# Does $\mathbb R^2$ contain more numbers than $\mathbb R^1$? [duplicate]

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?

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

• $\Bbb R^2$ 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
• @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

Indeed $\mathbb R^2$ has the same cardinality as $\mathbb R$, as the answers in this thread show.
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).
• The fact is that there is a bijection between $\Bbb R^2$ and $\Bbb R$. Since there is such a bijection, pick one, $f$. Then if $f(x,y)=z$ we say that $f$ encodes the pair $x,y$. Then if $g\colon\Bbb R^2\to X$ is any function, then $\bar g\colon\Bbb R\toX$ defined by $\bar g=g\circ f^{-1}$ is a function encoding $g$. All we need to do is who that such $f$ exists. And there are several more explicit examples, and other indirect proofs. – Asaf Karagila May 12 '15 at 22:20