[proofplan] We verify the three defining properties of a convex body: non-emptiness, convexity, and compactness. Non-emptiness is inherited by adding one point of $K$ to one point of $L$. Convexity follows by decomposing two points of $K+L$ into summands from $K$ and $L$, then applying convexity in each summand separately. Compactness is proved by taking an arbitrary sequence in $K+L$, extracting convergent subsequences from the compact sets $K$ and $L$, and passing the limit through addition in $\mathbb{R}^n$. [/proofplan] [step:Choose one point from each summand to prove non-emptiness] Since $K$ and $L$ are non-empty, choose $x_0 \in K$ and $y_0 \in L$. By the definition of the Minkowski sum, $x_0 + y_0 \in K + L$. Hence $K + L \neq \varnothing$. [/step] [step:Take convex combinations componentwise in $K$ and $L$] Let $z_1,z_2 \in K+L$ and let $\lambda \in [0,1]$. By definition of $K+L$, there exist $x_1,x_2 \in K$ and $y_1,y_2 \in L$ such that \begin{align*} z_1 &= x_1 + y_1, & z_2 &= x_2 + y_2. \end{align*} Then \begin{align*} \lambda z_1 + (1-\lambda)z_2 &= \lambda(x_1+y_1) + (1-\lambda)(x_2+y_2) \\ &= \bigl(\lambda x_1 + (1-\lambda)x_2\bigr) + \bigl(\lambda y_1 + (1-\lambda)y_2\bigr). \end{align*} Since $K$ is convex and $x_1,x_2 \in K$, we have \begin{align*} \lambda x_1 + (1-\lambda)x_2 \in K. \end{align*} Since $L$ is convex and $y_1,y_2 \in L$, we also have \begin{align*} \lambda y_1 + (1-\lambda)y_2 \in L. \end{align*} Therefore $\lambda z_1 + (1-\lambda)z_2 \in K+L$. Since $z_1,z_2 \in K+L$ and $\lambda \in [0,1]$ were arbitrary, $K+L$ is convex. [/step] [step:Extract convergent subsequences from the two compact summands] Let $(z_m)_{m=1}^{\infty}$ be a sequence in $K+L$. For each $m \in \mathbb{N}$, choose $x_m \in K$ and $y_m \in L$ such that \begin{align*} z_m = x_m + y_m. \end{align*} The sequence $(x_m)_{m=1}^{\infty}$ lies in the compact set $K$, so there exist a strictly increasing map $j:\mathbb{N}\to\mathbb{N}$ and a point $x \in K$ such that \begin{align*} x_{j(m)} \to x \end{align*} in $\mathbb{R}^n$ as $m \to \infty$. The sequence $(y_{j(m)})_{m=1}^{\infty}$ lies in the compact set $L$, so there exist a strictly increasing map $k:\mathbb{N}\to\mathbb{N}$ and a point $y \in L$ such that \begin{align*} y_{j(k(m))} \to y \end{align*} in $\mathbb{R}^n$ as $m \to \infty$. Since $(x_{j(k(m))})_{m=1}^{\infty}$ is a subsequence of $(x_{j(m)})_{m=1}^{\infty}$, we also have \begin{align*} x_{j(k(m))} \to x. \end{align*} Addition in $\mathbb{R}^n$ is continuous, hence \begin{align*} z_{j(k(m))} = x_{j(k(m))} + y_{j(k(m))} \to x+y. \end{align*} Because $x \in K$ and $y \in L$, the limit satisfies $x+y \in K+L$. Thus every sequence in $K+L$ has a convergent subsequence whose limit lies in $K+L$. By the Euclidean [sequential compactness](/page/Sequential%20Compactness) criterion, $K+L$ is compact. [/step] [step:Combine the defining properties of a convex body] We have shown that $K+L$ is non-empty, convex, and compact in $\mathbb{R}^n$. Therefore $K+L$ is a convex body in $\mathbb{R}^n$. [/step]