2
$\begingroup$

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 $\mathbb R$?

$\endgroup$

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

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
    $\begingroup$ Are you asking if there is a bijection $f : \mathbb{R}^2 \to \mathbb{R}$? Is that what $\cong$ means here? $\endgroup$ – JavaMan Mar 10 '13 at 2:56
  • 6
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Mar 10 '13 at 2:57
  • $\begingroup$ @ JavaMan: Yes. $\endgroup$ – A. Chu Mar 10 '13 at 2:57
  • $\begingroup$ @Camilo: Define elementary. Basic cardinal arithmetics is elementary. Explicit maps were given too. $\endgroup$ – Asaf Karagila Mar 10 '13 at 3:00
  • $\begingroup$ Well, to show an explicit bijection $f:\mathbb R^2\rightarrow \mathbb R$, cardinal arithmetic is not elementary if you're a freshman... $\endgroup$ – Camilo Arosemena-Serrato Mar 10 '13 at 3:02
2
$\begingroup$

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 $\mathbb{R}^3\rightarrow \mathbb{R}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.