# help with Cantor-Schroder-Berstein problem please!

• May 3rd 2008, 08:36 PM
ilovemath88

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!
• May 4th 2008, 09:20 AM
Plato
Quote:

Originally Posted by ilovemath88
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).
b) Define H as in the proof of the Cantor-Schroder-Berstein theorem and find H(2), H(8), H(13), and H(20).

It appears that this particular problem goes with a particular proof of Schroder-Berstein theorem.
So you are probability out of luck in getting help here unless some one has a copy of your textbook.
There are almost as many different proofs of the Schroder-Berstein theorem as there are textbooks that contain the theorem.
• May 24th 2008, 05:08 AM
PiperAlpha167
Quote:

Originally Posted by ilovemath88

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!

Here are some thoughts on part b).

Basically only two approaches to the proof.
I'll consider the so-called classical proof.
The other approach involves the use of AC.

In the classical approach, the notion of partitioning is central. Specifically, the partitioning of the sets A and B.

So, here's a plan.

First, confine G to, G:B->ran(G) (now a bijection).

Now we can construct H by combining parts of F and G^-1.

Partition both A and B by splitting A into two pieces (say X and Y), and splitting B into two pieces (say W and Z).

Take W = F[X] (the image of X under F) and Y = G[Z] (image of Z under G).

Consider the map H: A -> B.

H(x) = F(x) if x is in X,
____= G^-1(x) if x is in Y.

Once X and Y are well-defined, H will be well-defined.

Think about the following recursively defined indexed family of sets.
(I'll use A(1), A(2), ... to denote this family.)

A(1) = A - ran(G).
A(n+1) = F(G(A(n))), n>=1.

(I believe it would be a good idea to have an answer to the question, "why have we done this?".)

In your example, you should find that A(1) = {2,3,4,5,6}, A(2) = {13,14,15,16,17}, A(3) = ?, ... .
The pattern should be pretty clear.

Now let X = U(A(n)), where right side is simply the union of the family. And finally, let Y = A - X.

For H, I get H(2) = 1/8, H(8) = 1/3, H(13) = 1/19 and H(20) = 1/15.

To finish, you would need to demonstrate that H is indeed a bijection.