While working on convexity I stumbled upon a Theorem that started with:
Let $V_1, V_2 \subseteq \Bbb R^n$ be convex sets such that $V_1 \cap V_2 = \emptyset$ and $V_1-V_2$ is closed.
Note that in this case $V_1-V_2 := \{x \in \Bbb R^n:x=p-q, p\in V_1, q\in V_2\}$
Now I'm curious, does the notion that $V_1-V_2$ is closed imply that both $V_1$ and $V_2$ are closed? If not, what would be a counter example? Or if so, how could i proof that?
I managed to prove the theorem without solving this question, but I couldn't figure it out and now I can't seem to let it go.
Okay. Unfornately it is wrong. Here is the counter example:
Consider: $A = (-2, -1]$ and $B = [0,\infty)$. Then, $A,B$ are both convex, disjoint, $A$ not closed and $B - A = [1, \infty)$ is closed.
Thx to @gerw for tipping of.