Trouble understanding the proof of Schröder-Bernstein theorem in Zorich's book? I've seen this exercise on Zorich's Mathematical Analysis 1:



Doubts:


*

*It seems that from $\text{card }X = \text{card }Z$, it follows that $(\text{card }X\leq \text{card }Y)$ and $(\text{card }Y\leq \text{card }X)$ and then $(\text{card }X\ = \text{card }Y)$, It's not clear how. 

*I know I'd have to prove that $g$ is a bijection, probably showing that $g$ is injective and surjective? How do I do that? I lack the tools to deal with that function and prove that. 
 A: I reworded your questions. Hopefully I am not misunderstanding you.
Question 1: Why does it suffice to prove Proposition 1 below?
Proposition 1.
$$(X\supset Y\supset Z)\land(card X=card Z)\implies card X=card Y$$
Suppose we have proved this, and we want to prove the Schroder-Berstain theorem:
$$(card X\geq card Y)\land(card X\leq card Y)\implies card X=card Y$$
Proof. of the S-B thm using Prop. 1. Note that $(card X\geq card Y)\land(card X\leq card Y)$ is equivalent to saying there are injections $Y\to X\to Y$. We assume from the first injection $Y\subset X$. Note that we cannot simultaneously assume $X\subset Y$. However, we can let $Z$ be a copy of $X$ and assume that $Z\subset Y$. Then we have $X\supset Y\supset Z$. Also, we have $card X=card Z$ because $Z$ is a copy of $X$. Then by Prop. 1, $card X=card Y$.

Question 2: How does one show $g$ is bijective?
For simplicity let us write
$$C_1=\bigcup_{n=0}^\infty f^n(X)\backslash f^n(Y),\quad C_2=\overline{C}_1$$
Then $g(x)$ is defined as $g(x)=f(x)$ if $x\in C_1$ and $g(x)=x$ if $x\in C_2$.
Lemma 1. $x\in C_1\iff f(x)\in C_1$.
Proof. Easy, since $f$ is bijective.
Lemma 2. $g(X)\subset Y$.
Proof.  If $x\in C_1$, then $g(x)=f(x)\in Z\subset Y$. If $x\in C_2$, then $x$ satisfies
$$\forall n\geq0,\ x\notin f^n(X)\backslash f^n(Y)$$
which is equivalent to
$$\forall n\geq0,\ x\notin f^n(X)\text{ or }x\in f^n(Y)$$
Choose $n=0$ and we have "$x\notin X$ or $x\in Y$". The former is false, hence $x\in Y$, and $g(x)=x\in Y$.
Lemma 3. $g$ is surjective.
Proof. For any $y\in Y$. If $y\in C_2$, then $g(y)=y$. If $y\in C_1$, then $y\in f^n(X)\backslash f^n(Y)$ for some $n$. Clearly $n\geq1$. Hence $y\in f^n(X)\subset Z$. Therefore, $\exists x\in X$ s.t. $f(x)=y$. By Lemma 1, $x\in C_1$. Hence, $g(x)=f(x)=y$.
Lemma 4. $g$ is injective.
Proof. Suppose not, and there is some $y\in Y$ such that $g^{-1}(y)$ has more than one elements, say $x_1\neq x_2$. If $x_1,x_2$ are both in $C_1$ or $C_2$, then clearly $x_1=x_2$. If $x_1\in C_1$ but $x_2\in C_2$, then $f(x_1)=x_2=y\in C_2$. But by Lemma 1 $f(x_1)\in C_1$, contradiction.
By Lemma 2,3 and 4 we obtain the Schroder-Berstain theorem.
