Separate a special poset by function 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
 A: 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.
