proof of Hahn-Banach,second geometric form

In proof of Hahn-Banach,second geometric form: Let $A$ and $B$ be nonempty dis-joint convex subsets of a normed vector space E. If $A$ is closed and $B$ is compact,then there exists a hyperplane $H$ that strictly separates $A$ and $B$. We set $C=A-B$ then $C$ is closed but i don't know why. I think hypothesis "$A$ closed and $B$ compact" used there but i can't reach.

Next,another proof is we set $A_\epsilon=A+B(0,\epsilon)$,$B_\epsilon=B+B(0,\epsilon)$ then they both convex open but i can't prove. $A$ is closed and $B(0,\epsilon)$ is open so why $A_\epsilon$ is open?

• Disregard my first comment, I didn't see the second part of your question. – Aweygan Apr 29 '17 at 2:53

Let $V$ is a normed space (or more generally a topological vector space), $A\subset V$ be convex, and let $B\subset V$ be open and convex. Then $A+B$ is open and convex.
First observe that $A+B=\bigcup_{a\in A}(a+B)$ is a union of open sets, hence is open. To show $A+B$ is convex, let $a_1,a_2\in A$, $b_1,b_2\in B$, and $\lambda\in[0,1]$ be given. Then we have
$$\lambda(a_1+b_1)+(1-\lambda)(a_2+b_2)=(\lambda a_1+(1-\lambda)a_2)+(\lambda b_1+(1-\lambda)b_2)\in A+B,$$ and therefore $A+B$ is convex.