In Abbott's real analysis, I am asked to prove Schroder-Bernstein using steps that are given in the book. We have two sets $X$ and $Y$, and there are injections $f:X\rightarrow Y$and $g:Y\rightarrow X$. We are to prove that there exists a bijection $h:X\rightarrow Y$. The idea proposed by the author is to first define a set $A_1=X\backslash g(Y)=\lbrace x\in X | x\notin g(Y)\rbrace$, and then inductively define the sets $A_n=\lbrace x\in X | x=g(f(A_n))\rbrace$. Of course, if $A_1=\emptyset$, we are done with the proof as we have that $g(x)$ is bijective. If not, we let $A=\bigcup\limits_{n=1}^\infty A_n$ and $B=\bigcup\limits_{n=1}^{\infty}f(A_n)$, such that $f(A)=B$. Next we define sets $A'=X\backslash A$ and $B'=Y\backslash B$. I am asked to prove the following:
Show that $g: Y\rightarrow X$ maps $B'$ onto $A'$.
My attempt: My idea is to prove that $A'=g(B')$, i.e. show that $x\in A'\Leftrightarrow x\in g(B')$. First, letting $x\in g(B')$ we know that $x\notin A_1$ as $x\in g(Y)$. We also have that $x\notin A_i$ for all $i\geq 2$ as each $A_{n+1}=g(f(A_n))$ and $B'\notin A_n$ for all $n\in\mathbb{N}$.
I'm not sure how to prove that $x\in A'\Rightarrow x\in g(B')$.
Hint: Suppose $x \in A'$. To prove $x \in g(B')$, it suffices to prove the following two things:
Please let me know if this helps.