even and odd elements. – Thomas Andrews Feb 20 '13 at 18:16
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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: $$ (0.1415\dots,0.7172\dots) \mapsto (0.17411752\dots) $$
Emanuele Paolini
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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