[proofplan] Let $\alpha := \sup A$ and $\beta := \sup B$. First we show that $A+B$ is nonempty and that $\alpha+\beta$ is an upper bound for it. Then we prove that this upper bound is sharp: for every $\varepsilon>0$, choose elements of $A$ and $B$ within $\varepsilon/2$ of their respective suprema and add them. The epsilon characterisation of the supremum then identifies $\alpha+\beta$ as $\sup(A+B)$. [/proofplan] [step:Define the two suprema and verify that the Minkowski sum is nonempty] Define \begin{align*} \alpha := \sup A \end{align*} and \begin{align*} \beta := \sup B. \end{align*} These [real numbers](/page/Real%20Numbers) exist because $A$ and $B$ are nonempty and bounded above. Since $A$ is nonempty, choose $a_0 \in A$. Since $B$ is nonempty, choose $b_0 \in B$. Then $a_0+b_0 \in A+B$ by the definition of $A+B$, so $A+B$ is nonempty. [/step] [step:Show that $\alpha+\beta$ is an upper bound for $A+B$] Let $x \in A+B$. By definition of the Minkowski sum, there exist $a \in A$ and $b \in B$ such that \begin{align*} x = a+b. \end{align*} Since $\alpha=\sup A$, the number $\alpha$ is an upper bound for $A$, so $a \le \alpha$. Since $\beta=\sup B$, the number $\beta$ is an upper bound for $B$, so $b \le \beta$. Adding these two real inequalities gives \begin{align*} x = a+b \le \alpha+\beta. \end{align*} Thus $\alpha+\beta$ is an upper bound for $A+B$. In particular, $A+B$ is bounded above. [/step] [step:Approximate $\alpha+\beta$ from below by elements of $A+B$] Let $\varepsilon>0$. Since $A$ is nonempty and bounded above and $\alpha=\sup A$, the epsilon characterisation of the real supremum [citetheorem:8612] applied with $\varepsilon/2>0$ gives an element $a_\varepsilon \in A$ such that \begin{align*} \alpha-\frac{\varepsilon}{2} < a_\varepsilon \le \alpha. \end{align*} Similarly, since $B$ is nonempty and bounded above and $\beta=\sup B$, the same result applied with $\varepsilon/2>0$ gives an element $b_\varepsilon \in B$ such that \begin{align*} \beta-\frac{\varepsilon}{2} < b_\varepsilon \le \beta. \end{align*} Define \begin{align*} x_\varepsilon := a_\varepsilon+b_\varepsilon. \end{align*} Then $x_\varepsilon \in A+B$. Adding the two strict lower bounds gives \begin{align*} \alpha+\beta-\varepsilon < a_\varepsilon+b_\varepsilon = x_\varepsilon. \end{align*} Therefore, for every $\varepsilon>0$, there exists $x_\varepsilon \in A+B$ such that \begin{align*} \alpha+\beta-\varepsilon < x_\varepsilon \le \alpha+\beta. \end{align*} [guided] The goal is to prove that the upper bound $\alpha+\beta$ found above is not too large. The right tool is the epsilon characterisation of the real supremum: for a nonempty bounded-above set, its supremum can be approached from below by points of the set. Let $\varepsilon>0$ be arbitrary. Because $A$ is nonempty and bounded above, and because $\alpha=\sup A$, the epsilon characterisation of the real supremum [citetheorem:8612] applies to $A$. We use it with the positive number $\varepsilon/2$. It gives an element $a_\varepsilon \in A$ satisfying \begin{align*} \alpha-\frac{\varepsilon}{2} < a_\varepsilon \le \alpha. \end{align*} The same hypotheses hold for $B$: it is nonempty and bounded above, and $\beta=\sup B$. Applying the same theorem to $B$ with $\varepsilon/2$ gives an element $b_\varepsilon \in B$ satisfying \begin{align*} \beta-\frac{\varepsilon}{2} < b_\varepsilon \le \beta. \end{align*} Now add the two chosen elements. Define \begin{align*} x_\varepsilon := a_\varepsilon+b_\varepsilon. \end{align*} Since $a_\varepsilon \in A$ and $b_\varepsilon \in B$, the definition of the Minkowski sum gives $x_\varepsilon \in A+B$. Adding the two strict inequalities is valid by the order compatibility of real addition, and yields \begin{align*} \alpha+\beta-\varepsilon < a_\varepsilon+b_\varepsilon. \end{align*} Since $a_\varepsilon+b_\varepsilon=x_\varepsilon$, this is \begin{align*} \alpha+\beta-\varepsilon < x_\varepsilon. \end{align*} Also, because $a_\varepsilon \le \alpha$ and $b_\varepsilon \le \beta$, adding gives \begin{align*} x_\varepsilon = a_\varepsilon+b_\varepsilon \le \alpha+\beta. \end{align*} Thus for every $\varepsilon>0$, we have produced an element $x_\varepsilon \in A+B$ lying within $\varepsilon$ below $\alpha+\beta$. [/guided] [/step] [step:Apply the epsilon characterisation to identify the supremum] We have shown that $A+B$ is nonempty and bounded above, that $\alpha+\beta$ is an upper bound for $A+B$, and that for every $\varepsilon>0$ there exists $x_\varepsilon \in A+B$ such that \begin{align*} \alpha+\beta-\varepsilon < x_\varepsilon \le \alpha+\beta. \end{align*} By the epsilon characterisation of the real supremum [citetheorem:8612], applied to the set $A+B$ and the real number $\alpha+\beta$, it follows that \begin{align*} \sup(A+B)=\alpha+\beta. \end{align*} Since $\alpha=\sup A$ and $\beta=\sup B$, this is exactly \begin{align*} \sup(A+B)=\sup A+\sup B. \end{align*} [/step]