# Prove that if $A, B$ are convex , $B$ is closed, $C$ is bounded and $A+C \subset B+C$ then $A\subset B$

Given the sets $A,B,C \in \mathbb{R}^n$ such that: $$A+C \subset B+C$$ Show that if $A,B$ are convex, $B$ is closed and $C$ is bounded then $A\subset B$.

I kind of understand the geometrical interpretation of the problem but I fail to see where to start, too many properties for me to handle :(

• I don't understand what you mean by that. – Julio Cáceres Apr 14 '13 at 23:54
• Sorry, Julio, my message was a response to somebody else who then deleted their comment without telling me. – Sharkos Apr 14 '13 at 23:55
• I've attempted to solve the problem for the case where $C$ is a ball with radius $r$ and centered in the origin, making $A+C$ and $B+C$ inflated versions of $A$ and $B$. Then I tried to form a relationship between a line in $A$ and a line in $B$ but that wasn't possible, I still need to use the closure of $B$. – Julio Cáceres Apr 15 '13 at 0:00

It seems that the following plan will work, and not only for the case of $\mathbb{R}^n$, but for a general case of a locally convex space $X$, satisfying the Hahn-Banach Theorem.
The convexity of $A$ is not needed, because the question condition implies $a+C\subset B+C$ for each point $a\in A$. So it suffices to consider only a case when $A$ is one-point set. Without loss of generality, we may suppose that $A={0}$. Then we should prove that $C\subset B+C$ implies $0\in B$. Suppose the opposite. Since $B$ is closed, there exist a real number $\varepsilon>0$ and a linear functional $f$ on $X$ such that $f(0)=0$ and $f|B>\varepsilon$. Since $C$ is bounded, the value $c=\inf_{x\in C} f(x)<\infty$. Choose a point $x\in C$ such that $f(x)<c+\varepsilon/2$. Since $f(y+z)=f(y)+f(z)>\varepsilon+c$ for each $y\in B$ and $z\in C$, we see that $x\not\in B+C$, a contradiction.
• What guarantees the existence of the linear functional $f$. The only tools that I was given in class to demonstrate this were Caratheodorys Theorem, definition and some properties of closed set, convex hull, bound sets and convex sets. (I feel my teacher is cheating) – Julio Cáceres Apr 15 '13 at 3:51
• Such a linear functional exists by Hahn-Banach separation Theorem. Geometrically it should mean that there is a hyperplane strictly separating the zero from the set $B$. – Alex Ravsky Apr 15 '13 at 4:01
• Of course, there is an idea of the solution based on the spatial intuition and without an application of the Hahn-Banach theorem. We need to find a hyperplane $L$ strictly separating the set $B$ from the zero. Then the canonical hyperplane equation for $L$ should give to us the needed linear functional. Let $b$ be a point in $B$, which is closest to the zero. Such a point should exist, because $B$ is closed and should be unique, since $B$ is convex. As $L$ we may take a hyperplane passing through the midpoint of the segment $[0,b]$ and orthogonal to this segment. – Alex Ravsky Apr 15 '13 at 5:02