# Bijection from $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$? [duplicate]

Possible Duplicate:
bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$

I know it's possible to produce a bijection from $\mathbb{Z}$ to $\mathbb{Z}\times\mathbb{Z}$, but is it possible to do so from $\mathbb{R}$ to $\mathbb{R} \times \mathbb{R}$?

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## marked as duplicate by Brian M. Scott, Ross Millikan, EuYu, Rahul, Marvis Nov 18 '12 at 6:12

Yes. Describing one explicitly is mildly unpleasant, but it can be done. Showing there is a bijection is easier. –  André Nicolas Nov 18 '12 at 5:21
@Robert: It is possible. –  Brian M. Scott Nov 18 '12 at 5:21
The earlier question is slightly different, but this answer to it has everything that’s needed here in considerable detail. –  Brian M. Scott Nov 18 '12 at 5:26

Yes, one can produce such a bijection of $\mathbb{R} \times \mathbb{R}$ with $\mathbb{R}$. It is actually easier to do this with $[0,1]$ and $[0,1] \times [0,1]$. The general idea is as follows: take an element of $[0,1]$ and view it as a sequence of natural numbers. Then split that sequence into two subsequences: those of odd index and those of even index. Conversely, one can take two sequences and "weave" them together by alternating between the sequences, so that the terms of one sequence form the even-indexed terms of the new sequence, and the terms of the other sequence becomes the odd-indexed terms of the new sequence.

For example, if $0.a_1a_2a_3a_4 \cdots$ is a real number in $[0,1]$, we may produce the element

$$(0.a_1a_3a_5 \cdots, 0.a_2a_4a_6 \cdots) \in [0,1] \times [0,1]$$

Conversely, given

$$(0.a_1a_2a_3 \cdots, 0.b_1b_2b_3 \cdots) \in [0,1] \times [0,1]$$

we may produce the real number $0.a_1b_1a_2b_2a_3b_3\cdots \in [0,1]$.

Of course, problems arise related to the fact that a given real number has no fixed representation, i.e. $1 = 0.9999\cdots$, but these can be dealt with. Note that if one does not take care to fix these technical concerns, one still gets an injection from $[0,1]\times [0,1]$ into $[0,1]$, and so Schroeder-Bernstein gives us a bijection even if haven't we haven't explicitly defined it (which is a bit of a lame cop-out all the same).

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Consider $(a,b) \in \mathbb{R}^2$. Represent $a$ and $b$ in decimal as $$a = \sum_{n=-\infty}^{k} a_n \cdot 10^n ,\,\,\,\,\,\,\,\, b = \sum_{n=-\infty}^{k} b_n \cdot 10^n$$ where $k = \max \{\lfloor \log_{10} a \rfloor, \lfloor \log_{10} b \rfloor\}$ and $a_n,b_n \in \{0,1,2,3,4,5,6,7,8,9\}$.

Now map $(a,b)$ to $c = a_k b_k a_{k-1} b_{k-1}\ldots a_0 b_0 \cdot a_{-1} b_{-1} a_{-2} b_{-2} \ldots$ i.e. $$c= \sum_{n=-\infty}^{k} a_n \cdot 10^{2n} + \sum_{n=-\infty}^{k} b_n \cdot 10^{2n-1}$$

This is one such bisection.

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This doesn’t quite work; see this old answer for details. –  Brian M. Scott Nov 18 '12 at 5:27