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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?

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  • $\begingroup$ You don't use $b$ in the definition of "intersecting elements". $\endgroup$ – Olivier Bégassat Jun 21 '14 at 16:04
  • $\begingroup$ @OlivierBégassat: Thanks, corrected $\endgroup$ – porton Jun 21 '14 at 16:40
  • $\begingroup$ Isn't it the case that $a$ and $b$ are intersecting iff their meet is different from the bottom element? $\endgroup$ – Olivier Bégassat Jun 21 '14 at 16:46
  • $\begingroup$ @OlivierBégassat: Yes, but only for lattices with a bottom element, not for arbitrary posets. My definition is valid for any posets $\endgroup$ – porton Jun 21 '14 at 17:26
  • $\begingroup$ It's just that your question is phrased in terms of complete lattices, and those have top and bottom elements. $\endgroup$ – Olivier Bégassat Jun 21 '14 at 19:09
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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.

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  • $\begingroup$ "So in a lattice every element except $0$ is principal." It is wrong: non-principal filters are non-principal in this sense. It seems you confuse for-all-filters with for-all-principal-filters or for-all-ultrafilters. I haven't check the details, but I am sure in the lattice of filters on a set, not only $0$ is principal $\endgroup$ – porton Jun 22 '14 at 20:27
  • $\begingroup$ In fact a non-principal filter may be non-principal in a certain lattice. But that's true for every element of any lattice and out of the scope of this question. If you have a counterexample then you should revise your definitions. You don't provide a definition of a filter. There is a common definition of ultrafilters in topology, but not in order theory. So you are probably the one who confuses different notions. Regarding your last sentence, $0$ is not principal by your definition. I came to the conclusion that all other elements are principal. So where is the contradiction? $\endgroup$ – Keinstein Jun 24 '14 at 20:37

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