Help with Cantor-Schroder-Berstein theorem please

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The discussion revolves around applying the Cantor-Schroder-Berstein theorem to specific functions and sets. The user seeks clarification on calculating values for two functions, F and G, and expresses confusion about defining function H as required by the theorem. They provide initial calculations for f and g but are uncertain about their accuracy and how to proceed with H. The response emphasizes the need to define H and clarify the proof structure before further assistance can be provided. Understanding these elements is crucial for effectively applying the theorem.
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I really don't get this. Can someone please help me?

Apply the proof of the Cantor-Schroder-Berstein theorem to this situatuion:
A={2,3,4,5,...}, B={1/2,1/3,1/4,...}, F:A-->B where F(x)=1/(x+6)and G:B-->A where G(x)=(1/x)+5. Note that 1/3 and 1/4 are in B-Rng(F). Let f be the string that begins at 1/3, and let g be the string that begins at 1/4.

a)Find f(1), f(2), f(3), f(4).
For this one, f(1)=1/3, then f(2)=1/(1/3)+5=8, then f(3)=1/(8+6)=1/14, and f(4)=1/(1/14)+5=19? Is that right? and then I do the same for g? g(1)=1/4, g(2)=9, g(3)=1/15, and g(4)=20?
b) Define H as in the proof of the Cantor-Schroder-Berstein theorem and find H(2), H(8), H(13), and H(20).
I have no clue how to do this one at all. Can you help me please?


Thanks for the help!
 
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There are many ways to prove this theorem. You have to tell us what H is, and what YOUR proof looks like, before we can apply it.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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