# 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).

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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 $c_{2k+1} = b_k$).

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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.

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

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

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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 $\vert \mathbb{R} \vert = \vert \mathbb{R}^2 \vert$.

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