# least upper bound and greatest lower bound

• Sep 22nd 2007, 08:24 AM
BrainMan
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.
• Sep 22nd 2007, 09:52 AM
red_dog
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.
• Sep 22nd 2007, 09:59 AM
Plato
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.