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Math Help - What is the closure under F in this question?

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    What is the closure under F in this question?

    (This is not my question yet.)

    Suppose f:A \longrightarrow A. A set C is said to be closed under f if \forall x\in C (f(x) \in C) Now suppose B \subseteq A. The closure of B under f is the smallest set C such that B \subseteq C \subseteq A and C is closed under f, if there is such a smallest set.

    Let
    B_1 = B
    \forall n \geq 1, B_{n+1} = f(B_n) = {f(x)| x \in B_n}

    Then the closure of B under f is given by the set

    C = \cup_{n \in Z^+}B_n.

    Here is my question.

    Suppose \Im is a family of functions from A to A, and B \subseteq A. The closure of B under \Im is the smallest set C such that B \subseteq C \subseteq A and \forall f \in \Im, C is closed under \Im, if there is such a smallest set.

    What should be the closure of B under \Im in this case?

    I try to prove, but am unsure.

    Should it be C = \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right) or \cup_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right) ??

    Or could it be  \cup_{n \in Z^+} \cap_{f \in \Im} f \left(B_n\right) ?

    I tried to prove  \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right) is the closure of B under  \Im. Every property, closedness under f, and \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right) is the smallest set has been proved, except B \subseteq \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right) \subseteq A. (I'm kinda stucked.)


    I tried to use info from the first part. Let C_f be the closure of B under f Then I claimed that \cup_{f \in \Im}C_f is the closure of B under \Im, but my friend said it is not closed. For example,
    B={0}
    f(x)=x+2,
    g(x)=x+3

    C_f={0,2,4,6,8,10,...}
    C_g={0,3,6,9,12,15,...}

    C_f \cup C_g does not contain 5, but f\left(g(0)\right) = 5



    Thanks a lot!
    Last edited by armeros; April 26th 2009 at 02:45 AM.
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