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Math Help - [SOLVED] Cardinal Exponentiation

  1. #16
    Super Member Deadstar's Avatar
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    Quote Originally Posted by MoeBlee View Post
    For each f in {f | f is a function from BuC into A} you're going to choose an appropriate g and h where <g h> is in
    {g | g is a function from B into A} X {h | h is a function from C into A}.

    Now, think about f. It's a function whose domain is BuC. But B and C are disjoint. So f is the union of two functions, one function from B into A and another function other from C into A.

    So, for a given f, what should g and h be?
    g should be the function the takes B into A.

    h should be the function the takes C into A.

    If that's right then I truly do not understand this course.

    Edit... I dunno...
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  2. #17
    Super Member Deadstar's Avatar
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    Look Moe can you just write down the actual bijection and ill show you what part of it I don't understand. Swear to jebus we'll be here forever if you want me to discover it myself.
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  3. #18
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    I've as much as given it to you.

    To finish it off, first I'll introduce a notation:

    Let 'r' stand for restriction. I.e., for sets X and Y, we have

    XrY = {<p q> | <p q> in X and q in Y}.

    In other words, XrY is the set of all the ordered pairs <p q> in X such that p is in Y.

    Each f in {f | f is a function from BuC into A} is of this form:

    frB u frC.

    That's just another way of saying what I mentioned in my last post.

    So for f in {f | f is a function from BuC into A} let's map f to
    <frB frC>.

    So let J be this mapping:

    For f in {f | f is a function from BuC into A},
    let J(f) = <frB frC>.

    In other words, for f in {f | f is a function from BuC into A}, I'm mapping f to some g and some h such that
    <g h> is in
    {g | g is a function from B into A} X {h | h is a function from C into A}, and
    g and h are specifically frB and frC.

    Now, all that remains is to prove that this J as just defined IS a bijection from {f | f is a function from BuC into A} onto
    {g | g is a function from B into A} X {h | h is a function from C into A}.

    So, I'd like you now to do that.
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  4. #19
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    So I guess you figured out the proof that J is a bijection.
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  5. #20
    Super Member Deadstar's Avatar
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    Quote Originally Posted by MoeBlee View Post
    So I guess you figured out the proof that J is a bijection.
    No. This is taking too much of my time up. The proofs not as basic as I thought if would be and I have waaaay too much other stuff to be doing. 9 exams and this probably wont show up on the one exam it could be in. Not worth the time. I thank you for your effort.

    Thread marked as solved.
    Last edited by Deadstar; April 29th 2010 at 04:01 PM.
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