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Math Help - Combinatorics function

  1. #1
    Member oldguynewstudent's Avatar
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    Combinatorics function

    Q 28 from page 26 of Combinatorics A Guided Tour:

    Define g: 2^[ ^2 ^] \longrightarrow \mathbb Z by the rule g(S)=|S|, where S is any subset of [2]. Write g as a set of ordered pairs.

    S could be {1} {2} and {1,2}.

    So for S={1} would g(S) be 1? So would that be 2^1=2 and the ordered pair would be (2,1)?

    But then for S={2} the cardinality of S would again be 1 giving the ordered pair (4,1).

    I don't know how to do S={1,2}.

    I am confused to the nth power.

    Any clues towards unwarping my brain would be greatly appreciated.
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  2. #2
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    You have used the notation [2] before. But it is not standard.
    Does it mean \{1,2\}? If not please tell us how the textbook uses it.
    Note that \mathbb{P}(\{1\})=\{\emptyset,~\{1\}\}
    So g=\{(\emptyset,0),(\{1\},1\} in this case .
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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    You have used the notation [2] before. But it is not standard.
    Does it mean \{1,2\}? If not please tell us how the textbook uses it.
    Note that \mathbb{P}(\{1\})=\{\emptyset,~\{1\}\}
    So g=\{(\emptyset,0),(\{1\},1\} in this case .
    [n] as defined in the textbook is the set of all postive integers less than or equal to n.

    I don't believe that the problem involved the Power set.
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    Quote Originally Posted by oldguynewstudent View Post
    I don't believe that the problem involved the Power set.
    Quote Originally Posted by oldguynewstudent View Post
    Define g: by the rule g(S)=|S|, where S is any subset of [2].
    Well you yourself said it does in the OP.
    The function g maps the subsets of [2]
    to the integers. The power set of [2] is the set of all subsets of [2].

    From the OP the function g=\{(\emptyset,0),(\{1\},1),(\{2\},1),(\{1,2\},2).

    So it has everything to do with power sets.

    But once again you notation is confused.
    2^{[2]} is the set of all functions from the set [2] to the set \{0,1\}.
    However, that is not the way g was defined.

    If I were you, I would get a textbook that uses standard notation.
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  5. #5
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    Well you yourself said it does in the OP.
    The function g maps the subsets of [2]
    to the integers. The power set of [2] is the set of all subsets of [2].

    From the OP the function g=\{(\emptyset,0),(\{1\},1),(\{2\},1),(\{1,2\},2).

    So it has everything to do with power sets.

    But once again you notation is confused.
    2^{[2]} is the set of all functions from the set [2] to the set \{0,1\}.
    However, that is not the way g was defined.

    If I were you, I would get a textbook that uses standard notation.
    Yes, I see your point Plato. Thanks.

    When it says any subset, then null would be one of those subsets.

    I will have to ponder this and perhaps the professor will have time to expound on this. But since class hasn't started yet, perhaps he won't have time. Don't have a choice on the textbook however. Most of the text is very good, it's the unanswered problems that are a little fuzzy. I may try to pick up some extra material before the summer session starts.
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