Suppose we have a partial order $(X, \leq)$, and for some element $x\in X$ we find a set $S\subseteq X$ where
- No two elements of $S$ are comparable, so for any $s,t\in S$ we have $s\leq t$ if and only if $s=t$
- All elements of $S$ are strictly less than $x$, so for any $s\in S$ we have $s\leq x$ and $x\nleq s$
- All elements of $X$ which are less than $x$ are less than something in $S$, so for any $y \in X$ where $y\leq x$, either there is some $s\in S$ where $y\leq s$, or $y=x$
Note that for an element $x$, a set $S$ satisfying this need not exist. As a simple example, in the real numbers with the usual ordering, this set does not exist for any element. There is something to be said about infinite ascending chains here. Also, I'm pretty sure that if such a set $S$ exists, it would be unique.
I've asked this before on other platforms and been sent some interesting but ultimately unsatisfactory concepts. One was the idea of the covering relation, $x \lessdot y$ meaning $x<y$ and there is no $z$ where $x<z<y$. And yes, it is true that if for some $x$ this $S$ existed, then all the elements $s\in S$ would have $s\lessdot x$, but if the set $S$ doesn't exist, the set $\{s\in X\mid s\lessdot x\}$ would not satisfy the requirements. Another idea was finding sets whose meet was $x$, but that is also not sufficient, for example we could have two elements whose meet is $x$ and some infinite ascending chain whose elements are all less than $x$ and unrelated to the other two elements.
I'm tempted to just give this idea a name of my own choosing, maybe saying $x$ is depicted from below by $S$, but I'd rather use an existing term if it existed.