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Math Help - problem about reflexive relation

  1. #1
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    problem about reflexive relation

    Hi

    I have this problem in the topic of relations.

    Suppose R is a relation on A, and define a relation
    S on \mathscr{P} (A) as follows.

    S=\{(X,Y) \in \mathscr{P} (A) \times \mathscr{P} (A) \vert \forall x \in X \exists y\in Y(xRy)\}

    for each part ,give either a proof or a counterexample to justify your answer.

    a)If R is reflexive , must S be reflexive ?

    I was able to prove this part. I was just playing with the examples which
    satisfy the above theorem.

    Let A=\{1,2\}

    and lets define

    R=\{(1,2),(1,1),(2,2)\}

    clearly R is reflexive. So S would be

    S=\{ (\{1\},\{2\}),(\{1\},\{1,2\}),(\{1\},\{1\}),(\{2\}  ,\{2\}),\\ (\{2\}, \{1,2\}),(\{1,2\},\{1,2\}),(\{1,2\},\{2\}),\\ (\varnothing,\varnothing), (\varnothing, \{1\}),(\varnothing,\{2\}),(\varnothing,\{1,2\})\}

    is S correct ? I included \varnothing since , in the condition for
    S , we have ,

    \forall x \in X \exists y\in Y(xRy)

    which can be written as an implication,

    \forall x \left[ x \in X \Rightarrow \exists y \in Y(xRy) \right]

    so if

    X=\varnothing

    then the antecedent is true always , as can be seen from the truth table of the
    implication. But I couldn't include the ordered pairs like

    (\{1\},\varnothing)


    in S because if we let

    X=\{1\}

    and

    Y=\varnothing


    then x \in X is true but \exists y\in Y(xRy) is false

    Which means that, in the implication , antecedent is true and the consequent is false,
    which according to the truth table is false. So

    (\{1\},\varnothing)

    doesn't satisfy the condition for the inclusion in S.

    is my reasoning correct ? If so, then we can see that S is also reflexive,
    which is what the above theorem asserts

    I see one problem the way symbol for the power set P is displayed. I have used
    \mathscr{P} to display slanted P, but it just displays ordinary \mathscr{P}. Is maths package mathrsfs installed on this forum ?

    thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: problem about reflexive relation

    Quote Originally Posted by issacnewton View Post
    is my reasoning correct ?
    Yes,it is

    If so, then we can see that S is also reflexive, which is what the above theorem asserts
    Right.

    I see one problem the way symbol for the power set P is displayed. I have used \mathscr{P} to display slanted P, but it just displays ordinary \mathscr{P}. Is maths package mathrsfs installed on this forum ?
    With \mathcal{P}(A) we get \mathcal{P}(A) .
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  3. #3
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    Re: problem about reflexive relation

    ola fernando

    thanks for verifying.

    adios
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: problem about reflexive relation

    Quote Originally Posted by issacnewton View Post
    ola fernando
    Ola in Spanish means wave and hola means hello.

    thanks for verifying.
    You are welcome!.
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  5. #5
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    Re: problem about reflexive relation

    Quote Originally Posted by issacnewton View Post
    is my reasoning correct ?
    Of course, it's not a proof of the statement, it's just one example.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: problem about reflexive relation

    Quote Originally Posted by emakarov View Post
    Of course, it's not a proof of the statement, it's just one example.
    Of course, but isaacnewton said "if so we can see that S is reflexive" i.e. he didn't say "my example proves S is reflexive".
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