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 $\mathbb R$?
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1Are 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 02:56
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6Oh, 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 02:57
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@ JavaMan: Yes. – JSCB Mar 10 '13 at 02:57
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@Camilo: Define elementary. Basic cardinal arithmetics is elementary. Explicit maps were given too. – Asaf Karagila Mar 10 '13 at 03:00
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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 03:02
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@Camilo: Basic cardinal arithmetics are, and should be, covered in every intro course to set theory. For me that was first semester in the university. – Asaf Karagila Mar 10 '13 at 03:05
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@Camilo: Cantor's proof about the equipotency of the line with the plane was probably one of the first results of cardinal arithmetics. It's hard to get more elementary than that. – Asaf Karagila Mar 10 '13 at 03:18
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$\cong$ in what sense? Isomorphism, no, existence of a bijection yes. You must be careful with the context in which you use symbols. – ncmathsadist Mar 10 '13 at 03:05
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@ncmathsadist: Isomorphism as additive groups? Yes. But the comments indicate this means equicardinality. – Asaf Karagila Mar 10 '13 at 03:06
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@Asaf: In that case, yes. One must be careful in setting context. – ncmathsadist Mar 10 '13 at 03:34
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@Asaf: One of the closing users has informed me that he would like to withdraw his close vote, since this question is asking about whether a specific technique works. Reopening sounds reasonable to me. Do you think the questions are sufficiently distinct? – Zev Chonoles Mar 10 '13 at 15:52
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@Zev: I recall that casting closing votes allows cast reopening votes as well... Going through a moderator seems a bit... excessive. :-) But yeah, I suppose I'd vote to reopen after this was pointed out. – Asaf Karagila Mar 10 '13 at 17:12
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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 $\mathbb{R}^3\rightarrow \mathbb{R}$.
