Cancellation law for Minkowski sums Let $(X,\|\cdot\|)$ be a Banach space and $A,B,C\subset X$ closed bounded non-empty convex subsets. Let $+$ denote the Minkowski symbol for addition.
Does the $+$ satisfy: $$A+C\subset B+C\implies A\subset B$$
 A: It does. Suppose $a\in A$ with $a\notin B$. Since $B$ is closed and not empty, there is $b\in B$ with minimal distance from $A$. Project everything onto $a-b$, and denote the projected objects by primes. Since the sets are closed and convex, their projections are closed intervals. $B'$ is all on one side of $a'$, since the line segment from $b$ to a point whose projection is on the other side of $a'$ would contain points closer to $a$ than $b$. Thus $a'+C'\not\subset B'+C'$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$.
[Edit:]
As t.b. pointed out, this doesn't work in general, but as D. Thomine pointed out, it can be fixed using the Hahn–Banach theorem. Again, suppose $a\in A$ with $a\notin B$. Since $\{a\}$ is convex and compact and $B$ is convex and closed, there is a continuous linear map $\lambda:X\to\mathbb R$ strictly separating the two, so there is $s\in\mathbb R$ such that $\lambda(a)\lt s\lt\lambda(b)$ for all $b\in B$. Since $C$ is bounded and $\lambda$ is continuous, $\lambda(C)$ is bounded. Thus $\lambda(a)+\lambda(C)\not\subset\lambda(B)+\lambda(C)$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$.
Note that only the boundedness of $C$, not that of $A$ or $B$ has been used.
