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

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 $\exists i\in N, \alpha_i > \beta_i$.

$N=\{0,1,2,3,4,5,6\}$
$P=\{u,v,w,y,z\}$
$u=\{0,1,3,5\},\ v=\{0,2,4,6\},\ x=\{0,1,2\},\ y=\{0,3,4\},\ z=\{0,5,6\}$
$\alpha=\frac12(1_u+1_v),\ \beta=\frac13(1_x+1_y+1_z)$