# Bijection between $\mathbb{R}$ and $\mathbb{R}^2$ [duplicate]

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)?

-
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
show 1 more comment

## marked as duplicate by Asaf Karagila, David Mitra, Clayton, 5pm, tomaszFeb 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.

## 1 Answer

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

-
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
add comment