Difference of Closed Convex sets in Banach Space Let $A$ be a closed, convex, set in a Banach space $X$, and let $B$ be a closed, bounded, convex set in $X$. Assume that $A \cap B = \emptyset$. Set $C = A- B$.  Prove that $C$ is closed, and convex.
So proving $C$ is convex is not too hard, however I am having issues proving it is closed. According to existing convex geometry literature, this is a well-known result, that requires the assumption of $B$ being bounded. However, I am not sure how to use it. I assume the proof would begin by picking $c_n \in C$ such that $c_n \to c$, $c \in X$, and showing that in fact that $c \in C$. We would write:
$$
c_n = a_n - b_n
$$
and somehow show that $c = a -b$ for some $a,b \in C$, presumably $a = \lim_{n \to \infty} a_n$, $b = \lim_{n \to \infty} b_n$. But the existence of the limit for $c$ says nothing about the limits of $a_n,b_n$ (it can be deduced that $a_n$ is norm bounded using the norm bound on $b_n$), but I am truly at a loss on how to proceed. A hint would be appreciated.
 A: If $X$ is reflexive, then the theorem would hold. We find that $b_n$ must have a weak point of accumulation $b$ by the Banach-Alaoglu theorem and the Eberlein-Šmulian theorem. Then $b\in B$ and $c+b\in A$ by A convex subset of a Banach space is closed if and only if it is weakly closed and $c=c+b-b$.
That $B$ is bounded is really necessary for this theorem. Consider the following example.
Let $X=\mathbb{R}^2$ and $A=\{(x,y):x\geq0,xy\geq1\}$ and $B=\{(x,y):x\geq0,xy\leq-1\}$. Then $A$ and $B$ are disjoint, closed, convex and bounded. However $A-B=\{(x,y):y>0\}$ is not closed.
That $X$ is reflexive is also necessary for this theorem. Consider the following example.
Let $X=\ell^1$ and $A=\{x\in X:x_n\geq0,\sum x_n=1\}$. Define $T\in B(X)$ by $(Tx)_n=(1+\frac1n)x_n$ and let $B=T(A)$. Then $A$ and $B$ are disjoint, closed, convex and bounded. So $0\not\in A-B$. However $e_n-T(e_n)\to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is also necessary. Consider the following example.
Let $X=\ell^2$ and let $A$ be the closed unit ball and let $B=\{(1+\frac1n)e_n\}$. Then $A$ and $B$ are disjoint, closed and bounded, and $A$ is convex. So $0\not\in A-B$. However $e_n-(1+\frac1n)e_n\to0$, so $A-B$ is not closed.
That $A$ and $B$ are convex is, however, not necessary in the finite dimensional case. We then find that $b_n$ must have a point of accumulation $b$ by the Bolzano-Weierstrass theorem. Then $b\in B$ and $c+b\in A$ and $c=c+b-b$.
