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.

normedlinear space; in that case my example below obviously doesn't work. But my impression has always been that the interior point condition is to rule out the non locally convex case. Can you actually state the version of the separation theorem you are looking at and be even more precise about what you want? (In particular, in the locally convex case, to get any example you will need also $K_1$ and $K_2$ to be non-compact and non closed at the very least.) $\endgroup$