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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).

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    $\begingroup$ Reflexivity is $\color{red}{\forall x}\quad x \preccurlyeq x$, and is required to hold for $\preccurlyeq$ to be a partial order. $\endgroup$ Commented Apr 8 at 12:18
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    $\begingroup$ 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.) $\endgroup$ Commented Apr 8 at 12:25
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    $\begingroup$ Then you can delete "but every such $x$ has at least some $y$ with $x \preccurlyeq y$" from your first sentence. $\endgroup$ Commented Apr 8 at 12:29
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    $\begingroup$ Better now, without the "such" ;-) +1 $\endgroup$ Commented Apr 8 at 12:30
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    $\begingroup$ 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! $\endgroup$ Commented Apr 8 at 12:37

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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.

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  • $\begingroup$ 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?) $\endgroup$ Commented Apr 20 at 12:17
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An ungrounded order $<$ is connected if $E=\{\{u,v\}:u<v\}$ 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.

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