Convex set with no interior contained in hyperplane? 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.
 A: Let $T\colon\mathcal{X}\to\mathcal{X}$ be a compact operator with a dense range on an infinite-dimensional separable Banach space $\mathcal{X}$ $.^{*}$ Then $K:=\overline{T(B_1(0))}$, where $B_1(0)$ is the unit ball, is a convex set with an empty interior (otherwise, ${\rm{span} }K=\mathcal{X}$, which contradicts the infinite-dimensionality of $\mathcal{X}$ because $K$ is compact). But via the dense range of $T$, we have
$$ \mathcal{X}=\overline{\rm{span}} K=\overline{\rm{aff}}K \,, $$
and therefore $K$ cannot lie in any closed hyperplane $.^{**}$
$.^*$ Such operators always exist on (infinite-dimensional) separable Banach spaces.
$.^{**}$ $\rm{span}$ and $\rm{aff}$ denote, respectively, the (not necessarily closed) linear span and affine hull; for the latter case, allow complex scalars when $\mathcal{X}$ is over $\mathbb{C}$, otherwise simply assume $\mathcal{X}$ is over $\mathbb{R}$.
A: 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
$$
\left\{\sum_{b \in B} t_b b : t_u = 0\right\}
$$
