Let $P$ be a partial order. The up-set $Up(x)$ generated by an element $x \in P$, is the set of all elements in $P$ greater than or equal to $x$. Down-sets $Down(x)$ are defined analogously. Let $x$ and $y$ be distinct elements of $P$. $x$ is said to be retractable to $y$ iff both $(Up(x) - \{x\}) \subseteq Up(y)$ and $(Down(x) - \{x\}) \subseteq Down(y)$. If $x$ and $y$ are comparable and $x$ is retractable to $y$, then $x$ is said to be irreducible to $y$. My two questions are:
- In the definition of retractable, is it the case that in fact at least one of the subset relations are equality relations, and both are equality relations when and only when $x$ and $y$ are incomparable?
- If $x$ and $y$ are irreducible, is it the case that one of them is an immediate predecessor of the other?
No, neither needs to be an equality: Let $P = \{ (x,x), (y,y), (z,y), (y,w), (z,w) \}$ where all variables are distinct. In the general situation, equality implies the elements are incomparable, but incomparability need not imply equality; simply consider $P = \{ (x,x), (y,y) \}$.
Yes. If $x \le y$, then $\text{Up}(x) - \{x \} \subseteq \text{Up}(y)$ says that $z \ge x, z \ne x$ implies $z \ge y$, i.e. $y$ is the successor to $x$. On the other hand, if $y \le x$, then you can conclude $x$ is the successor to $y$ using $\text{Down}(x) - \{x \} \subseteq \text{Down}(y)$.