How to prove that $\sup(A-B) = \sup(A) - \inf(B)$?

Question

How to prove that $\sup(A-B) = \sup(A) - \inf(B)$?

My attempt: Let $c \in A-B$ and define $c= a-b$, where $a \in A$ and $b \in B$. Then $a-b \leq \sup(A) - \inf(B)$. Hence, $\sup(A-B) \leq \sup(A) - \inf(B)$.

Moreover, for any $\epsilon >0$, $\sup(A) \leq a + \epsilon$ and $\inf(B) \geq b - \epsilon$. This implies that $$\sup(A) + b - \epsilon \leq \inf(B) + a + \epsilon$$ $$\sup(A) - \inf(B) \leq a-b + 2\epsilon \leq \sup(A-B)$$

Is this proof correct or logical?

Answer 1

Let $z \in A-B, z=a-b$ for $a \in A, b \in B$. Since $\sup (A)$ is the least upper bound for A, inf(B) is the greatest lower bound for B, we know $$ a \leq \sup (\mathrm{A}), \mathrm{b} \geq \inf (\mathrm{B}) $$ so $z=a-b \leq \sup (A)-\inf (B)$. Therefore $\operatorname{sub}(A)$-inf(B) is an upper bound for A-B.

Fix $\epsilon>0$, then there exists $a \in A$ such that $\sup (A)-\frac{\epsilon}{2}<a$, and there exists $b \in B$ such that $\inf (\mathrm{B})-\frac{\epsilon}{2}>\mathrm{b}$. Let $k$ be another upper bound for $\mathrm{A}-\mathrm{B}$, then $a-b \leq k$ for $a \in A, b \in B$. Therefore, $$ \begin{array}{c} \sup (A)-\inf (B)<k+\epsilon \\ \sup (A)-\inf (B) \leq k \end{array} $$ Therefore, $\sup (A)-\inf (B)=\sup (A-B)$.

Answer 2

$A-B=\{a-b : a\in A,b\in B\}$

Then $\forall a\in A, b\in B$

$a\le \sup(A) $ and $b\ge \inf(B) $

Implies $a-b\le \sup(A) -\inf(B) $

Hence, $\sup(A) -\inf(B) $ is an upper bound of the set $A-B$ and $\sup(A-B) $ is the least upper bound.

Implies $\sup(A-B) \le \sup(A) -\inf(B) $