Assume $A = \prod\_{i=1}^n\{0,1\}$, i.e. element $(a\_1,\cdots,a\_n)=a\in A$ is n-tuples like $(1,0,1,\cdots)$. 


There is an obvious partial order on the $A$: say $a < b$ for $a,b\in A$ if and only if $\forall i\in \{1,\cdots,n\}$ $a\_i\leq b\_i$. Note this is only a partial order.


We define the order-preserved partition $(A\_1,A\_2)$ of $A$: $A\_1 \cup A\_2 = A$, $A\_1\cap A\_2 = \emptyset$, and for any $a\in A\_1,b\in A\_2$, we have $a\ngeq b$, i.e. there is no element in $A\_1$ bigger than any element in $A\_2$. We also denote the pair as $A\_1 < A\_2$.


We define an increasing function on the partial ordered set $A$:
$f((a\_1,\cdots,a\_n))=\prod\_{i=1}^np\_i^{1-a\_i}$, where $0 < p\_1,p\_2\cdots,p\_n <1$.


The question is that for any fixed order-preserve partition $(A\_1,A\_2)$ of $A$ whether we can find a positive number $\theta >0$ and $f$ of above form such that $A\_1 \subset\{ f<\theta\}$ and and $A\_2 \subset \{f>\theta\}$ or not.


In some sense, I want to know whether order-preserve partition can be obtained geometrically by above certain type function $f$.


I tried $n\leq 3$ case, above question has easy positive answer.


Thanks a lot



Separate a special poset by function
No. There is a counterexample for each $n\geq4$.


I will only show a counterexample for $n=4$; you can get one for any higher
$n$ by appending $0$'s.


Let $A\_{1}=\left\{ \left( x,y,z,w\right) \in\left\{ 0,1\right\}
^{4}\ \mid\ x+y\leq1\text{ and }z+w\leq1\right\} $. Let $A\_{2}=\left\{
0,1\right\} ^{4}\setminus A\_{1}$. Then, $\left( A\_{1},A\_{2}\right) $ is
what you call an order-preserved partition (I have seen this being called an
"admissible partition"). Your conjecture claims that there exist $p\_{1}
,p\_{2},p\_{3},p\_{4}\in\left( 0,1\right) $ and $\theta\in\mathbb{R}\_{+}$ such
that the function $f:\left\{ 0,1\right\} ^{4}\rightarrow\mathbb{R},\ \left(
x,y,z,w\right) \mapsto p\_{1}^{1-x}p\_{2}^{1-y}p\_{3}^{1-z}p\_{4}^{1-w}$
satisfies $A\_{1}\subseteq f^{-1}\left( \left( -\infty,\theta\right)
\right) $ and $A\_{2}\subseteq f^{-1}\left( \left( \theta,+\infty\right)
\right) $.


Assume that this were true. Then, there would exist $q\_{1},q\_{2},q\_{3}
,q\_{4}\in\left( 0,1\right) $ and $\eta\in\mathbb{R}$ such that the function
$g:\left\{ 0,1\right\} ^{4}\rightarrow\mathbb{R},\ \left( x,y,z,w\right)
\mapsto q\_{1}x\_{1}+q\_{2}x\_{2}+q\_{3}x\_{3}+q\_{4}x\_{4}$ satisfies $A\_{1}\subseteq
g^{-1}\left( \left( \eta,\infty\right) \right) $ and $A\_{2}\subseteq
g^{-1}\left( \left( -\infty,\eta\right) \right) $. (Indeed, take
$q\_{i}=\ln p\_{i}$ and $\eta=\ln p\_{1}+\ln p\_{2}+\ln p\_{3}+\ln p\_{4}-\ln\theta
$. Then, the inequalities $f\left( x,y,z,w\right) \lesseqgtr\theta$ become
$g\left( x,y,z,w\right) \gtreqless\eta$ after taking natural logarithms and
slightly rewriting.)


Now, the two points $\left( 1,0,1,0\right) $ and $\left( 0,1,0,1\right) $
both belong to $A\_{1}$ and therefore to $g^{-1}\left( \left( \eta
,\infty\right) \right) $ (since $A\_{1}\subseteq g^{-1}\left( \left(
\eta,\infty\right) \right) $). Since $g^{-1}\left( \left( \eta
,\infty\right) \right) $ is a convex set, the midpoint between these two
points (that is, the point $\left( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}
{2},\dfrac{1}{2}\right) $) must therefore also belong to $g^{-1}\left(
\left( \eta,\infty\right) \right) $.


But the two points $\left( 1,1,0,0\right) $ and $\left( 0,0,1,1\right) $
belong to $A\_{2}$ and therefore to $g^{-1}\left( \left( -\infty,\eta\right)
\right) $ (since $A\_{2}\subseteq g^{-1}\left( \left( -\infty,\eta\right)
\right) $). Since $g^{-1}\left( \left( -\infty,\eta\right) \right) $ is a
convex set, the midpoint between these two points (that is, the point $\left(
\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2}\right) $) must therefore
also belong to $g^{-1}\left( \left( -\infty,\eta\right) \right) $.


So the point $\left( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}
{2}\right) $ must belong to $g^{-1}\left( \left( \eta,\infty\right)
\right) $ and to $g^{-1}\left( \left( -\infty,\eta\right) \right) $ at
the same time. But this is absurd, since $g^{-1}\left( \left( \eta
,\infty\right) \right) \cap g^{-1}\left( \left( -\infty,\eta\right)
\right) =\varnothing$. So we have a contradiction.


I am wondering whether it is possible to fix your conjecture by (essentially)
assuming that the kind of counterexamples above (viz., two points in $A\_{1}$
having the same midpoint as two points in $A\_{2}$) does not exist.


A stronger requirement that definitely makes your conjecture true is the following:


**(1)** no convex combination of the points in $A\_1$ can be a convex combination of the points in $A\_2$ at the same time.


Indeed, if **(1)** is true, then the convex hull of $A\_1$ is disjoint from the convex hull of $A\_2$; but (from basic linear optimization theory) we know that this entails that there exists a hyperplane separating $A\_1$ from $A\_2$; this immediately translates into the existence of $g$ and $\eta$ as above.


