Show $\sup{A}-\inf{B}=\sup\{a-b:a\in A, b\in B\}$ Let $A, B \subset\mathbb{R}$ be bounded sets.  Show $$\sup{A}-\inf{B}=\sup\{a-b:a\in A, b\in B\}$$
 A: We have
$$\sup A\geq a\quad,\quad\forall a\in A\qquad;\qquad \inf B\leq b\quad,\quad\forall b\in B$$
so 
$$\sup A-\inf B\geq a-b\quad,\quad\forall a\in A,b\in B$$
hence we deduce 
$$\sup A-\inf B\geq S:=\sup\{a-b:a\in A, b\in B\}$$
Now let $\epsilon>0$ there's $a\in A$ s.t. $a\geq \sup A-\epsilon/2$ and there's $b\in B$ s.t. $b\leq \inf B+\epsilon/2$ then
$$\sup A-\inf B-\epsilon=(\sup A-\epsilon/2)-(\inf B+\epsilon/2)\leq a-b$$
hence
$$\sup A-\inf B= S$$
[edit: conclusion is that sup A - inf B ≤ a-b, not a+b]
A: Just for fun, although this is an old and contextless question, here is an alternative proof which (in my opinion) is simpler than that of the first answer, because it uses simpler definitions, and is presented in a more 'calculational' way.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\inf}[1]{\text{inf}(#1)}
\newcommand{\sup}[1]{\text{sup}(#1)}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
First, here are the definitions: for all $\;z\;$, $$
\tag{1}
\sup{A} \leq z \;\equiv\; \langle \forall a : a \in A : a \leq z \rangle
$$ and $$
\tag{2}
z \leq \inf{B} \;\equiv\; \langle \forall b : b \in B : z \leq b \rangle
$$ for any set $\;A\;$ with an upper bound, and any lower-bounded $\;B\;$.  And as a proof principle, the following is very useful when dealing with upper and lower bounds: $$
\tag{3}
x = y \;\equiv\; \langle \forall z :: x \le z \;\equiv\; y \le z \rangle
$$ which says that two numbers are equal iff they have the same upper bounds.

Now we can prove $$
\tag{4}
\sup{A} - \inf{B} \;=\; \sup{A-B}
$$ by calculating as follows for any $\;z\;$, essentially "upper bounding" the most complex left hand side, and then expanding and simplifying, working towards the right hand side:
$$\calc
    \sup{A} - \inf{B} \le z
\op=\hint{arithmetic -- prepare for the next step}
    \sup{A} \le z + \inf{B}
\op=\hint{'expand' $\;\sup{\cdots}\;$ using definition $\ref{1}$}
    \langle \forall a : a \in A : a \le z + \inf{B} \rangle
\op=\hint{arithmetic -- prepare for the next step}
    \langle \forall a : a \in A : a - z \le \inf{B} \rangle
\op=\hint{'expand' $\;\inf{\cdots}\;$ using definition $\ref{2}$}
    \langle \forall a : a \in A : \langle \forall b : b \in B : a - z \le b \rangle \rangle
\op=\hints{logic: combine quantifications; bring $\;a\;$ and $\;b\;$ together}\hint{-- simplifying, and working towards $\;A-B\;$ in RHS of $\ref{4}$}
    \langle \forall a : a \in A \land b \in B : a - b \le z \rangle
\op=\hint{definition of $A-B$ -- working towards the RHS of $\ref{4}$}
    \langle \forall x : x \in A - B : x \le z \rangle
\op=\hint{introduce $\;\sup{\cdots}\;$ by definition $\ref{1}$}
    \sup{A-B} \le z
\endcalc$$
By principle $\ref{3}$ this completes the proof.
