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Math Help - least upper bound and greatest lower bound

  1. #1
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    least upper bound and greatest lower bound

    How would you prove this question: Let A and B be nonempty sets of the real numbers. Define A-B={a-b:a in A,b in B}. Show that if A and B are bounded, then sup(A-B)=supA - infB and inf(A-B)=infA - supB. Thanks a lot for any help.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let m_1=\inf A, \ M_1=\sup A, \ m_2=\inf B, \ M_2=\sup B.
    Let M=M_1-m_2
    Let a\in A, \ b\in B\Rightarrow a\leq M_1,b\geq m_2\Rightarrow a-b\leq M_1-m_2, so M is a majorant for A-B.
    Let \epsilon >0. Then M-\epsilon=M_1-\frac{\epsilon}{2}-\left(m_2-\frac{\epsilon}{2}\right).
    Exist a\in A and b\in B such that a>M_1-\frac{\epsilon}{2} and b<m_2-\frac{\epsilon}{2}.
    Then a-b>M_1-\frac{\epsilon}{2}-m_2-\frac{\epsilon}{2}=M-\epsilon, so M-\epsilon is not a majorant.
    So, M=\sup A-\inf B is the supremum for A-B.

    Proove in the same way the second part.
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    Suppose that U_A  = \sup (A)\quad \& \quad L_B  = \inf (B) then \left( {x \in A} \right)\left[ {x \le U_A } \right]\quad \& \quad \left( {y \in B} \right)\left[ {y \ge L_B } \right].
    Now this means that \left( {x \in A} \right)\left[ {x \le U_A } \right]\quad \& \quad \left( {y \in B} \right)\left[ {y \ge L_B } \right]\quad \& \quad - y \le  - L_B
    So U_A  - L_B is an upper bound for the set A - B.

    Now show that U_A  - L_B is the least upper bound.
    \left( {\varepsilon  > 0} \right)\left[ {\exists a \in A:U_A  - \frac{\varepsilon }{2} < a \le U_A } \right]\left[ {\exists b \in B:L_B  \le b < L_B  + \frac{\varepsilon }{2}} \right].
    From which we see that \left( {a - b} \right) \in \left[ {A - B} \right]\quad \& \quad \left( {U_A  - L_B } \right) - \varepsilon  < a - b \le U_A  - L_B.
    Thus no number less than U_A- L_B is an upper bound for A - B making U_A - L_B the least upper bound for A - B.

    You need to fill many details. But this is the basic idea.
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