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?