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?
Find a bijection function from $\mathbb{R}\to\mathbb{R}\backslash\mathbb{Z}$
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elementary-set-theory
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0Cantor-Bernstein *gives* an explicit function! – 2012-12-02
2 Answers
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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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0I see! Thanks, very elegant solution. – 2012-12-02
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- 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.
- Find a proof of the Cantor-Bernstein theorem which doesn't use the axiom of choice.
- Follow the proof using $f$ and $g$ to produce a bijection $h:(0,1)\to(0,1]$.
- Using translations of $h$ you get a bijection $\Bbb R\setminus\Bbb Z\to\Bbb R$.
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0Yes, I didn't think to use the algorithm of the proof to find the explicit function. Thanks! – 2012-12-02