Is $\mathbb R^2$ equipotent to $\mathbb R$? [duplicate]

This question already has an answer here:

I know that $\mathbb N^2$ is equipotent to $\mathbb N$ (By drawing zig-zag path to join all the points on xy-plane). Is this method available to prove $\mathbb R^2$ equipotent to $\mathbb R$?

marked as duplicate by MJD, Asaf Karagila♦, cardinal, Pedro Tamaroff♦, QuixoticMar 10 '13 at 3:46

• Are you asking if there is a bijection $f : \mathbb{R}^2 \to \mathbb{R}$? Is that what $\cong$ means here? – JavaMan Mar 10 '13 at 2:56
• Oh, please please please search the site before think your question has never been asked before. This particular one has been asked about a zillion times now. – Asaf Karagila Mar 10 '13 at 2:57
• @ JavaMan: Yes. – A. Chu Mar 10 '13 at 2:57
• @Camilo: Define elementary. Basic cardinal arithmetics is elementary. Explicit maps were given too. – Asaf Karagila Mar 10 '13 at 3:00
• Well, to show an explicit bijection $f:\mathbb R^2\rightarrow \mathbb R$, cardinal arithmetic is not elementary if you're a freshman... – Camilo Arosemena-Serrato Mar 10 '13 at 3:02

That method is not available to prove the equipotency of $\mathbb R$ and $\mathbb R^2$. The geometry is different. When the zigzag finishes a diagonal line in $\mathbb Z^2$, it can move over and do the next diagonal line. But in $\mathbb R^2$ there is no "next" line.
In fact it turns out that there is no continuous bijection between $\mathbb R$ and $\mathbb R^2$.
For ideas that do work, see Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$.