I got doubt with the proof of this theorem.
Let $X$ be a tvs, $A,B \subset X$ with $A$ an open convex set and $B$ convex such that $A \cap B = \emptyset$. Then there exists $f \in X^*$ (where $X^*$ is the topological dual of $X$) that satisfies $$ f(a)< \inf_{b \in B} f(b) $$
proof:
We already know that, given a tvs $X$, a convex set $A\subset X$ with $A$ open, and a point $x_0$ such that $ x_0 \not \in A$, then there exists $f \in X^*$ with $$ f(a)<f(x_0) \quad \forall a \in A $$ With this, in order to prove the above theorem, we can take $C\dot{=} A-B$, which is convex and open, and separate $C$ from $0$, i.e., exists $f \in X^*$ such that $$ f(a-b)< f(0)=0 \quad \forall a,b \in A,B $$ and, since $A$ is open we conclude that the last inequality implies $$ f(a)<\inf_{b \in B} f(b) $$ My doubt is: Can we, with the same hypothesis (i.e. $A$ open and convex, $B$ just convex and disjoint from $A$) of the above theorem, change the roles of $A$ and $B$ in order to obtain: $$ f(b)< \inf_{a \in A} f(a) \quad \forall b \in B $$ Any hint would be welcome. Thanks in advance.