Let $N$ be a finite subset of the naturals. Let $P$ be a set of subsets of $N$ such that:


1) $P\neq \varnothing$,


2) $\forall x\in P, |x| >1$,


3) $\forall x,y\in P,$ if $x\neq y$, then $x\not\subseteq y$, and


4) $\forall x,y\in P,$ $x\cap y \neq \varnothing$.


For each $x\in P$, let $1\_x\in \mathbb R\_+^N$ be such that


$
1\_{xi} = 
\left\{\begin{array}{ll}
 1 & \textrm{if }i\in x\\
0& \textrm{if } i\notin x.
\end{array}\right.
$


In other words, $1\_x$ is the indicator function for the set $x$.


Let $C\equiv conv\{1\_x|x\in P\}$. That is, $C$ is the convex hull of the indicator functions for each set in $P$.


Question: Is it true that $\not \exists \alpha, \beta\in C$ such that $\alpha \gneq \beta$?


By $\alpha \gneq \beta$, I mean that $\forall i\in N, \alpha\_i \geq \beta\_i$ and $\exists i\in N, \alpha\_i > \beta\_i$.



Can convex combinations of indicator functions for pairwise non-disjoint sets unordered by inclusion dominate one another?
**Counterexample.**


$N=\{0,1,2,3,4,5,6\}$


$P=\{u,v,w,y,z\}$


$u=\{0,1,3,5\},\ v=\{0,2,4,6\},\ x=\{0,1,2\},\ y=\{0,3,4\},\ z=\{0,5,6\}$


$\alpha=\frac12(1\_u+1\_v),\ \beta=\frac13(1\_x+1\_y+1\_z)$


