Convex analysis: relative interior in finite and infinite dimension Let $X$ be a normed space. Given a nonempty convex set $C \subset X$, the relative interior of $C$, denoted by $\text{ri} C$, is the set of the interior points of $C$, considered as a subset of $\text{aff} C$ (this is the smallest affine set which contains $C$). In other words, 
$$
\text{ri} C = \left\{x \in X: \,  \exists \varepsilon > 0 \text{ s.t. } B(x; \varepsilon ) \cap \text{aff} C \subseteq C\right\}.
$$
Question n.1. How can I show that $\text{ri} C \ne \emptyset$? I have no idea on how to start, sincerely. The book gives some hint: for example, up to translations (which are allowed, since $\text{ri}(x+C)=x+\text{ri}C$ as one can easily verify), we can suppose $0 \in \text{aff} C$, i.e. the affine of $C$ is a subspace. Now we can consider a basis of this subspace, $(e_1, \ldots ,e_k)$: now how can I conclude? 
Question n.2. Secondly, my book says that the concept of relative interior is important in finite dimension, since in infinite dimensions it can happen that a convex nonempty set has empty relative interior. Could you please provide an example of this? My book suggest considering a dense hyperplane with no interior points, but this is clearly a mistake (at least, it is a misleading counterexample, I cannot understand it).
Thanks.
 A: *

*Suppose the dimension of aff $C$ is $k-1$, and choose $k$ affinely independent points of $C$. Denote them by $v_1,\ldots,v_k$. Since $C$ is convex, it contains the convex hull conv$(v_1,\ldots,v_k)$, that is, all points of the form $a_1v_1+\cdots a_kv_k$ for $a_1,\ldots,a_k\geq 0$ and $a_1+\cdots+a_k=1$. In particular it contains the point $x=(v_1+\cdots+v_k)/k$. Now, aff $C$ consists of all points of the form $a_1v_1+\cdots+a_kv_k$ for arbitrary $a_1,\ldots,a_k$. We can find a sufficiently small $\epsilon>0$ such that all points in aff $C$ at a distance at most $\epsilon$ from $x$ are contained in conv$(v_1,\ldots,v_k)$. Since conv$(v_1,\ldots,v_k)$ itself is contained in $C$, we are done. We have shown that $x\in$ Ri $C$.


*I think the following works. Take $l^2(\mathbb R)$ to be your infinite dimensional normed space. Write $e_k=(0,0,\ldots,0,1,0,0,\ldots)$ for all $k$, where the $1$ is in the $k$th place, and let $e_0=(0,0,\ldots)$. Let $C$ be the convex hull of all the vectors $e_0,e_1,\ldots$. Then $C$ consists of all vectors of the form $(x_1,\ldots,x_m,0,\ldots)$ where $x_i\geq 0$ and $x_1+\cdots+x_m\le1$, for all $m\geq 0$, whereas aff $C$ consists of all vectors of the form $(x_1,\ldots,x_m,0,\ldots)$ for arbitrary $x_i$. No matter which such vector we pick, it will not be in the relative interior of $C$, because any ball centered at $x$ will contain the point $(x_1,\ldots,x_m,-\delta,0,\ldots)$ for some small $\delta>0$, which is in aff $C$, but it is not in $C$.
