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How to form a bijection from $(0,1]$ to $\mathbb{R}$:

$f(x) = \left\{\begin{array}{ll} 2-\frac{1}{x}&\text{if }x\in(0, .5]\\ \frac{2x-1}{1-x}&\text{if }x\in(.5, 1]. \end{array}\right.$

So, to go from $\mathbb{R}$ to $\mathbb{R}^4$ shouldn't be so hard... First we convert all of $\mathbb{R}$ in to a decimal representation. The numbers then have the form: $a_1a_2a_3a_4\ldots$  Where the $a_i$s are the digits $0$, $1$, $2,\ldots,9$

At some point there is a decimal point, suppose it precedes the $a_j$th digit (could be the first)

Eliminate all duplicate representations:  $3.41=3.4099999\ldots$ and $0002 = 2$, by choosing the one with the fewest digits. 

Now map the remaining representations to ( a_1a_5a_9\ldots, a_2a_6a_{10}\ldots, a_3a_7a_{11}\ldots, a_4a_8a_{12}\ldots)

Put a decimal point in each one preceding the $a_j$, $a_{j+1}$$a_{j+2}$ and $a_{j+3}$ digits.

This is not bijective, though! I know such a mapping exists, but I don't want in existence proof I want a scalable mapping I can use.

Is there some modification I can make to make it bijective?

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    Your $f$ is not well defined for $x=1$.2012-03-22

1 Answers 1

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Let $f$ be any bijection from $\mathbb{R}$ to $P(\mathbb{N})$ (the simplest one I could think of uses continued fractions, e.g. see here). To construct a bijection from $\mathbb{R}$ to $\mathbb{R}^n$ take $g_i(A) = \left\{ \left.\frac{x-i}{n} \right|\ x\in A, x =i\ (\mathrm{mod}\ n) \right\}$ and set $h_i = f^{-1} \circ g_i \circ f$ and then $h(x) = \langle h_0(x), h_1(x), \ldots, h_{n-1}(x) \rangle$ will be your bijection.