Showing some set is open in a normed vector space.

Question

Given $A,B$ two convex subsets of a normed vector space $E$, such that $A\cap B = \emptyset$. I’m trying to show that if $A$ is open then the set $A-B = \left\lbrace a-b ,\; a\in A, \; b\in B\right\rbrace$ is also open. I’ve tried to find some way to write this set as a union of open sets but I couldn’t come up with something. I also thought that maybe since $A$ is open, I could find a open ball around $a-b$ given that I have one around $a$ but it doesn’t seem to work, or at least I can’t seem to see how it should work. Any help or hint is appreciated. (For context, I’m trying to show this as part of the proof of the geometric version of Hahn-Banach’s theorem)

Answer 1

Sketch:

I'm assuming that the topology is the induced metric topology on $E$. Let $B(a,r)$ be a ball around $a$ in $A$. Now, consider the ball $B(a-b,r)$, which is a ball around $a-b$. Let $c\in B(a-b,r)$. Let $a'=c+b$, then, $\|a'-a\|=\|c-(a-b)\|<r$. Therefore, $a'\in B(a,r)$ and $c=a'-b\in A-B$.

Answer 2

Note that, fixed $x=a-b\in A-B$, where $a\in A$ and $b\in B$, there exists some $\varepsilon>0$ such that $\operatorname{B}(a,\varepsilon)\subset A$, hence the set $$\operatorname{B}(x,\varepsilon)=\operatorname{B}(a,\varepsilon)-b=\{\alpha-b : \alpha\in \operatorname{B}(a,\varepsilon)\}$$ is contained in $A-B$.