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I am trying to prove that $ ^\mathbb{Q}\mathbb{R}$~$\mathbb{R}$ .

I want to use Cantor Schroder Bernstein Theorem rather than coming up with a bijection. Any suggestions on how to get started with using this theorem?

Should I begin with a function from $ ^\mathbb{Q}\mathbb{R}$ to $\mathbb{R}$ or the other way around first? I was also thinking maybe we do not need to explictly find an injection for one direction, instead, we could justify an injection by saying if we take $f\in ^\mathbb{Q}\mathbb{R}$. Then $f\subset\mathbb{Q}$ x $\mathbb{R}$ and continuing from here somehow.

Would it be best to consider $ ^\mathbb{N}\mathbb{R}$ instead?

Asaf Karagila
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user376239
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  • @mucciolo I saw that link, but it did not help to answer my question. This is the part I am more focused on. – user376239 Nov 18 '17 at 04:21

1 Answers1

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Since $\Bbb R$ and the interval $(0,1)$ have the same cardinality, and $\Bbb Q$ has the cardinality of $\Bbb N,$ it suffices to find an injective $f:(0,1)\to ^\Bbb N (0,1)$ and an injective $g:^\Bbb N (0,1)\to (0,1).$

(I). For $x\in (0,1)$ let $f(x)$ be the function $\{(n,x): n\in \Bbb N\}$.

(II). For $v\in ^\Bbb N (0,1)$, for each $n\in \Bbb N$ let $v(n)=\sum_{j=1}^{\infty}v(n,j)2^{-j}$ such that (i) $v(n,j)\in \{0,1\}$ and (ii) $v(n,j)=0$ for infinitely many $j.$ This assures the uniqueness of the sequence $(v(n,j))_{j\in \Bbb N}$ for each $n.$

Arrange the double-sequence $(v(n,j))_{n,j\in \Bbb N}$ into the following sequence $(A^v(k))_{k\in \Bbb N}$ of finite sequences:

$A^v(1)=(v(1,1)).$

$A^v(2)=(v(2,1),v(1,2)).$

$A^v(3)=(v(3,1),v(2,2),v(1,3)).$

$A^v(4)=(v(4,1),v(3,2),v(2,3),v(1,4)).\quad$ ET CETERA.

Now concatenate all the $A^v(k)$ together into one sequence $A^v(1)\land A^v(2)\land A^v(3)\land ....$. Denote this sequence as $(v^*(m))_{m\in \Bbb N}$. Finally let $$g(v)=\sum_{m=1}^{\infty}v^*(m)2^{-m}.$$

Since $v(n)\ne 0$ for any $n\in \Bbb N$ there exists $m$ with $v^*(m)\ne 0$ so $g(v)>0.$ And for any $n$ the set $\{j: v(n,j)=0\}$ is infinite, so $v^*(m)=0$ for infinitely many $m,$ so $g(v)<1.$ So $g(v)\in (0,1)$, as desired.

Finally to show that $g$ is injective: If $g(v_1)=g(v_2)$ then $v^*_1(m)=v^*_2(m)$ for all $m$ . (Because if two base-2 representations represent the same number and each representation has infinitely many $0$'s then they are the SAME representation.) So by the definition of the sequences $v_1^*,v_2^*,$ we have $v_1(n,j)=v_2(n,j)$ for all $n$ and all $j,$ implying $v_1=v_2.$

DanielWainfleet
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  • I am not too sure what you are saying for (I). This would be sending $x\in\mathbb{R}$ to what in $ ^\mathbb{N}\mathbb{R}$? – user376239 Nov 25 '17 at 06:07
  • Also, I am not clear on what you are doing for the function $g$. Are you suggesting a diagonalization? The use of series here is confusing me. – user376239 Nov 25 '17 at 06:10
  • In (I), $ f_x(n)=x$ for $n\in \Bbb N$ and $x\in (0,1) .$ To get an injection $h:\Bbb R\to ^{\Bbb N}\Bbb R,$ take a bijection $j:\Bbb R\to (0,1)$... (e.g. $j(x)= (1/2)+ (1/\pi) \arctan (x-1/2).$...)... and let $h(y)=f_{j(y)}.$....... In (II) for a function $v:\Bbb N\to (0,1)$ we take the binary-digit sequence of each of $v(1), v(2), v(3),$ etc. and shuffle theses sequences together into one binary sequence $v^(1),v^(2),$ etc.,, which we interpret as the digits in base 2 of a number $g(v).$ – DanielWainfleet Nov 25 '17 at 06:57
  • Wait, I thought (I) was your injection from R to functions from N to R? – user376239 Nov 26 '17 at 01:55
  • Well, yes......In (II) the function $g$ is an injection from $^{\Bbb N}(0,1)$ into $(0,1).$ – DanielWainfleet Nov 26 '17 at 17:38
  • I do not really understand your answer. Are you using a diagonalization method? – user376239 Dec 05 '17 at 03:14
  • The idea is that for $v:\Bbb N\to (0,1)$ we treat each $v(n)$ as a binary sequence, and we take the sequence of sequences $v(1),v(2),v(3),$... and re-arrange it into a single binary sequence $v(1,1),v(2,1),v(1,2),v(3,1),v(2,2), v(1,3), v(4,1),..$. which we call $v^*. $ It would be better if I wrote $v'(n,j)$ for $v(n,j)$. – DanielWainfleet Dec 05 '17 at 06:31