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Math Help - Set Theory Problem

  1. #1
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    Set Theory Problem

    A is a convex, nonempty set. A is not bounded below and is bounded above. B* = sup A.
    Prove:
    Claim 1: (-inf, b*) is a subset of A
    Prove:
    Claim 2: A is a subset of (-inf, b*]

    Case 1: b* is not an element of A
    Prove:
    Claim 2.1: A = (-inf, b*)

    Case 2: b* is an element of A
    Prove:
    Claim 2.2: A = (-inf, b*)

    (There are 4 proofs here)
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  2. #2
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    Re: Set Theory Problem

    Quote Originally Posted by Kiefer View Post
    A is a convex, nonempty set. A is not bounded below and is bounded above. B* = sup A.
    Prove:
    Claim 1: (-inf, b*) is a subset of A
    Prove: Claim 2: A is a subset of (-inf, b*]

    Case 1: b* is not an element of A
    Prove: Claim 2.1: A = (-inf, b*)

    Case 2: b* is an element of A
    Prove:Claim 2.2: A = (-inf, b*)
    Don't you have a typo in that?
    Surely you mean A=(-\infty,b^*]~?
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  3. #3
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    Re: Set Theory Problem

    (I'm assuming A is a subset of the reals.)

    Claim: A is a subset of (-inf, b*]
    Proof: b* = sup A.

    Claim: (-inf, b*) is a subset of A
    Proof: Because b* = sup A, there exists a_{n} \in A  s.t.  a_{n}\nearrow b*. Because A not bounded below, there exists c_{n} \in A s.t. c_{n}\searrow -\infty.

    Let x \in (-\infty, b*). Then there exists m_{1} s.t. c_{m_{1}} < x and m_{2} s.t. x < a_{m_{2}} <= b*.

    Since c_{m_{1}}, a_{m_{2}} \in A, c_{m_{1}} < x < a_{m_{2}}, it follows by A convex that x in A. Thus (-\infty, b*) \subset A.

    Claim: If b* in A, then A = (-\infty, b*]. If b* not in A, then A = (-\infty, b*).

    Proof: From previous two claims. Know that (-\infty, b*) \subset A \subset (-\infty, b*].

    If b* in A, then (-\infty, b*) \subset A, and b* in A, so that their union, (-\infty, b*], is a subset of A.

    But then (-\infty, b*] \subset A \subset (-\infty, b*], thus A = (-\infty, b*].

    If b* not in A, then  A \subset (-\infty, b*] and A \subset \{b*\}^{c}, so A is contained in their intersection, (-\infty, b*).

    Thus (-\infty, b*) \subset A \subset (-\infty, b*), thus A = (-\infty, b*).
    Last edited by johnsomeone; September 10th 2012 at 07:50 PM.
    Thanks from Kiefer
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