least upper bound and greatest lower bound

**BrainMan** · September 22nd 2007, 08:24 AM

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.

**red_dog** · September 22nd 2007, 09:52 AM

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.

**Plato** · September 22nd 2007, 09:59 AM

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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