Order polytope

Remarks

Clause 2 says that $f$ is order-preserving. The vertices of $\mathcal{O}(P)$ are the indicator functions of the upper sets of $P$. Its volume is $e(P)/p!$, where $p = \abs{P}$ and $e(P)$ is the number of linear extensions. For an integer $m \ge 1$, the dilate $m \, \mathcal{O}(P)$ has $\Omega(P, m + 1)$ points with integer coordinates, where $\Omega$ is the order polynomial.

Example: for the chain $x < y$, $\mathcal{O}(P)$ is the triangle $0 \le f(x) \le f(y) \le 1$. Check: its vertices $(0, 0)$, $(0, 1)$, $(1, 1)$ are the indicators of the upper sets $\emptyset$, $\set{y}$, $\set{x, y}$, and its area is $1/2 = e(P)/2!$.

Non-example: the point $f(x) = 1$, $f(y) = 0$ satisfies clause 1 but not clause 2, so it is not in $\mathcal{O}(P)$.

For an antichain of $p$ elements, $\mathcal{O}(P)$ is the unit cube $[0, 1]^p$.