Thread: least upper bound and greatest lower bound

1. 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.

2. 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.
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.

3. 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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prove sup(a-b)=supa-infb

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