# Questions about the Schroeder-Bernstein Theorem

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 $$i-1$$?

• 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. – DanielWainfleet Aug 14 at 15:30
• @Daniel: Don't forget the Cantor–Schroeder–Bernstein. And the variations of Schröder, Schroder, and Schroeder as well. – Asaf Karagila Aug 20 at 13:17
• @DanielWainfleet Which proof does Simmons use? Is it the one that constructs $C_0:=A\setminus g[B],\,C_{n+1}:=g[f[C_n]],\,C:=\bigcup_{n\ge0}C_n$? – J.G. Aug 20 at 13:35
• – Asaf Karagila Aug 20 at 13:37
• @J.G. It is the proof in Wikipedia (attributed to Julius Konig) but I think Simmons' presentation is clearer. – DanielWainfleet Aug 21 at 7:26

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.
• 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? – K.M Aug 14 at 18:22
• No, they can also be empty. Think e.g. when $f$ is bijection and $g=f^{-1}$. – Berci Aug 14 at 18:24