I'm trying to understand the necessity for the assumption in the Hahn-Banach theorem for one of the convex sets to have an interior point. The other way I've seen the theorem stated, one set is closed and the other one compact. My goal is to find a counter example when these hypotheses are not satisfied but the sets are still convex and disjoint. So here is my question:


**Question:** I would like a counter example to the Hahn-Banach separation theorem for convex sets when the two convex sets are *disjoint* but neither has an interior point. It is trivial to find a counter example for the *strict* separation but this is not what I want. I would like an example (in finite or infinite dimensions) such that we *fail* to have any separation of the two convex sets at all. 


In other words, we have $K\_1$ and $K\_2$ with $K\_1 \cap K\_2 = \emptyset$ with both $K\_1$ and $K\_2$ convex belonging to some normed linear space $X$. I would like an explicit example where there is **no** linear functional $l \in X^\*$ such that $\sup\_{x \in K\_1} l(x) \leq \inf\_{z \in K\_2} l(z)$.


I'm quite sure that a counter example cannot arise in finite dimensions since I think you can get rid of these hypotheses in $\mathbb{R}^n$. I'm not positive though.



A counter example to Hahn-Banach separation theorem of convex sets.
Here is a simple example of a linear space and 2 disjoint convex sets such that there is no linear functional separating the sets. Note that the notions of convexity and linear functional *do not* require any norm or whatever else. You can introduce them, if you want, but they are completely external to the problem.


The usual trick with taking the difference of the sets shows that it is enough to assume that one set is a point, say, the origin. Now we want to design a convex set $K$ not containing the origin such that the only linear functional $\ell$ that is non-negative on this set is $0$. To this end, take the space $X$ to be the space of all real sequences with finitely many non-zero terms and let $K$ be the set of all such sequences whose last non-zero element is positive. Now, if $x\in X$, choose $y$ to be any sequence whose last non-zero element is $1$ and lies beyond the last non-zero element in $x$. Then, for every $\delta>0$, both $x+\delta y$ and $-x+\delta y$ are in $K$, so $\pm \ell(x)+\delta \ell(y)\ge 0$ with any $\delta>0$ whence $\ell(x)=0$. Thus $\ell$ vanishes identically. 


