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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0How do you know the sets have the same cardinality? You should be able to use that to find an explicit bijection. – 2012-12-02
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2If you can find a bijection between $[0, 1)$ and $(0, 1)$ then you are done since you can glue these pieces together to form $\mathbb{R}$ and $\mathbb{R} \setminus \mathbb{Z}$. – 2012-12-02
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0You are right, but the exercise is to prove that the sets have the same cardinality, not using the Cantor–Bernstein theorem, but to find an explicit function. – 2012-12-02
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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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0Thanks Asaf! But to fix $a_{n}$, irrational numbers should be countable. How do I define the sequence properly? – 2012-12-02
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0@Denis: The irrational numbers are not countable, of course. But does that mean that there is no countable sequence of irrational numbers?? $k+\pi$ for $k\in\mathbb N$, for example? Square roots of square-free integers? And so on and so forth... – 2012-12-02
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0OK, you mean not to map all the irrational numbers as sequence, but to choose one specific irrational sequence like $k+\pi$ ? – 2012-12-02
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0@Denis: Hence the term "Fix $a_n$ as a sequence **of** irrational numbers", and not "Fix $a_n$ as a sequence **of the** irrational numbers". – 2012-12-02
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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