# Partial order where only some elements are reflexive

Are there interesting examples of "almost" partial orders $$\preccurlyeq$$, where only some elements $$x$$ satisfy the reflexivity axiom $$x \preccurlyeq x$$, but every $$x$$ has at least some $$y$$ with $$x \preccurlyeq y$$? Is there any study of such structures or are they uninteresting?

Such a partial order can be motivated by a philosophical view in which some propositions explain themselves (are "necessary") and some propositions need further explanation (are "contingent") but still can always be explained. For example one could imagine a simple universe in which all states of the world $$\{P_j\}_{j \in \mathbb{Z}}$$ are causally connected but contingent, i.e. $$\cdots \preccurlyeq P_{-1} \preccurlyeq P_0 \preccurlyeq P_1 \preccurlyeq \cdots$$, and there is some absolute reason $$G$$ which explains all of them, i.e. $$P_{j} \preccurlyeq G$$ but $$G \preccurlyeq G$$ as well.

EDIT: To clarify, this would be a transitive, antisymmetric relation (as noted by Izaak van Dongen in the comments) with the additional property, that every element is bounded by some element above. In fact I would be interested in the stronger case of a transitive, antisymmetric relation which admits arbitrary joins. This would translate to the philosophical example above by requiring that any collection of states of the world $$\{P_{\lambda}\}_{\lambda \in \Lambda}$$ can be explained by a single proposition (namely the "sum" of the individual explanations which exist by definition).

• Reflexivity is $\color{red}{\forall x}\quad x \preccurlyeq x$, and is required to hold for $\preccurlyeq$ to be a partial order. Commented Apr 8 at 12:18
• Your example seems to contradict your first sentence: AFAIU, $x:=G$ satisfies $x\preccurlyeq x$ but there is no other $y$ such that $x\preccurlyeq y$? Also, could you please provide inside your post a definition of what you call "almost partial order"? (I guess you mean: a binary relation which is antisymmetric and transitive.) Commented Apr 8 at 12:25
• Then you can delete "but every such $x$ has at least some $y$ with $x \preccurlyeq y$" from your first sentence. Commented Apr 8 at 12:29
• Better now, without the "such" ;-) +1 Commented Apr 8 at 12:30
• If by "almost partial order" you mean "transitive antisymmetric relation", I think all such relations can be obtained by starting with a partial order and "deleting" the reflexivity of some elements. Conversely if you "add back in" the reflexivity to an almost partial order you get a partial order. So it's essentially "a partial order equipped with a distinguished subset" or "a two-coloured partial order". Your condition means something like "no element from the distinguished subset is maximal". I don't know if these structures are studied! Commented Apr 8 at 12:37

As suggested Izaak van Dongen in his comment, it is easy to check that a binary relation $$\preccurlyeq$$ on a set $$X$$ is an almost partial order if the relation $$\le: = \preccurlyeq\cup \{(x,x):x\in X\}$$ is a partial order on a set $$X$$ and the distinguished set $$Y=\{x\in X:x\not\preccurlyeq x\}$$ contains no maximal elements with respect to the order $$\le$$.
In particular, we always can choose $$Y=X$$ and we can choose $$Y=\varnothing$$ iff $$(X,\le)$$ has no maximal elements.
Moreover, it is easy to see that $$(X,\preccurlyeq)$$ admits arbitrary joins iff it admits the join of all its elements iff it has the largest element that is there exists $$G\in X$$ such that $$x\preccurlyeq G$$ for each $$x\in X$$. That is iff $$G$$ is the largest element of $$(X,\le)$$ and $$G\not\in Y$$. This can be theologically interpreted that there exists God, Who is the ultimate cause of all things (including Himself), Who is unique, and not a derived thing. A relevant discussion, can be found, for instance, in Copleston–Russell debate.
• Hi Alex. I am not sure I follow your reasoning in "$(X, \preceq)$ admits arbitrary joins ... iff there exists $G \in X$ such that $x \preceq G$ for each $x \in X$". Particularly I don't understand the "if" direction. Did you just mean "$\implies$"? Admittedly, it's not totally clear what OP means by arbitrary joins. (Also, are $Y = X$ and $Y = \emptyset$ meant to be the other way around?) Commented Apr 20 at 12:17
An ungrounded order $$<$$ is connected if $$E=\{\{u,v\}:u is an edge set of a connected graph on atleast two vertices. While your our object: $$A=(X,\preceq)$$ can always be formed via taking some indexed family of connected orders $$\{<_i\}_{i\in I}$$ and any $$S\subseteq \bigcup_{i\in I}[\bigcup_{u<_iv}\{u,v\}]$$, then defining $$\preceq$$ as any relation on $$A$$ satisfying $$\small a\preceq b\iff (a=b\in S)\lor \exists i\in I:a<_i b$$, thus I would say there is not much interesting here - as these are basically just posets with loops removed and a condition no connected component be a loop.