Let $E$ be a Banach space. A set $C \subseteq E$ is called *ideally convex* if for every bounded sequence $(x\_n)$ in $C$ and for every sequence $(\lambda\_n)$ in $[0,1]$ that sums up to $1$ the vector $\sum\_n \lambda\_n x\_n$ is also in $C$.


(So a bounded set is ideally convex if and only if it is *$\sigma$-convex* in [this](https://mathoverflow.net/q/56161/102946) sense. But in the following question, $C$ is automatically unbounded.)


**Question.** Let $C \subseteq E$ be ideally convex and dense in $E$. Does it follows that $C = E$?


**Remark.**
Every open convex set is ideally convex (by the Hahn-Banach separation theorem). In this special case it is not particularly difficult to prove that the question has a positive answer: an open, convex, and dense set $C \subseteq E$ is automatically equal to $E$.



Large ideally convex sets
Just take your favorite decreasing sequence $\lambda\_k$ of positive numbers with sum $1$ and inductively construct the vectors $x\_k\in C$ such that $\left \|\sum\_{k=1}^n\lambda\_k x\_k-x\right\|\le\lambda\_{n+1}$ (then automatically $\|x\_n\|\le \frac{\lambda\_{n}+\lambda\_{n+1}}{\lambda\_n}\le 2$ for $n\ge 2$) to get a series converging to $x$. I use that the density of $C$ implies the density of $\lambda C$ for any $\lambda>0$, of course.


