Let $A$ be a set equipped with a binary reflexive and symmetric relation $\uparrow$ (such relations are often called tolerances, see also "Are there real-life relations which are symmetric and reflexive but not transitive?"). Now consider the following property for an element $a \in A$: $$ \forall_{b,b' \in A} (b \uparrow a \land a \uparrow b' \rightarrow b \uparrow b') \quad (\star)\ . $$ So $a$ plays the role of a "transition go-between" for the otherwise not necessarily transitive relation $\uparrow$.

My questions: Is the property $(\star)$ known and studied? Is there a name for it? Are there natural (preferably mathematical) examples?

I personally came across this in a quite involved context of a domain-theoretic nature (consistency between pieces of data can be seen as a tolerance relation in Scott information systems), but this would take time and details to explain. A somewhat simpler example would be the following from geometry, due to a friend of mine.

Let $A$ be the area on the (euclidean) plane defined by two rays with shared origin $O$ (a "planar cone"?... not sure how I should call this). Define $X \uparrow_r Y$ by $$ |X-Y| \leq r \ , $$ for a fixed radius $r$; the relation $\uparrow_r$ is reflexive and symmetric, but not transitive. Provided that the angle between the two rays is at most $\pi/3$, an example of a point which satisfies $(\star)$ is the origin $O$: indeed, assume that for $X,Y \in A$ it is $X \uparrow_r O$ and $O \uparrow_r Y$, that is, $|O-X| \leq r$ and $|Y-O|\leq r$; the worst case, with angle $\pi/3$, would be that $X$ lies on one ray, $Y$ on the other, and $|O-X|=|O-Y|=r$, so then $O$, $X$, and $Y$ would form an equilateral triangle, and this would mean that $X \uparrow_r Y$ as well.

  • $\begingroup$ I thought there might be people who exclusively visit MO, and I've just asked about this there as well. I hope this doesn't cause any problems. $\endgroup$ – Basil Jun 12 '14 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.