# Directed subposet of a poset containing the minimal elements

The following appears naturally in a certain context:

Let $P$ be a graded partially ordered set. Let $M$ be the subset of minimal elements of $P$. Define subsets $E_i$ inductively as follows: First, let $E_0:=M$. Then, if $|E_i|\leq 1$, set $E_{i+1}=\emptyset$. Otherwise, for each incomparable pair $x\neq y$ in $E_i$, consider the minimal elements $z$ with $x<z>y$ and put them into the set $E_{i+1}$. This defines $E_{i+1}$ out of $E_i$. Finally, set $E=E_0\cup E_1\cup E_2\cup ...$.

Questions: Is there a more conceptual definition of the subposet $E$? Does it have a universal property making somehow clear why it is defined like above? Is it a well-known construction in the theory of posets? Does it have a name?

• Are you allowing the possibility that, for certain $x$ and $y$, the set $\{z:x<z>y\}$ has no minimal elements? You assumed only that the whole $P$ has plenty of minimal elements, not that its subsets do. – Andreas Blass Aug 17 '14 at 14:55
• You don't seem to need to treat the case $|E_i|\leq 1$ separately, since in that case there are no $x\neq y$ in $E_i$ and so $E_{i+1}$ ends up empty anyway. – Joel David Hamkins Aug 17 '14 at 14:58
• A somewhat related construction is as follows: $E'_0=\emptyset$, $E'_{\alpha+1}$ is the set of minimal elements of $P\smallsetminus E'_\alpha$, $E'_\gamma=\bigcup_{\alpha<\gamma}E'_\alpha$ for limit $\gamma$, and $E'=\bigcup_{\alpha\in\mathrm{Ord}}E'_\alpha$. Then $E'$ is the largest well-founded downward closed subset of $P$. – Emil Jeřábek supports Monica Aug 17 '14 at 16:25
• Andreas, yes, there might be no minimal elements in $\{z:x<z>y\}$ and this set even might be empty. However, in my application, there is always at least one minimal element in that set. – Werner Thumann Aug 17 '14 at 16:30
• Werner, if you had said that the background to your question was combinatorial group theory, to which you had applied operads and other category theory, then you would have got much more informative answers. If you reduce these things to order theory you probably lose the entire conceptual content. Please edit your question so that I can remove my down-vote. – Paul Taylor Aug 20 '14 at 11:58

In order to have any hope of getting a universal property you have to think in terms of order-preserving functions (ie categorically) and not picking elements one at a time (graph theoretically).

The subject that would have results like this is called domain theory. It assumes that the poset already has (honest) joins of directed subsets.

The most likely kind of universal property is that $E$ is the image of the least co-closure.

A co-closure is an order-preserving function $f:X\to X$ such that $f(x)=f(f(x))\leq x$ for each $x\in X$.

In the example where you construction arose, can you construct such a function?

If so, that's your universal property. If not, you're at sea.

Edit in response to Werner's: Look up SFP or bifinite domains, on whom the main authors are Gordon Plotkin, Mike Smyth and Achim Jung (though the word bifinite was mine). However, you still haven't told us where the question came from.

• Whoever you are, why have you down-voted me? – Paul Taylor Aug 17 '14 at 14:46
• Paul, suppose that $P$ consists of the nonempty subsets of a fixed infinite set, ordered by $\subseteq$. In this case, the only co-closure $f$ seems to be the identity map, whose image would be the whole of $P$, but $E$ would consist precisely of the finite subsets. – Joel David Hamkins Aug 17 '14 at 16:03
• It seems to me more generally that in any separative partial order, the only co-closure is the identity map, but such orders can have nontrivial $E$. – Joel David Hamkins Aug 17 '14 at 16:12
• I had in mind that $E$ would get closed under directed joins too. But Werner gave us no idea of where his construction came from and his presentation suggested that is was not definitive but open to experimentation. I tried to make what sense of it I could. – Paul Taylor Aug 17 '14 at 16:36
• Thanks Paul for the keywords, I will have to look them up. I used the definition of E in the context of arxiv.org/abs/1407.5171 Definition 6.3. But I have the feeling that there is more behind that definition and that's why I'm asking. – Werner Thumann Aug 20 '14 at 11:37

After editing the above construction a little bit, the following holds: $E$ is the smallest subposet of $P$ which contains $M$ and satisfies the following property: If $x\in P\setminus E$ then there exists a greatest element $y\in E$ with the property $y<x$.