= Solution
The sum is nonempty because $A$ and $B$ are nonempty. It is a <bounded set>: if $|a|\leq M_A$ and $|b|\leq M_B$, then $|a+b|\leq M_A+M_B$.
To prove that it is a <closed set>, let $x_j=a_j+b_j\in A+B$ be a <convergent sequence> with $x_j\to x$. By the <Heine-Borel theorem>, $A$ is a <compact set>, so $(a_j)$ has a <convergent subsequence> $a_{j_r}\to a\in A$. Then
$$
b_{j_r}=x_{j_r}-a_{j_r}\longrightarrow x-a.
$$
Since $B$ is closed, $x-a\in B$, and hence $x=a+(x-a)\in A+B$. Therefore $A+B$ is nonempty, closed and bounded, so $A+B\in\mathcal K$.
Back to article page
