name for a subset of a binary relational structure which is "closed downward"?

Question

Let $X$ be a set and suppose that $R$ is a binary relation on $X$. Suppose further that $S\subseteq X$ has the property that whenever $y\in S$ and $xRy$ ($x\in X$), then also $x\in S$. Does such an $S$ have a name? Of course, if $R$ happens to be a partial order (or even preorder), then $S$ is often called either a "down set", "lower set", or sometimes an "ideal" (usually one requires that the preorder be directed in the definition of ideal). But I am not aware of any terminology in the case that $R$ is assumed only to be a binary relation. If anyone can help, I would be very grateful. Thank you.

Answer 1

I don't see any reason not to use the same terminology as one uses in the case of a partial order (or a pre-order), since a set $S$ is closed in your sense with respect a relation if and only if it is downward-closed with respect to the reflexive transitive closure of the relation, which is a partial pre-order.

Such sets $S$ are variously called downward closed, or down sets, or lower sets, or closed under predecessors. Sets like this are also precisely the open sets in the lower-cone topology, whose basic open sets consist of the predecessors of a single node (for the transitive closure of the relation).

Answer 2

Since $R$ could be a $\le$, or $\ge$, or all kinds of other possibilities, I kind of agree that "lower set" is not ideal (excuse the pun). "Leftward closed set" seems somewhat more reasonable, since no matter what $R$ is, if you write $xRy$ then you are writing $x$ on the left.

Answer 3

For the property $x \in S$ and $x \mathrel{R} y$ imply $y \in S$, I would say that $S$ is saturated by $R$. Thus, in your case, saturated by $R^{-1}$ might be an option.