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Math Help - [SOLVED] Cantor-Schroder-Bernstein

  1. #1
    Super Member Deadstar's Avatar
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    [SOLVED] Cantor-Schroder-Bernstein

    Consider the injections f : \mathbb{N} \rightarrow \mathbb{N} \sqcup \mathbb{N} : n \rightarrow (0, n) and g : \mathbb{N} \sqcup \mathbb{N} \rightarrow \mathbb{N} given by
    (i, n) \rightarrow 3n + i, where i = 0 or i = 1.
    (a) From the proof of Cantor-Schroder-Bernstein compute F(n) directly for n =
    0, 1, 2, 3, 4 and find F^{-1} (0,4).

    Is this just simply finding g o f? Like for n=2 it would be f(2) = 2 then g(0,2) = 6?

    Or is it... Find x such that (g o f)(x) = 0,1,2,3,4?

    OR! Is it something to do with the fact that... if i=0 then F(n) can be either 0,3,6, ... (3n + 0 basically). And if i=1 then F(n) can be either 1,4,7, ... (3n+1)..?

    Since F isn't actually defined I'm not sure what it is.

    AND. Since the CSB theorem is involved, there will be bijection A -> B so it makes me think that some values of n will equal 0,3,6, ... or 1,4,7 ... and the rest will have to be defined as something else...
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Consider the injections f : \mathbb{N} \rightarrow \mathbb{N} \sqcup \mathbb{N} : n \rightarrow (0, n) and g : \mathbb{N} \sqcup \mathbb{N} \rightarrow \mathbb{N} given by
    (i, n) \rightarrow 3n + i, where i = 0 or i = 1.
    (a) From the proof of Cantor-Schroder-Bernstein compute F(n) directly for n = 0, 1, 2, 3, 4 and find F^{-1} (0,4).
    Do you realize that there are as many difference proofs of the Cantor-Schroder-Bernstein theorem as there are places where a proof is published?
    So there is no way for us to know how your F is defined.
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  3. #3
    Super Member Deadstar's Avatar
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    Bleh im dumb.

    Proof is the one with the zig-zig diagram that has A on one side, B on the other, f is an injection from A to B and B is an injection from B to A.
    Then you repeatedly apply f and g to an element a \in A.
    Now, we must be in one of two situations: either

    (a) we can go up indefinitely from a or end on the A side, or,
    (b) we end on the B side.

    Let A_1 = a \in A, a is type a, A_2 = a \in A, a is type b. Then A = A_1 \cup A_2 and A_1 \cap A_2 = \emptyset.

    From this we have a well defined map F : A \rightarrow B given by...

    F(a) = f(a) if a \in A_1, g^{-1}(a) if a \in A_2
    Taken from the notes!

    Indeed I probably shoulda noticed this bit first...

    So, Does this mean that F(0), F(1), etc = f(0), f(1), ... respectively since they are single integers and F^{-1}(0,4) = g(0,4) = 12 since its of the form (i,n), so we apply g to it..?
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  4. #4
    Super Member Deadstar's Avatar
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    Bump!

    In before ban/infraction/mod madness
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