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Thread: 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 $\displaystyle A=(-\infty,b^*]~?$
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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 $\displaystyle a_{n} \in A s.t. a_{n}\nearrow b*$. Because A not bounded below, there exists $\displaystyle c_{n} \in A$ s.t. $\displaystyle c_{n}\searrow -\infty$.

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

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

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

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

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

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

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

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