I have the following problem:
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
- for each $x\in \hat{S}$ only at most finitely many coordinates are non-zero, denote the set of those coordinates $N(x)$;
- for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\cap N(x')=\emptyset$.
Claim: No $x\in \hat{S}$ can be represented as a convex combination of other elements of $\hat{S}$.
I find the result rather obvious, but I do not know how to write the proof formally.
The set $\hat{S}$ is linearly independent. To see this assume: $$ v = a_1 x_1 + \ldots + a_n x_n = 0 $$ where $x_1, \ldots, x_n \in \hat{S}$. Let $\pi_i : S \to \Bbb{R}^\infty$ be the projection onto the span of the $e_i$ with $i \in N(x_i)$, where the $e_i$ are the standard basis for $\Bbb{R}^\infty$. Then, by condition 2: $$ \pi_i(x_j) = \left\{ \begin{array}{l@\quadl} 0 & \mbox{if $i \neq j$}\\ x_j & \mbox{if $ i = j$} \end{array}\right. $$ whence, for $i = 1, \ldots, n$, $\pi_i(v) = a_i x_i = 0$ so that $a_i = 0$ (since, by definition of $S$, $x_i \neq 0$). As $\hat{S}$ is linearly independent, no $x\in \hat{S}$ can be written as a convex combination of other elements.