Proof checking about cardinality of all Real numbers

Question

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,

$$\operatorname{card}_{\mathbb{R}}(0,1)=\operatorname{card}_{\mathbb{R}}(1,2)=\operatorname{card}_{\mathbb{R}}(2,3)=\cdots$$

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

$$\alpha\longmapsto\beta\iff\alpha\longmapsto\alpha+1$$

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!

Answer 1

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.