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The following is the Schroeder-Bernstein Theorem in Real Analysis with Real Applications by Donsig and Davidson p.$63$:

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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 $i-1$?

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    $\begingroup$ There are other proofs. Search this site. Caution: It is also called the Cantor-Bernstein or Schroeder-Cantor-Bernstein theorem. There is one proof which I gave as an Answer on this site, which can also be found in Introduction To Topology And Modern Analysis, by Simmons, which is the simplest proof I know of. $\endgroup$ – DanielWainfleet Aug 14 at 15:30
  • $\begingroup$ @Daniel: Don't forget the Cantor–Schroeder–Bernstein. And the variations of Schröder, Schroder, and Schroeder as well. $\endgroup$ – Asaf Karagila Aug 20 at 13:17
  • $\begingroup$ @DanielWainfleet Which proof does Simmons use? Is it the one that constructs $C_0:=A\setminus g[B],\,C_{n+1}:=g[f[C_n]],\,C:=\bigcup_{n\ge0}C_n$? $\endgroup$ – J.G. Aug 20 at 13:35
  • $\begingroup$ Related math.stackexchange.com/questions/3320794/…? $\endgroup$ – Asaf Karagila Aug 20 at 13:37
  • $\begingroup$ @J.G. It is the proof in Wikipedia (attributed to Julius Konig) but I think Simmons' presentation is clearer. $\endgroup$ – DanielWainfleet Aug 21 at 7:26
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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.

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  • $\begingroup$ Since the process is infinite and the subsets $A_{n}$ and $B_{n}$ are disjoint, does that necessarily mean that $A$ and $B$ are infinite sets? $\endgroup$ – K.M Aug 14 at 18:22
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    $\begingroup$ No, they can also be empty. Think e.g. when $f$ is bijection and $g=f^{-1}$. $\endgroup$ – Berci Aug 14 at 18:24

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