Let $\mathfrak{A}$ is a poset. For $a, b \in \mathfrak{A}$ we will denote $a
\curlyvee b$ if only if there is a non-least element $c$ such that $c
\leqslant a \wedge c \leqslant b$.


I call a poset $\mathfrak{A}$ *separable* if and only if $\forall x
\in \mathfrak{A}: \left( x \curlyvee a \Leftrightarrow x \curlyvee b \right)
\Rightarrow a = b$.


Let $\mathfrak{A}$ is a family of posets indexed by some set $n$. We introduce
a partial order (called *product order*) on $\prod \mathfrak{A} = \prod\_{i \in n} \mathfrak{A}\_i$ by
the formula (for every $a, b \in \prod \mathfrak{A}$)
$$ a \leqslant b \Leftrightarrow \forall i \in n : a\_i \leqslant b\_i . $$
It is easy to prove that
$$ a \curlyvee b \Leftrightarrow \exists i \in
 \operatorname{dom}\mathfrak{A}: a\_i \curlyvee b\_i . $$
Let $\mathfrak{A}$ is an indexed family of separable posets. Can we infer that
$\prod\mathfrak{A}$ is a separable poset?


I think the answer is ``no'', because I have failed to prove it. But I don't
know a counter-example.



A property of a product of posets
Here's a counterexample, taking advantage of the option to have or not have a 0 in a poset. Take the index set to be $\{1,2\}$. Take $\mathfrak A\_1$ to be an atomless Boolean algebra minus its 0 element. It is well-known and easy to see that this is separable; if $a\neq b$, then one of $a-b$ and $b-a$ in the Boolean algebra is non-0 and serves as the $x$ required in separability. Let $\mathfrak A\_2$ be the poset consisting of three elements, namely a least element 0 and two larger, incomparable elements $p$ and $q$. It is easy to check that this is separable. I claim that the product of these two separable posets is not separable. Let 1 be the top element of $\mathfrak A\_1$. Then $(1,p)$ and $(1,q)$ are distinct elements of the product, so separability would require the existence of a pair $x=(x\_1,x\_2)$ such that $x\curlyvee(1,p)$ but $x\not\curlyvee(1,q)$ or vice versa; by symmetry, I can ignore the "vice versa" possibility. Since $x\curlyvee(1,p)$, there must be $y=(y\_1,y\_2)$ such that $y\_1\leq x\_1$, $y\_2\leq x\_2$, and $y\_2\leq p$ (the remaining conditions, that $y\_1\leq 1$ and $y$ is not least in the product, are automatic, the first because 1 is the top of $\mathfrak A\_1$ and the second because $\mathfrak A\_1$ has no least element). Then $y'=(y\_1,0)$ satisfies $y'\leq y\leq x$ and $y'\leq(x\_1,0)\leq(1,q)$, and $y$ is not least in the product (again because $y\_1$ is not least in $\mathfrak A\_1$). This contradicts the fact that $x\not\curlyvee(1,q)$.


EDIT: In response to a comment from Porton, suggesting that I misinterpreted a quantifier, I admit that this is a possibility, but I think the result I claimed, that a product of separable posets with least elements is separable, holds under either interpretation. First, the interpretation I had in mind (because it seems the more reasonable concept) is that separability means that $a=b$ follows from $(\forall x)(x\curlyvee a\iff x\curlyvee b)$. As in the question, I'll write $n$ for the index set, $\mathfrak A\_i$ for the factor with index $i$, and $\prod\mathfrak A$ for the product. I'll also write 0 for the least element in any poset. To prove separability, I'll assume $a,b\in\prod\mathfrak A$ with $a\neq b$, and I'll find an $x\in\mathfrak A$ such that either $x\curlyvee a$ and $x\not\curlyvee b$ or vice versa. Since $a\neq b$, there is an index $i$ with $a\_i\neq b\_i$; fix such an $i$. Since $\mathfrak A\_i$ is separable, choose some $y\in\mathfrak A\_i$ such that either $y\curlyvee a\_i$ and $y\not\curlyvee b\_i$ or vice versa; without loss of generality, assume the former, and fix a non-0 $z\leq a\_i$ with $z\leq y$. Define $x\in\prod\mathfrak A$ by setting $x\_i=z$ and $x\_j=0$ for all $j\neq i$. Then $x\neq0$ (because $x\_i=z\neq0$) and $x\leq a$, so $x\curlyvee a$. Yet $x\not\curlyvee b$, because any $w\in\prod\mathfrak A$ that is below both $x$ and $b$ would have $w\_i$ below both $x\_i=z$ and $b\_i$, which is absurd as $z\leq y$ and $y\not\curlyvee b\_i$.


The alternative reading of the definition of separability is that the colon after the quantifier makes the scope of the quantifier the whole implication rather than the antecedent. Under this interpretation, the implication is true for whatever value of $x$ one chooses. I choose the value 0 (since we're assuming the posets have least elements 0) and infer the implication $(0\curlyvee a\iff0\curlyvee b)\implies a=b$. But by definition of $\curlyvee$, we never have $0\curlyvee$ anything, so it is true that $(0\curlyvee a\iff0\curlyvee b)$ because both sides of the biconditional are false. So we infer from our implication that $a=b$. Thus, under this reading of the definition, the only way for a poset with a least element to be separable is to have only one element. And that's preserved by products.


