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Thread: prove or disprove: sup of two sets

  1. #1
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    prove or disprove: sup of two sets

    Hi,
    i need to prove or disprove the following:

    $\displaystyle Sup(A\cup B)=max\left \{ SupA, SupB \right \}$

    thanks in advanced!

    *edit:
    both sets are bounded from above.
    Last edited by Stormey; Jan 21st 2013 at 04:51 AM.
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  2. #2
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    Re: prove or disprove: sup of two sets

    Prove (1) $\displaystyle \sup(A\cup B)\le\max(\sup A,\sup B)$ and (2) $\displaystyle \sup(A\cup B)\ge\max(\sup A,\sup B)$.

    For (1), show that $\displaystyle \max(\sup A,\sup B)$ is an upper bound of $\displaystyle A\cup B$.

    For (2), use the fact that $\displaystyle x\ge y$ and $\displaystyle x\ge z$ imply $\displaystyle x\ge\max(y,z)$.
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  3. #3
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    Re: prove or disprove: sup of two sets

    I've found the following to be useful:
    supremums of larger sets are larger and infimums of larger sets are smaller

    Formally, (assuming the sets are bounded):
    1. If $\displaystyle C\subseteq D$ then $\displaystyle sup(C)\leq sup(D)$
    2. If $\displaystyle C\subseteq D$ then $\displaystyle inf(C)\geq inf(D)$

    Proof of 1 - Let $\displaystyle u=sup(D)$. Let $\displaystyle x\in C$, then $\displaystyle x\in D$ and so $\displaystyle x\leq u$; i.e. $\displaystyle u$ is an upper bound of $\displaystyle C$. So $\displaystyle sup(C)\leq u$. The proof for infs is exactly similar.

    For your original question, since $\displaystyle A\subseteq A\cup B$, $\displaystyle sup(A)\leq sup(A\cup B)$, and similarly for B.
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  4. #4
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    Re: prove or disprove: sup of two sets

    Thanks guys.
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