Injective map from $\mathbb{R}^2$ to $\mathbb{R}$ 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.
 A: 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. 
A: 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 $c_{2k+1} = b_k$). 
A: 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 $\vert \mathbb{R} \vert = \vert \mathbb{R}^2 \vert$.
A: let x_n be the n'th term any unique continued fractional expansion for x.
Then $z(a,b)=\sum (1/(2n)^{1/a_n}+1/(2n+1)^{1/b_n}$)
bonus:
By the Lindemann–Weierstrass theorem
$z=\sum (e^{-pi/a_n}+\pi^{-e/b_n}$)
