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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.

DanielG
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  • Did you mean to write $\Bbb R\setminus\Bbb Q$? – José Carlos Santos Jun 27 '21 at 18:30
  • yes I'm sorry.. – DanielG Jun 27 '21 at 18:30
  • 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 $x\in\mathbb{R}$. – Cornman Jun 27 '21 at 18:31
  • I was asked to prove it specific with Schröder–Bernstein theorem. – DanielG Jun 27 '21 at 19:02

2 Answers2

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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 $$0.\, a_1 \, 00\dots 0\, a_2\, 00\dots 0\, a_3\, 00\dots 0\,\dots$$ where the string of zeroes after $a_k$ is of length $a_k$. 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?

Sayan Dutta
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  • “Clearly, this is an injection…”. Clearly it is not. What are you going to do about $\frac1{12+}\frac1{3+}\cdots$ and $\frac1{1+}\frac1{23+}\cdots$? There is probably a way around this but it requires an argument more detailed than “clearly”. – MJD Dec 22 '24 at 14:53
  • @MJD You are absolutely right, I am surprised nobody pointed this out before! But your solution also doesn't seem fine since [1,2,...] and [102,3,...] gets mapped to the same place :( – Sayan Dutta Dec 22 '24 at 15:09
  • @MJD Maybe $[a_1, a_2, \dots ]\mapsto 0., a_1 , 00\dots 0, a_2, 00\dots 0, a_3\dots$ works where the string of zeroes after $a_k$ is of length $a_k$. – Sayan Dutta Dec 22 '24 at 15:15
  • I think there are a lot of ways around this. For example, map $[a_1, a_2, \ldots]$ to $.0^{a_1}10^{a_2}1\ldots$ where $0^i$ means a string of $i$ zeroes. Or: $0.\bar{a_1}9\bar{a_2}9\ldots$ where $\bar{a_i}$ is the base-9 representation of the number $a_i$. – MJD Dec 22 '24 at 17:57
  • @MJD yes, you're right. I'll place an edit. – Sayan Dutta Dec 23 '24 at 06:20
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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 $f(n\sqrt 2)=(2n-1)\sqrt 2.$

DanielWainfleet
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