Let $E$ be a finite-dimensional space and let $C \subseteq E$ be a nonempty convex set. Then, the relative interior of $C$, which we denote by $\mbox{ri}(C)$, is nonempty.
If space $E$ is infinite-dimensional, does the result above still hold?
$$ ri(C)= \{x \in \operatorname{aff}(C): B[x,\epsilon] \cap \operatorname{aff}(C) \subseteq C \text{ for some }
\epsilon>0\}$$
Best Answer
If I understood your notation right then a subset $C=\prod_{n=1}^{\infty} [-2^{-n};2^{-n}]$ of the space $\ell_2$ has empty relative interior, because $\operatorname{aff} C$ contains $x+e_n$ for each $x\in C$ and each $n$, where $e_n$ is the standard unit vector of $\ell_2$, but $x+2^{2-n} e_n\not\in C$.