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Math Help - The cardinality of set

  1. #1
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    The cardinality of set

    I've got a problem, I need to find the cardinality of set of every choice function for P(N)-\{\emptyset\} and i have no idea how to start. Can anyone help me,please?
    Last edited by Plato; December 4th 2011 at 03:56 AM.
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  2. #2
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    Re: The cardinality of set

    it should be "P(N)-{empty set}" sorry i'm new here i don't know exactly how to write it.
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  3. #3
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    Re: The cardinality of set

    Quote Originally Posted by infinity003217 View Post
    it should be "P(N)-{empty set}" sorry i'm new here i don't know exactly how to write it.
    I edited your LaTeX here is the code:
    [tex]P(N)-\{\emptyset\} [/tex] gives P(N)-\{\emptyset\}

    Here is a website that discusses the axiom of choice.

    Now, I for one find the wording of this question a bit odd.
    Have you posted the exact wording of the question as given to you?
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    Re: The cardinality of set

    Ok, i'll try to translate it one more time, I have the set of choise functions P(N)-\{\emptyset\} and i need to find the cardinality of this set. Sorry but I'm not English so probably I make a lot of mistakes.
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  5. #5
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    Re: The cardinality of set

    Quote Originally Posted by infinity003217 View Post
    Ok, i'll try to translate it one more time, I have the set of choise functions P(N)-\{\emptyset\} and i need to find the cardinality of this set.
    Is it possible that you are translating characteristic function as choice function?
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  6. #6
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    Re: The cardinality of set

    no, i don't think so. it is the choice function, related to the axiom of choice. I don't know if it helps.
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  7. #7
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    Re: The cardinality of set

    P(N) - {} will not give you a choice function. It will give you every possible non-empty subset of natural numbers.

    A function is generally defined as a set of ordered pairs where the first element is unique (more precisely, a set of pairs (x,z) where if you have (x,z) and (y,z) it must be true that x = y). Also, in the absence of qualification, functions are left-total: every object in the domain must be defined. A choice function is a function defined with domain: a collection of non-empty sets, and target: elements of those sets (the union), with the special restriction that each set must map to one of its own elements.

    Now P(N) - {} definitely has a choice function: you don't even need the Axiom of Choice for it! For each subset of N, simply take the smallest element. This works because the naturals are well-ordered. There are {2^\aleph^_0} (or \beth_1) non-empty subsets of N, so that would be the cardinality of any choice function on N. I'd suspect there are \beth_2 many such choice functions on P(N) - {}, but that's just a wild guess.
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  8. #8
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    Re: The cardinality of set

    ok, now I understand. Thank you.
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