A subset of an infinite-dimensional simplex where no element is in the convex hull of other elements of the subset.

Question

Let

$$S=\{x\in R_{+}^\infty \mid \sum_{n=1}^\infty x_n=1\}$$

be an infinite simplex.

Suppose $\hat{S}$ is a countable subset of $S$ such that

1. for each $x\in \hat{S}$ only at most finitely many coordinates are non-zero, denote the set of those coordinates $N(x)$;
2. for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\neq N(x’)$.

How could one prove that no $x$ is in a convex hull of other elements of $\hat{S}$, i.e., no $x\in \hat{S}$ can be represented as a convex combination of other elements of $\hat{S}$?

EDIT: Condition (2) should be read: "for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\triangle (\cup_{x'\in \hat{S}}N(x’)) \neq \emptyset$". Here $\triangle$ stands for a symmetric difference of sets.

Thank you! (These two infinities intimidate me...)

Answer 1

No countable set is convex [unless it is empty] (since in a convex set, an interval between two points is contained in the set, and an interval has uncountably many points. However, your statement after the i.e. has nothing to do with convexity.