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, $A_0 \sim B$.
 A: The equality does not hold: as you've stated, it could be that $\bigcap A_n$ is nonempty. However, we can show that there is a bijection between $A_0$ and $B$ using this idea. 
Consider the following picture of the situation:

We see there are three types of regions: the region $\mathcal A=\bigcup A_n\setminus B_n$ (yellow), the region $\mathcal B=\bigcup B_n\setminus A_{n+1}$ (blue), and then the remaining elements, $\mathcal C=\bigcap A_n=\bigcap B_n$ (grey). 
Note that $A_0\supseteq B_0\supseteq A_1\supseteq B_1\dots\supseteq\mathcal C$.
We define a map $h:A_0\to B$ by letting 
\begin{align}
h:x\mapsto\begin{cases}
f(x)&x\in\mathcal A\\
x&x\in \mathcal B\cup\mathcal C
\end{cases}
\end{align}
Clearly $h$ is a bijection on $\mathcal B\cup\mathcal C$, since it is the identity function. Since $B=B_0=\mathcal B\cup \mathcal C\cup (\mathcal A\setminus (A_0\setminus B_0))$ (i.e. the set $A_0$ without its outer ring), all we have left to show is that $h$ is also a bijection from $\mathcal A$ to $\mathcal A\setminus(A_0\setminus B_0)$.
I claim that $f\restriction (A_n\setminus B_n)$ is a bijection between $A_n\setminus B_n\to A_{n+1}\setminus B_{n+1}$ for all $n$. It follows from this that $f\restriction \mathcal A$ is a bijection from $\mathcal A\to \mathcal A\setminus (A_0\setminus B_0)$, and thus that $h$ is a bijection $A_0\to B$. Clearly since $f$ is injective, its restriction is also injective.
Let $x\in A_{n}\setminus B_n$, then $f(x)\notin B_{n+1}$, since $f[B_n]=B_{n+1}$ and $x\notin B_n$ and $f$ is injective. Since $x\in A_n$, we see that $f(x)\in A_{n+1}=f[A_n]$. Therefore $f[A_n\setminus B_n]\subset A_{n+1}\setminus B_{n+1}$.
Finally $f\restriction A_n\setminus B_n$ maps surjectively to $A_{n+1}\setminus B_{n+1}$: if $y\in A_{n+1}\setminus B_{n+1}$, then $f^{-1}(y)\in A_n$. We also see $f^{-1}(y)\notin B_{n}$, since $f[B_n]=B_{n+1}$.
A: 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 $\bigcap B_i$).
