Defining principal elements of every poset. Is this a new idea? Fix an arbitrary complete lattice $\mathfrak{A}$ with order $\sqsubseteq$. I call elements $a,b\in\mathfrak{A}$ intersecting and denote $a\not\asymp b$ iff there is a non-least element $c\in\mathfrak{A}$ such that $c\sqsubseteq a \wedge c\sqsubseteq b$.
I call full star of element $a$ the set $\star a=\{c\in\mathfrak{A} \,|\, c\not\asymp a\}$.
I call a complete free star such a subset $S\subseteq\mathfrak{A}$, not containing the least element, that $\bigsqcup T \in S \Leftrightarrow T \cap S \neq \emptyset$ for every set $T\subseteq\mathfrak{A}$ (by $\bigsqcup T$ I denote the join (=supremum) of the set $T$.)
Now we can define principal elements of the complete lattice $\mathfrak{A}$:
An element $a\in\mathfrak{A}$ is principal iff its full star $\star a$ is a complete free star.
Question: Has anyone researched principal elements of complete lattices before me today?
Note: I call such elements principal because for the lattice $\mathfrak{F}$ of filters on a set (ordered reverse to set-theoretic inclusion of the filters), this coincides with the customary definition of principal filters.
Moreover, my idea can be generalized from complete lattices to arbitrary posets just replacing the formula $\bigsqcup T \in S \Leftrightarrow T \cap S \neq \emptyset$ with more general formula
$$\forall Z \in \mathfrak{A}: ( \forall X \in T : Z \sqsupseteq X \Rightarrow Z
\in S) \Leftrightarrow T \cap S \neq \emptyset.$$
Has anyone researched this case (for arbitrary posets, not just complete lattices) before me?

Hm, maybe the set of principal elements of a (distributive) lattice is the same as the center of the lattice?
 A: To avoid confusion, you should provide a definition of a filter. I'm using the more general definition of an order filter that is sometimes called an upset.
But your suggestion looks not very promising to me as your can see below.
Here, a filter $F⊆\mathfrak A$ is always an order filter: 
$$x∈F⇒(∀y∈\mathfrak A:x≤y⇒y∈F)$$
In any lattice the equivalence $c\sqsubseteq c\wedge a ⇔ c \sqsubseteq a$ holds. 
Thus, $a≭b$ iff there exists a principal filter that contains $a$ and $b$ that is not generated by a minimal element. This is some kind of orthogonality relation. It reminds me to the orthogonality of $\ell$-groups and to residuated lattices. 
$*a$ is the union of all principal filters of the atoms below $a$ (in case of their existence) or the infinitely descending chains just above the minimal elements.
A complete free star $S$ is an order filter as for $a,b∈S$ the relations $\{a,b\}⊆\mathfrak A$ and $a\vee b∈S$ hold (for arbitrary orders consider $a≤b$ and $a∈S$). Furthermore if $a\vee b∈S$ then $a∈S$ or $b∈S$ holds. This is a prime filter or – in the dual order – a prime ideal. Furthermore, it is related to irreducible elements. In a finite lattice the set $S$ is the union of principal filters of $\vee$-irreducible elements. So it is something like an orthogonality relation, again.
So in a lattice every element except $0$ is principal. In an arbitrary ordered set an element $x$ is principal if the filter $↑(↓x\setminus \min\mathfrak A)$ generated by its principal ideal $↓x$ without the minimal elements is something like a prime filter i.e. if each upper bound of any nontrivial subset of its complement $\mathfrak A\setminus \bigl(↑(↓x\setminus \min\mathfrak A)\bigr)$ is in that complement. For this consideration the don't play an important role. It's much easier if you keep them in the filters.
BTW.: Have you ever heard of Formal Concept Analysis? Their view of lattice theory might be helpful for you.
