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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I show that the set of reals and the set of pairs of reals have the same cardinality?

I know that since reals are uncountable infinite, I can't create a list of reals and talk about the $i^{th}$ real mapping to the $i^{th}$ real pair. So how can I construct a one-to-one and onto mapping $f: \mathbb R \to \mathbb R^2$?

Thank You

share|cite|improve this question
Try playing around with decimal expansion. – Alex B. Oct 12 '11 at 4:05
It is easier to see with reals in $(0,1)$ and $(0,1) \times (0,1)$. Do you know that all the reals have the same cardinality as the reals in $(0,1)$? – Ross Millikan Oct 12 '11 at 4:14
Rather than explicitly constructing a bijection between $\mathbb R$ and $\mathbb R^2$, which can get a bit tricky, it may be easier to construct surjections in each direction (one is trivial) and apply the law of trichotomy (which holds for cardinality under AC). – Ilmari Karonen Oct 12 '11 at 5:12
This question is closely related:… – Martin Sleziak Oct 12 '11 at 6:51

If $a$ is the cardinality of $\mathbb N$, then we have $$2^a\cdot2^a=2^{a+a}=2^a.$$

share|cite|improve this answer
shouldn't 2^a+a = 2^(2a) ? – Jeff Oct 12 '11 at 4:54
Yes, we have $a+a=2a=a$ if $a$ is an infinite cardinal number. – Pierre-Yves Gaillard Oct 12 '11 at 4:55
alright, I see but how do i apply this concept to Reals and pairs of reals? – Jeff Oct 12 '11 at 6:15
@Pierre: To be exact, it is not always true that $x+x=x$ for infinite cardinals. However for $\aleph_0$ it is true that $\aleph_0+\aleph_0=\aleph_0$, even without the axiom of choice. – Asaf Karagila Oct 12 '11 at 7:12
Dear @Asaf: Thank you very much for your comment. I always take the axiom of choice for granted. (In fact I adhere to Bourbaki's set theory, and, as you certainly know much better than I, in this theory, the axiom of choice is not a separate axiom, but is "built in".) – Pierre-Yves Gaillard Oct 12 '11 at 8:52

Let the binary expansions of $(x,y)\in[0,1)\times[0,1)$ be $$ \begin{array}{} x=\sum_{k=1}^\infty x_k2^{-k}&\text{and}&y=\sum_{k=1}^\infty y_k2^{-k} \end{array} $$ (finite where possible) where $(x_k,y_k)\in\{0,1\}\times\{0,1\}$. Define $f:[0,1)\times[0,1)\mapsto[0,1)$ by $$ f(x,y)=\sum_{k=1}^\infty x_k2^{1-2k}+y_k2^{-2k} $$ that is, $f(x,y)$ interleaves the bits of $x$ and $y$. It is easy to see that $f$ is injective, which means the cardinality of $[0,1)\times[0,1)$ is less than or equal to that of $[0,1)$.

Define $g:[0,1)\mapsto[0,1)\times[0,1)$ by $g(x)=(x,0)$. $g$ is injective.

Use the Cantor-Bernstein-Schroeder Theorem to get the existence of a bijection between $[0,1)\times[0,1)$ and $[0,1)$.

share|cite|improve this answer
so R^2 doesn't literaly mean R^2, by R^2 it refers to pairs of Reals. I need to construct the mapping between the reals to pairs of reals. – Jeff Oct 12 '11 at 5:57
@Jeff: what is the difference between $\mathbb{R}^2$ and pairs of reals? $(x,y)\in[0,1)\times[0,1)$ is another way of saying $x\in[0,1)$ and $y\in[0,1)$. – robjohn Oct 12 '11 at 6:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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