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I need to find bijection $ f:\mathbb{R}\to\mathbb{R}\backslash\mathbb{Z} $ Such a function exists, because the two sets have the same cardinality, but I can't find an explicit one, any ideas?

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    Cantor-Bernstein *gives* an explicit function!2012-12-02

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Hint: Fix $a_n$ as a sequence of irrational numbers, and write $\mathbb Z=\{z_n\mid n\in\mathbb N\}$. Define a function which sends $a_n$ to $a_{2n}$; $z_n$ to $a_{2n+1}$; and $x$ to itself otherwise.

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    I see! Thanks, very elegant solution.2012-12-02
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  1. Take $f:(0,1)\to(0,1]$ to be the inclusion map and define $g:(0,1]\to(0,1)$ by $g(x)=x/2$. These are injections.
  2. Find a proof of the Cantor-Bernstein theorem which doesn't use the axiom of choice.
  3. Follow the proof using $f$ and $g$ to produce a bijection $h:(0,1)\to(0,1]$.
  4. Using translations of $h$ you get a bijection $\Bbb R\setminus\Bbb Z\to\Bbb R$.
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    Yes, I didn't think to use the algorithm of the proof to find the explicit function. Thanks!2012-12-02