Cantor theorem

Consider a subset $S$$S$ of $\mathbb{R}$$\mathbb R$ , those number look like $0.100101...$$0.100101...$, each decimal point comes from $\{0,1\}$$\{0,1\}$.

That's just the binary system of $\mathbb{R}$$\mathbb R$.

We can consider a isomorphism from $S$$S$ to $2^{\mathbb{N}}$$2^{\mathbb N}$, that is, the free module ${\mathbb{F}}_2^{\mathbb{N}}\cong P(\mathbb{N})$$\mathbb F_2^{\mathbb N}\cong P(\mathbb N)$.

Then $S$$S$ is not countable $⟺2^{\mathbb{N}}$$\iff 2^\mathbb N$ is not countable.

 

If $2^{\mathbb{N}}$$2^\mathbb N$ is countable, write them as $\begin{array}{c}0.11001110101010....\\ 0.01111110010001....\\ 0.01111000010001....\end{array}$$0.11001110101010....\\0.01111110010001....\\0.01111000010001....$

View it as a matrix, and consider the $(b_0{b}_1{b}_2...=(a_{00}{a}_{11}{a}_{22}...+(0.1.1...$$(b_{0}b_{1}b_{2}...=(a_{00}a_{11}a_{22}...+(0.1.1...$ This will turn $0\mapsto 1,1\mapsto 0$$0\mapsto 1,1\mapsto 0$.

For example, $(010...+(0.11...=(0.01...$$(010...+(0.11...=(0.01...$

And $(b_{00}{b}_{11}{b}_{22}...$$(b_{00}b_{11}b_{22}...$ is not in this table, since $\mathrm{\forall }i\in \mathbb{N},b_i\ne a_{ii}\Rightarrow \mathrm{\forall }i\in \mathbb{N},(b_0{b}_1{b}_2...\ne (a_{i0}{a}_{i1}{a}_{i2}...$$\forall {i\in\mathbb N},b_{i}\ne a_{ii}\Rightarrow\forall i\in\mathbb N,(b_0b_1b_2...\ne(a_{i0}a_{i1}a_{i2}...$.

In general, $|X|\ne |2^X|$$|X|\ne|2^X|$, since if there exists a bijection $f:X\to 2^X$$f:X\to 2^X$

Consider $A:=\{x\in X:x\notin f(x)\}$$A:=\{x\in X:x\notin f(x)\}$ Remark. $f(x)\subseteq X$$f(x)\subseteq X$, thus $A=X-\bigcup _{x\in A}f(x)=\bigcap _{x\in A}\overline{f(x)}$$A=X-\bigcup_{x\in A} f(x)=\bigcap_{x\in A} \overline{f(x)}$

But since $f$$f$ is a bijection, $\mathrm{\exists }a\in X,f(a)=A$$\exists a\in X,f(a)=A$, thus $A\ne \bigcap _{x\in A}\overline{f(x)}$$A\ne\bigcap_{x\in A} \overline{f(x)}$.

Since $A\cap \bigcap _{x\in A}\overline{f(x)}=A\cap \bar{A}\cap \bigcap _{x\in A,x\ne a}\overline{f(x)}=\mathrm{\varnothing }$$A\cap\bigcap_{x\in A}\overline{f(x)}=A\cap\overline{A}\cap\bigcap_{x\in A,x\ne a}\overline{f(x)}=\empty$. $◻$$\square$