Given a semilattice $S$, a subset $E$, and a positive integer $n$, let $E^{[n]}$ be the set of all products of $n$-tuples in $E$. Thus $\bigcup_{n\geq 1} E^{[n]}$ is nothing but the subsemigroup of $S$ generated by $E$, which I'll denote by $\langle E\rangle$.

The following definition arose in some work I am writing up, as a technical condition needed to make a theorem work.

**Definition.** $S$ has "generation depth $\leq n$" if there exists $n$ such that $E^{[n]} = \langle E\rangle$ for *every* subset $E\subseteq S$.

The terminology is my own, because I don't know if there is existing terminology that I should be using instead. So my questions are: has anyone seen this definition before, and do they have a reference where this condition is given an explicit name?

**Some remarks.**
It is clear that for each $n$, I can find a finite semilattice which does not have generation depth $\leq n$ (a free semilattice on at least $n+1$ generators, for instance). On the other hand, easy pigeon-hole arguments show that a semilattice of width $n$ has generation depth $\leq n$, as does a semilattice of height $n$.

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**UPDATE**
Some more context, in case it helps or is suggestive. The condition arises from the following question:

> Given a semilattice $S$ and a weight $\omega$ on $S$, that is to say, a submultiplicative function $\omega: S \to [1,\infty)$, suppose $\psi:S \to {\mathbb C}$ is approximately multiplicative, in the sense that
$$ \sup_{x,y\in S} \omega(x)^{-1}\omega(y)^{-1} |\psi(x)\psi(y)-\psi(xy)| \hbox{  is small.} $$
Does this force $\psi$ to be a small perturbation of a multiplicative function $S\to\{0,1\}$?

It turns out that the answer is YES if $S$ has "generation depth $\leq n$" for some $n$, regardless of the choice of $\omega$ -- roughly speaking, if I know what $\psi$ does on some subset $E$, then the condition allows me to control what $\psi$ does on the filter generated by $E$.  
As a partial converse, I can find a semilattice $S$ and a weight $\omega$ such that the answer is NO (the counter-example is what motivated the definition).