Schröder-Bernstein, check proof

Question

Theorem: if $r:A\to B, s:B\to A$ are injections, there is a bijection $t:A\to B$.

I will prove this lemma: "if exist $f:A\to B\subset A$ injective, then exist $h:A\to B$ bijective.

With this lemma, consider the injective function $f=s\circ r:A\to s(B)\subset A$, and exist $h:A\to s(B)$ bijective. Now, $s_{|B}:B\to s(B)$ is bijective. Finally, $t= s_{|B}^{-1}\circ h:A\to B$ is a bijection.

I want check the proof of lemma: Put $Y=A-B$, $X=Y\cup (\bigcup_{i\in \mathbb{N}}f^i(Y))$, where $ f^i(Y)=f(f(...f(Y)...)), i$ times.

Because $Y\cap B=\emptyset$, $ f^k(Y)\subset B$ and $Y$ are disjoint. Because $f$ is injective, $Y\cap f^k(Y)=\emptyset \to f^m(Y)\cap f^{m+k}(Y)=\emptyset $ for all $k,m$. Note in the definition of $X$ that the union is disjoint, and $X=Y\cup f(X)$.

Finally, Note that $A-X=(Y\cup B)-(Y\cup f(X))=B-f(X)$.

If we define $h:A\to B$ as $h=f$ in $X$ and $h=id$ in $A-X$, is bijective.

Answer 1

It’s almost right, and the one error may just be a typo. It should be

$$A\setminus X=(Y\cup B)\setminus(Y\cup f[X])=B\setminus f[X]\;,$$

not

$$A\setminus X=(Y\cup B)\setminus(Y\setminus f[X])=B\setminus f[X]\;.$$