I am learning a real analysis textbook in Chinese by myself and my question arising from seeing the proof of statement:
$A, B$ are two sets. If $\exists A^*\subset A, B^*\subset B,$ s.t. $A\sim B^*, A^*\sim B\Rightarrow A\sim B$.
($A\sim B$ means $A$ and $B$ have the same cardinality or there exists 1-1 correspondence $\phi$ between A and B.)
The proof is done as follow:
If $A\sim B^* \subset B,B\sim A^*\subset A$,let $\phi$ be a 1-1 function between $A$ and $B^*$, $\psi$ be a function between $A^*$ and $B$. Let $A_0=A^*, B_0=B^*, A_1=A-A_0.$ Define:
$$B_1=\phi(A_1)\equiv\{y|y=\phi(x),x\in A_1\}$$ $$A_2=\psi(B_1)\equiv\{x|x=\psi(y),y\in B_1\}$$
(1) $\color{red}{\text{Since}\ A_2\subset A_0}$, we have $A_1\cap A_2=\emptyset$.
(2) Also, let $B_2=\phi(A_2)$, since $\phi$ is 1-1, $B_1\cap B_2=\emptyset.$
(3) $\color{red}{\text{In general, if we have constructed } A_1, A_2,..., A_n\ \textbf{pairwise disjoint}, B_1,B_2,...,B_n \ \textbf{pairwise disjoint,}} $$A_{i+1}=\psi(B_i),B_i=\phi(A_i),i=1,2,...,n-1,$ let $$A_{n+1}=\psi(B_n),B_{n+1}=\phi(A_{n+1}).$$ (4) Becuase $\psi$ is 1-1, from $B_1,...B_n$ pairwise disjoint, we know that $\color{red}{A_{n+1}\text{ and } A_2,...,A_n\text{ are pairwise disjoint.}}$
(5) Also, since $A_{n+1}\subset A_o$, $A_{n+1}$ and $A_1$ are pairwise disjoint.
(6) Now, since $\phi$ is 1-1, $A_1,...,A_{n+1}$ are pairwise disjoint, $B_{n+1}$ and $B_1,...,B_n$ are pairwise disjoint.
(7) We obtain two sequences of pairwise disjoint sets $\{A_n\}^\infty_{n=1}$, $\{B_n\}^\infty_{n=1}$， $A_{i+1}=\psi(B_n),B_{n+1}=\phi(A_{n+1}),i=1,2,3...$. Therefore $\bigcup^\infty_{n=1}A_n\sim^\phi\bigcup^\infty_{n=1}B_n$.
(8) Also, through $\psi$, $B\sim A_0, B_k\sim A_{k+1}$, therefore
$$B-\bigcup^\infty_{k=1}B_k\sim^\psi A_0-\bigcup^\infty_{k=1}A_k=A_0-\bigcup^\infty_{n=2}A_n.$$
(9) $A_1=A-A_0$, $A_0\subset A\Rightarrow A_0=A-A_1$. Therefore $$A_0-\bigcup^\infty_{n=2} A_n= A-\bigcup^\infty_{n=1}A_n,$$ therefore \begin{align} A & = (A-\bigcup^\infty_{n=1} A_n)\cup(\bigcup^\infty_{n=1}A_n)\\ & = (A_0-\bigcup^\infty_{n=2} A_n)\cup (\bigcup^\infty_{n=1}A_n)\\ & \sim (B-\bigcup^\infty_{n=1}B_n)\cup(\bigcup^\infty_{n=1}B_n) \\ & = B \end{align}
Starting from (1) and (2). When I first attempted to follow the proof myself, I wrote something:
$A_1$ is a subset of $A$ where the one-to-one correspondence does not hold. Therefore, $B_1=\phi(A_1)$ may or may not be in $B_0$ .
But then I don't see why $A_2\cap A_1=\emptyset$. I think I am not quite sure if $\phi$ is the 1-1 function between $A_0$ and $B$, what exactly is $\phi(A_1)$. Where will $\phi$ map $A_1$ onto? In my understanding, $A_2\cap A_1=\emptyset$ only if $B_1\subset B_0$, if so, why? Did I misunderstand some very important concepts or I misunderstand the proof?.
Proceed to line (3), if we iterate the process, we could obtain $A_1\cap A_2=\emptyset$，$A_2 \cap A_3=\emptyset$, so on and so force. It is a weaker condition as compared to pairwise disjoint. Are we simply suppose $A_1,A_2,...,A_n$ are pairwise disjoint? (Similarly, to $B_n$, 1,...,n.) If that's the case, what makes it supposition legitimate?
It first occur to me line (3) to (7) is proof by induction, but I tried to follow the iterative procedure but cannot see why $A_1,A_2,...,A_n$ and $B_1,B_2,...B_n$ are pairwise disjoint. I am pretty lost at the end of the proof. It seems each step other than (1) and (3) are true but I don't know what exactly this proof is showing in each step. Could someone please, aside from answering my two questions stated above, also give me some idea what is the rough idea of the proof? In other words, are there are general concepts or ideas behind the proof as a whole?
