Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Can anyone give an example of an injective map from $\mathbb{R}^2$ to $\mathbb{R}$? Clearly, such a map cannot be continuous (for instance by Borsuk-Ulam Theorem).

Thanks in advance.

share|cite|improve this question
up vote 11 down vote accepted

How about this construction: Express $(x,y)\in\mathbb{R}^2$ in decimals ($x=\sum a_k 10^k$, $y=\sum b_k 10^k$) and define the image of $(x,y)$ as the real number which you obtain by interlacing the decimals (i.e. take $c_{2k} = a_k$ and $c_{2k+1} = b_k$).

share|cite|improve this answer
Very good idea, thanks! – Henry Wegener Oct 23 '11 at 14:36
I think a bit of care needs to be taken in that you'll want to take decimal expansions that don't end in a repeated trail of nines, otherwise $(1,1)$ will go to 1, but $(1,0.\overline{9})$ would go to $1.\overline{90}$. – user5137 Nov 4 '11 at 20:31
I rolled back because: (1) this edit was far from being an actual edit. It was a comment at best, I personally found it content altering; (2) the notation in the post was that $(a,b)$ is an ordered pair, adding $(0,1)$ as an interval makes it harder and much less clearer. – Asaf Karagila Jun 30 '12 at 21:49

Hopefully this answer complements Dirk's direct answer. I use the fact that $\mathbb R$ has the same cardinality as $2^{\mathbb N}$ (and hence $\mathbb R^{2}$ has the same cardinality as $2^{\mathbb N} \times 2^{\mathbb N}$). Our goal, then, is to establish an injective function from $2^{\mathbb N} \times 2^{\mathbb N}$ to $2^{\mathbb N}$.

Given $S, T \subseteq \mathbb N$, consider the set: $$ R := \{2x \,:\, x \in S\}\ \cup \ \{ 2y+1 \,:\, y \in T \}. $$ Given the set $R$, it is easy to "decode" the sets $S$ and $T$. (Hint: Consider the odd and even elements of $R$.) Therefore the mapping $(S,T) \mapsto R$ as above is injective.

share|cite|improve this answer
This map is also surjective, and continuous at least under the trivial topology :-) – Asaf Karagila Oct 23 '11 at 15:21

let x_n be the n'th term any unique continued fractional expansion for x.

Then $z(a,b)=\sum (1/(2n)^{1/a_n}+1/(2n+1)^{1/b_n}$)

By the Lindemann–Weierstrass theorem
$z=\sum (e^{-pi/a_n}+\pi^{-e/b_n}$)

share|cite|improve this answer
I am unable to follow the example. Can you explain why this function is injective? – Srivatsan Oct 23 '11 at 14:38
@SrivatsanNarayanan because these quantities are linearly independent over the rationals – GM2001 Oct 23 '11 at 14:53

One can also consider space-filling curves. Since there's an onto map from $[0,1]$ to $[0,1]\times [0,1]$, it follows that $\vert[0,1] \vert \geq \vert [0,1] \times [0,1] \vert$. Of course, the other inequality clearly holds (via the mapping $x \mapsto (x,0)$), hence $\vert [0,1] \vert = \vert [0,1] \times [0,1] \vert$. Finally, since $\vert [0,1] \vert = \vert \mathbb{R} \vert$, we have $\vert \mathbb{R} \vert = \vert \mathbb{R}^2 \vert$.

share|cite|improve this answer

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.