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 $\mathfrak{A}$ \prod\mathfrak{A}$is a separable poset? I think the answer is no'', because I have failed to prove it. But I don't know a counter-example. 1 # A property of a product of posets Let$\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$\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.