About a lemma to prove the Cantor-Bernstein-Schroeder theorem. I am reading "Logic in mathematics and set theory" by Kazuyuki Tanaka and Toshio Suzuki.  
In this book, there is a lemma to prove the Cantor-Bernstein-Schroeder theorem.
I cannot understand why the equality $$A_0 = (A_0 - B_0) \cup (B_0 - A_1) \cup (A_1 - B_1) \cup (B_1 - A_2) \cup \cdots \cup (A_n - B_n) \cup (B_n - A_{n+1}) \cup \cdots$$ holds.
Maybe there exists an element $x$ such that $x \in A_i$(and $x \in B_i)$ for all $i$.
For example I think if $A_0 = B = A_1$ and $f = id$, then $x \in A_i$ for all $i$ if $x \in A_0$.

Lemma 1.12
  Let $A_0, B, A_1$ be sets such that $A_1 \subset B \subset A_0$ and $A_0 \sim A_1$.
  Then, $A_0 \sim B$.  

Proof:
Let $f : A_0 \to A_1$ be a bijection.
Let $B_0 := B$.
Let $A_{n+1} := f[A_n], B_{n+1} := f[B_n]$ for $n \in \{0, 1, 2, \cdots \}$.
Then, $A_0 = (A_0 - B_0) \cup (B_0 - A_1) \cup (A_1 - B_1) \cup (B_1 - A_2) \cup \cdots \cup (A_n - B_n) \cup (B_n - A_{n+1}) \cup \cdots.$
Let $g : A_0 \to B_0$ be a mapping such that $g(x) = f(x)$ if $x \in A_n - B_n$ for some $n$ and $g(x) = x$ for $x \in B_n - A_{n+1}$ for some $n$.
Then, it is easy to prove $g : A_0 \to B_0$ is a bijection.
So, $A_0 \sim B$.

It doesn't. If the equality is true for some $A_1\subseteq B_0\subseteq A_0$ and $f:A_0\to A_1$, we can simply extend the sets using a new set $C$, and also extend $f$ by making it the identity on $C$, in which case $C$ will be contained within all sets $A_i$ and $B_i$. However you can augment the argument by making $g$ act as the identity on $C$ too, where we define $C$ to be the intersection $\bigcap A_i$ (which equals $\bigcap B_i$).
