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Find bijection $\Bbb R\setminus\Bbb Q \to \Bbb R $
I've tried using Schröder–Bernstein theorem to show a 1-1 function in both directions. But I only succeed to prove one direction. Explicit function seems much harder to prove. thanks for any help.

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  • $\begingroup$ Did you mean to write $\Bbb R\setminus\Bbb Q$? $\endgroup$ Jun 27, 2021 at 18:30
  • $\begingroup$ yes I'm sorry.. $\endgroup$
    – DanielG
    Jun 27, 2021 at 18:30
  • $\begingroup$ Maybe you should try to find a bijection $\mathbb{R}\setminus \mathbb{N}\to\mathbb{R}$ first, and see if that helps, which I do not know, but I think it helps and should be easier. Maybe you can then generalize that bijection. Or even simpler: Start by finding a bijection $\mathbb{R}\setminus\{x\}\to\mathbb{R}$ for some arbitrary $x\in\mathbb{R}$. $\endgroup$
    – Cornman
    Jun 27, 2021 at 18:31
  • $\begingroup$ I was asked to prove it specific with Schröder–Bernstein theorem. $\endgroup$
    – DanielG
    Jun 27, 2021 at 19:02

2 Answers 2

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We know that all irrational numbers have a unique infinite continued fraction. So, take any irrational number $r$ and consider its continued fraction \begin{equation*} r=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\dots}}}} \end{equation*} Map $r$ to the real number $0.a_1a_2a_3\dots$. Clearly, this is an injection from $\mathbb{R}\backslash\mathbb{Q}\to \mathbb{R}$.

Now, consider the famous injection from $\mathbb{R}\to \mathbb{R}\backslash\mathbb{Q}$ which maps $q+n\sqrt{2}$ to $q+(n+1)\sqrt{2}$ for $q\in \mathbb{Q}$, $n\in \mathbb{N}$ and maps the rest to itself.

Now, apply Schröder–Bernstein theorem.

Does that help?

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Take a countably infinite $\Bbb S\subset \Bbb R\setminus \Bbb Q .$ Now $\Bbb S\cup\Bbb Q$ and $\Bbb S$ are countably infinite, so take a bijection $f:\Bbb S\cup\Bbb Q \to \Bbb S.$ Extend the domain of $f$ to $\Bbb R$ by letting $f(x)=x$ for $x\in \Bbb R\setminus (\Bbb S\cup \Bbb Q).$ Then $f:\Bbb R\to \Bbb R\setminus \Bbb Q$ is a bijection.

E.g. let $g:\Bbb Z^+\to \Bbb Q$ be a bijection and let $\Bbb S=\{n\sqrt 2\,:n\in\Bbb Z^+\}.$ For $n\in\Bbb Z^+$ let $f(g(n))=2n\sqrt 2$ and let $f(n\sqrt 2)=(2n-1)\sqrt 2.$

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