Separation of two disjoint convex closed sets

Question

Assume that $A,B\subset \mathbb{R}^n$ are two disjoint closed convex sets. Without using that $A$ and $B$ are closed sets, it follows already, that there is a non zero element $v$ and a real number $c$ such that

$$ \inf_{x\in A} v^Tx \ge c \ge \sup_{y\in B} v^Ty$$

Can one use the fact that $A$ and $B$ are closed to show that there is a $v'$ and $c'$, such that either

$$ v'^Tx \ge c' > v'^Ty \quad \mbox{or} \quad v'^T y \ge c' >v'^T x$$

for all $x\in A$ and $y\in B$? Without the closedness assumption this is clearly not true.

Answer 1

The sharper inequality may not hold for the original $v$ (which is not unqique). However, for the closed case, you can find points $a\in A$, $b\in B$ that minimize $|a-b|$ and can try $v=a-b$.

Answer 2

The above is not true see the countererxample given in Distance between a cone and a disjoint hyperplane: "Take U to be the closed convex hull of points of the form (0,y,0) and (1,y,ey) with y≤0, and take V to be the closed convex hull of points of the form (0,y,−ey) and (1,y,0) with y≤0. The only separating hyperplane is z=0, which has points from both."