Separation theorem Is it true that given a real vector space $X$ and two disjoint convex sets $A,B\subseteq X$, there is always a linear functional that (weakly) separates them? I.e., is there a non-zero linear functional $\phi\colon X\to \mathbb R$ and a $\gamma\in\mathbb R$, such that $\phi(a)\le \gamma \le \phi (b)$ for all $a\in A$ and $b\in B$? If not, can you give a counterexample?
I know the statement is true if you add the aditional hypothesis that at least one of the sets has an internal point. This follows from the Hahn-Banach theorem using the Minkowski functional, but I was wandering whether this hypothesis is necessary.
 A: The answer is no. here is an elaborated counterexample:
Let $X=c_{00}$ (the space of real valued sequences which are eventually zero). 
Let $A$ tbe he subset of $X$ of all sequences whose last non-zero entry is positive (the zero sequence is not in $A$). 
Let $B= \{0\}$. 
Clearly $A\cap B=\varnothing$ and both $A$ and $B$ are convex, so the hypothesis is satisfied.
Now, suppose we have a separating linear functional $\phi:X\to \mathbb R$. It is clearly zero on $B$. Suppose we had $a\in A$, such that $\phi (a)>0$. Let $N$ be the index of the last non zero entry of $a$. By modifying slightly the $N+1$ entry of $-a$ we will get an element $a'\in A$, such that $\phi (a')<0$ contradicting the separation property. The same holds for if we had $a\in A$ with $\phi (a)<0$. Therefore, $\phi$ must be zero on $A$ and hence on $X$.
A: The answer is no in general. Put $X:=\ell^1(\mathbb N^*)$, 
$$\begin{align} A_0 &:=\{x=\{x_n\}\in X, \forall n\geq 1, x_{2n}=0\},\\ B &:=\{x=\{x_n\}\in X, \forall n\geq 1, x_{2n}=2^{-n}x_{2n-1}\}.\end{align}$$
Let $c\in X$ given by $c_{2n-1}=0$ and $c_{2n}=2^{-n}$. Then $c\notin A_0+B$, and putting $A:=A_0-c$, we note that $A$ and $B$ are two closed disjoint subset which cannot be weakly separated.
A: Consider $X=l_2(\mathbb{N})$, take 
$$
A=c_{00}(\mathbb{N})
$$
$$
B=A+(2^0,2^{-1},\ldots,2^{-n},\ldots)
$$
It is easy to check that $A$, $B$ are disjoint convex dense subsets with empty interior. Since they are dense in $X$, they can not be separated.
A: You can find a counterexample in $\mathbb R^2$. Let $A=\{(x,y): y\leq 0\}$ and let $B$ be the closed convex set above the graph of $e^x$. Since the $d(A,B)=0$ they can not be separated. 
