Let $K$ be a convex set in a normed space $X$. Assume that $int(K)=\emptyset$ (norm topology). Must $K$ be contained in some (affine) hyperplane?
It's fairly easy to see that this is true in $ℝ^n$, but i couldn't generalize.



Convex set with no interior contained in hyperplane?
Based on Jack's comment.


The "Hilbert cube" in Hilbert space $l^2$. $$C :=\{(x\_1,x\_2,\dots) : |x\_k| \le 2^{-k}\;\forall k\}$$
$C$ is convex, compact (so it has empty interior) but has dense span (so it not contained in a closed hyperplane).




---


However, $C$ is contained in a (non-closed) hyperplane. (Axiom of Choice required.)  

The linear span of $C$ is not the whole of $l^2$. Indeed, if $(a\_1,a\_2,\dots)$ is in the span of $C$, then
$\limsup\_k 2^k |a\_k| < \infty$, but $(1,1/2,1/3,\cdots) \in l^2$ fails that property.  

Now we define a Hamel basis for $l^2$ as follows: first choose any vector $u$ not in the span of $C$; next add to {u} a Hamel basis of the span of $C$; then extend that to a Hamel basis $B$ of $l^2$. Then we can see that $C$ is contained in the hyperplane
$$
\left\{\sum\_{b \in B} t\_b b : t\_u = 0\right\}
$$


