The following is the Schroeder-Bernstein Theorem in Real Analysis with Real Applications by Donsig and Davidson p.$63$:
There are certain parts of the proof that I'm having trouble understanding. When I try drawing a diagram using Figure $2.6$ as a reference, for example, I'm unable to complete the diagram using a finite number of points. That is, using Figure $2.6$ I replaced the "ellipsis block" in both $A$ and $B$ by $A_{4}$ and $B_{4}$. Then I assigned exactly one point to each of $A_{1},A_{2},A_{3},A_{4},$ and $A_{5}$. I repeated the process with $B$ and assigned exactly one point to each $B_{1},B_{2},B_{3},B_{4},$ and $B_{5}$. But then applying the recursive process in the given proof, I can't apply the function $f$ to $A_{4}$ since this would imply the existence of some subset $B_{5}$ of $B$ which doesn't exist in the diagram that I've constructed. I'd like to know where I'm going wrong or what important assumption I'm missing.
Also, I'm unsure what $(fg)^{i-1}$ is supposed to represent. Is it supposed to be some sort of composite function made up of both $f$ and $g$, raised to the power $i-1$?
Best Answer
It's an infinite process, we have $B_1=B\setminus f(A)$, then $$A_1=g(B_1),\ B_2=f(A_1),\ A_2=g(B_2),\ \dots, B_5=f(A_4),\ A_5=g(B_5),\ \dots$$ Yes, $(fg)^{i-1}$ is the $i-1$-fold composition of the composite $fg$ with itself. If $i-1=0$, we mean the identity function.