Cardinality of real numbers in the interval $[0,1]$ equals to $2^{\aleph_0}$. Now, I want to show that cardinality of all real numbers is equal to cardinality of real numbers in the interval $[0,1]$.

It is obvious,


By example: For every $\alpha\in(0,1)$ and for every $\beta\in(1,2)$ we have a bijection


Finally, we have

$$\operatorname{card}_{\mathbb{R}}(-\infty,+\infty)={\underbrace{\operatorname{card}_{\mathbb{R}}(0,-1)+\operatorname{card}_{\mathbb{R}}(-1,-2)+\operatorname{card}_{\mathbb{R}}(-2,-3)+\cdots}_{\aleph_0}}+\\+\underbrace{\cdots+\operatorname{card}_{\mathbb{R}}(0,1)+\operatorname{card}_{\mathbb{R}}(1,2)+\operatorname{card}_{\mathbb{R}}(2,3)+\cdots}_{\aleph_0}+\\ +\underbrace {{\operatorname{card}_{\mathbb Z}}(-\infty,+\infty)}_{\aleph_0}=\\=\aleph_0×2^{\aleph_0}+\aleph_0×2^{\aleph_0}+\aleph_0=2 \aleph_0×2^{\aleph_0}+\aleph_0=\aleph_0×2^{\aleph_0}+\aleph_0=2^{\aleph_0}+\aleph_0=2^{\aleph_0}$$

Is the method I use correct?

Thank you!

  • $\begingroup$ Yes it is! Technically, if you want the cardinality of a large set to be the sum of the cardinalities of smaller sets, those smaller sets shouldn't overlap at all. But you can use intervals of the form $[n,n+1)$ instead of $[n,n+1]$ and solve that problem (or argue that the double-counting of integers doesn't actually cause a problem). $\endgroup$ – Greg Martin Jul 27 '19 at 6:28
  • $\begingroup$ If you deem obvious the fact that $\aleph_0\times 2^{\aleph_0}=2^{\aleph_0}$, then this looks sound. I'd say that you are basically going to extreme lengths to avoid a proof that $[a,b)\cong [a,b]$. Yet, the decomposition you are doing of $\Bbb R$ implicitly assumes that. $\endgroup$ – Gae. S. Jul 27 '19 at 6:29
  • $\begingroup$ I would shun graphic proofs. If anyone tells me anything is "obvious" I'd immediately ask him to prove it just on general principals (and I'd state that the entire result I am proving is just as obvious). Also not sure I'd buy that an infinite $\sum_{\alpha\in Set_1}|Set_2| = |Set_1||Set_2|$ and, just for general purposes, I'd ask to prove that $2^{\aleph_0} + \aleph_0 = 2^{\aleph_0}$. I'd actually think $|[a,b]|=|\mathbb R|$ is more "obvious" than many of the assumptions made. But, in all, your proof is not incorrect. $\endgroup$ – fleablood Jul 27 '19 at 15:49
  • $\begingroup$ Also, if you know that $|(0,1)|=|\mathbb R|=2^{\aleph_0}$ then why don't you just do $[0,1] = (0,1)\cup \{0,1\}$ so $|[0,1]| = 2^{\aleph_0} + 2 = 2^{\aleph_0}$. $\endgroup$ – fleablood Jul 27 '19 at 15:57
  • $\begingroup$ BTW, when was it ever proven that $\aleph_0\times 2^{\aleph_0} = 2^{\aleph_0}$? $\endgroup$ – fleablood Jul 27 '19 at 15:57

Another method.

$f:\Bbb R\to (0,1),$ where $f(x)=\frac {1}{2}+\frac {1}{\pi}\arctan (x),$ is a bijection.

To get a bijection $g:(0,1)\to [0,1],$ take $S=\{x_n:n\in \Bbb N\}\subset (0,1)$ with $m\ne n \implies x_m\ne x_n.$ E.g. $x_n=1/(n+1)$.

Let $g(x_1)=0$ and $g(x_2)=1.$ For $n> 2$ let $g(x_n)=x_{n-2}.$ For $x\in (0,1)\setminus S$ let $g(x)=x.$

Of course now $(g\, f):\Bbb R\to [0,1]$ is a bijection.


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