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Consider a fixed but arbitrary preordered set $X$. Is there a name for those binary relations $R$ on $X$ satisfying the following? They seem to show up a lot. (Note that every such $R$ is necessarily transitive.)

  1. $R(x,y) \rightarrow x \lesssim y$
  2. $R(x,y) \wedge (y \lesssim z) \rightarrow R(x,z)$
  3. $(x \lesssim y) \wedge R(y,z) \rightarrow R(x,z)$

Examples.

  • The strict $(<)$ and non-strict $(\lesssim)$ order relations on any preordered set $X$.

  • Let $I$ denote an arbitrary non-empty set. Then the set $\mathbb{R}^I$ of functions $I \rightarrow \mathbb{R}$ is partially ordered in the obvious way via the pointwise order: $$f \leq g \iff (\forall i \in I)(f(i) \leq g(i)).$$ Now define a binary relation $R$ on $\mathbb{R}^I$ as follows. $$R(f,g) \iff (\forall i \in I)(f(i) < g(i)).$$ Then $R$ satisfies the conditions of interest. More generally, we can replace $\mathbb{R}$ with any preordered set $X$, and we can replace $<$ with any binary relation on $X$ satisfying 1,2 and 3.

  • (This is the example I'm most interested in.) Let $X$ denote a metric space. Then its powerset $\mathcal{P}(X)$ is partially ordered by inclusion. Define $R$ on $\mathcal{P}(X)$ by asserting that for all $A,B \in \mathcal{P}(X)$, we have $R(A,B)$ iff $A$ is not only contained in $B$, but in fact, it can be "swelled" by some non-zero positive number while remaining contained in $B$. See here for a precise definition; the relation of interest is denoted "$\lhd$" in the link.

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goblin GONE
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  • I don't know a name for these. Looks like the idea is: $R(x,y)$ means that $x\lesssim y$ with "some room to spare". – Greg Martin Nov 30 '14 at 00:14
  • @GregMartin, yep, that's basically the idea. In the original question, I included an irreflexivity axiom, namely $\neg R(x,x).$ I ended up removing it, because the category whose objects are preordered sets equipped such a relation $R$ ends up being better behaved if we omit irreflexivity. But yes, the "room to spare" intuition is definitely where I'm coming from. – goblin GONE Nov 30 '14 at 00:21
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    Another example would be where $x\lesssim y$ is the inclusion relation on subsets of some topological space, and $R(x,y)$ if and only if the closure of $x$ is contained in the interior of $y$. – Greg Martin Nov 30 '14 at 00:23
  • @GregMartin, nice example. Generalizing: given any preordered set $X$ and monotonic functions $\mathrm{cl},\mathrm{int} : X \rightarrow X$ satisfying the axiom $\mathrm{int}(x) \lesssim x \lesssim \mathrm{cl}(x),$ we get such a $R$ relation on $X$ by defining that $R(x,y)$ iff $\mathrm{cl}(x) \lesssim \mathrm{int}(y)$. I wonder if assuming that one or both elements of ${\mathrm{cl},\mathrm{int}}$ are idempotent gives us anything further about $R$. – goblin GONE Nov 30 '14 at 00:30
  • @GregMartin, okay, if $\mathrm{cl}$ is idempotent, then we furthermore have $R(x,y) \rightarrow R(\mathrm{cl}(x),y)$. Similarly with $\mathrm{int}$ on the other side. – goblin GONE Nov 30 '14 at 00:35

1 Answers1

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It would be reasonable to call such things "coreflexive profunctors." Here's why.

The notion of a profunctor comes from category theory. When we specialize to the $\mathrm{Bool}$-enriched case, the following definition is obtained.

Definition 0. Let $P$ and $Q$ denote preordered sets. Then a relation $R : P \nrightarrow Q$ is said to be a ($\mathrm{Bool}$-enriched) profunctor iff the following two axioms hold.

  • $p' \lesssim p, R(p,q) \rightarrow R(p',q)$
  • $R(p,q), q \lesssim q' \rightarrow R(p,q')$

Now we just have to explain what "coreflexive" means:

Definition 1. Let $P$ denote a preordered set, and $R : P \nrightarrow P$ denote a relation. Then:

  • $R$ is reflexive iff $p \lesssim p' \rightarrow R(p,p')$
  • $R$ is coreflexive iff $R(p',p) \rightarrow p \lesssim p'$

Note that if $P$ is just a set, then viewing it as a (discrete) poset, the above notion of reflexivity reduces the usual notion.

goblin GONE
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