Results 1 to 3 of 3

Thread: least upper bound and greatest lower bound

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    38

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Let $\displaystyle m_1=\inf A, \ M_1=\sup A, \ m_2=\inf B, \ M_2=\sup B$.
    Let $\displaystyle M=M_1-m_2$
    Let $\displaystyle a\in A, \ b\in B\Rightarrow a\leq M_1,b\geq m_2\Rightarrow a-b\leq M_1-m_2$, so $\displaystyle M$ is a majorant for $\displaystyle A-B$.
    Let $\displaystyle \epsilon >0$. Then $\displaystyle M-\epsilon=M_1-\frac{\epsilon}{2}-\left(m_2-\frac{\epsilon}{2}\right)$.
    Exist $\displaystyle a\in A$ and $\displaystyle b\in B$ such that $\displaystyle a>M_1-\frac{\epsilon}{2}$ and $\displaystyle b<m_2-\frac{\epsilon}{2}$.
    Then $\displaystyle a-b>M_1-\frac{\epsilon}{2}-m_2-\frac{\epsilon}{2}=M-\epsilon$, so $\displaystyle M-\epsilon$ is not a majorant.
    So, $\displaystyle M=\sup A-\inf B$ is the supremum for $\displaystyle A-B$.

    Proove in the same way the second part.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Suppose that $\displaystyle U_A = \sup (A)\quad \& \quad L_B = \inf (B)$ then $\displaystyle \left( {x \in A} \right)\left[ {x \le U_A } \right]\quad \& \quad \left( {y \in B} \right)\left[ {y \ge L_B } \right]$.
    Now this means that $\displaystyle \left( {x \in A} \right)\left[ {x \le U_A } \right]\quad \& \quad \left( {y \in B} \right)\left[ {y \ge L_B } \right]\quad \& \quad - y \le - L_B$
    So $\displaystyle U_A - L_B $ is an upper bound for the set $\displaystyle A - B$.

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

    You need to fill many details. But this is the basic idea.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Greatest Lower Bound project
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Mar 30th 2011, 03:14 AM
  2. Replies: 0
    Last Post: Feb 19th 2010, 01:06 AM
  3. greatest least bound and least upper bound proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Nov 4th 2009, 04:44 PM
  4. Greatest lower bound and lower bounds
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 13th 2009, 02:26 PM
  5. Upper bound/Lower bound?
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Sep 13th 2009, 10:48 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum