Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$
Do the real numbers and the complex numbers have the same cardinality?

Does $\mathbb R^2$ contain more numbers than $\mathbb R^1$? I know that there are the same number of even integers as integers, but those are both countable sets. Does the same type of argument apply to uncountable sets? If there exists a 1-1 mapping from $\mathbb R^2$ to $\mathbb R^1$, would that mean that 2 real-valued parameters could be encoded as a single real-valued parameter?

share|improve this question
3  
$\Bbb R^2$ has the same cardinality as $\Bbb R$. This has been dealt with quite a few times at MSE; the first answer to this question is very thorough. –  Brian M. Scott Nov 30 '12 at 0:21
    
There are plenty of answers for this question on this site. In short, the answer is yes. –  Asaf Karagila Nov 30 '12 at 0:24
    
NO! ${\aleph _1} = {2^{{\aleph _0}}} = \left| {\mathbb R} \right| = \left| {{{\mathbb R}^n}} \right|$ –  glebovg Nov 30 '12 at 0:36
1  
@glebovg: That is false. $\aleph_1$ does not have to be equal to the cardinality of the continuum. –  Asaf Karagila Nov 30 '12 at 1:02
    
@AsafKaragila Yes. I meant $\mathfrak{c} = {2^{{\aleph_0}}} = \left| {\mathbb R} \right| = \left| {{{\mathbb R}^n}} \right|$. –  glebovg Nov 30 '12 at 2:39
add comment

marked as duplicate by Brian M. Scott, Asaf Karagila, EuYu, Berci, Henning Makholm Nov 30 '12 at 0:31

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

up vote 0 down vote accepted

Indeed $\mathbb R^2$ has the same cardinality as $\mathbb R$, as the answers in this thread show.

And indeed it means that functions of two variables can be encoded as functions of one variable. However do note that such encoding cannot be continuous, but can be measurable.

Lastly, to extend this result to all infinite sets one needs the axiom of choice. In fact the assertion "For every infinite $A$ there is a bijection between $A$ and $A^2$" is equivalent to the axiom of choice. If one requires that $A$ is well-ordered then this is true without the axiom of choice, but for many "sets of interest" (e.g. the real numbers) one cannot prove the existence of a well-ordering without some form of choice.

Despite the last sentence, the existence of a bijection between $\mathbb R$ and $\mathbb R^n$ does not require the axiom of choice (for $n>0$, of course).

share|improve this answer
    
That's amazing! Thanks for the response. –  Jeff Nov 30 '12 at 21:34
    
You're welcome, Jeff. –  Asaf Karagila Nov 30 '12 at 22:30
add comment

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