Suppose that $A$ and $B$ are two nonempty convex closed sets in $\mathbb{R}^n$, with $A \cap B = \emptyset$. Further, define $A - B = \{a - b \space | \space a\in A, b \in B\}$. Prove that $A - B$ is closed if $A$ or $B$ is bounded (or both).
I've started by noting that the image of $A \times B$ under the linear function $f(a,b) = a - b$ is just $A - B$, but I'm not sure if this is a good start.
Of course we can assume $B$ to be bounded. Let $x_n=a_n-b_n$ be a converging sequence in $A-B$.
$(b_n)$ is bounded, hence there is a converging subsequence $b_{n_k} \to b$ and we have $b \in B$, since $B$ is closed.
We deduce $a_{n_k} = x_{n_k}+b_{n_k}$ is also convergent as the sum of two convergent sequences. Since $A$ is closed, the limit $a$ is in $A$.
Finally we have
$$\lim_{n \to \infty} x_n = \lim_{k \to \infty} x_{n_k} = \lim_{k \to \infty} a_{n_k} - \lim_{k \to \infty} b_{n_k}=a-b \in A-B$$
and we are done. Note that for the first equality, we have used that the sequence $(x_n)$ is a priori convergent, thus the limit can be computed at any subsequence.
It is also conventient to give a counterexample, if neither $A$ nor $B$ are bounded:
Take $A = \{(x,y) | xy=0\} \subset \mathbb R^2$ to be the coordinate axes and $B = \{(x,y) | xy=1\} \subset \mathbb R^2$ be the hyperbola. We have $0 \notin A-B$, but points of $A$ and $B$ get arbitrary close to each other, hence $0$ is a limit point of $A-B$.
