1
$\begingroup$

I have two relations R and R' satisfying following property -

R(a,b) & R'(a,a') & R(a',b') => R(a,b')

Pictorially, it looks like this -

 a --(R)-> b
 | \
(R') \ (* new edge being added *)
 |     \
 v       \
 a'--(R)-> b'

Is there any standard name for this property? I am calling it "Closure of R w.r.t R'" for time being.

$\endgroup$
3
  • $\begingroup$ It is just me, or does $b$ have no purpose? (Except that $a$ is related to something.) $\endgroup$
    – Jason Rute
    Commented May 25, 2013 at 0:44
  • $\begingroup$ Also this should be in the logic or category theory labels, not set theory. Although, it seems like this is better suited for math.stackexchange.com. You should also provide a bit of motivation/context if you have it. Is there a natural example of this property. It seems fairly contrived to me. $\endgroup$
    – Jason Rute
    Commented May 25, 2013 at 0:51
  • $\begingroup$ Changed tag to logic. R can be interpreted as path relation and R' as edge relation, but with an exception - R'(a,b) does not imply R(a,b). Therefore, R is not a transitive closure of R'. Instead, I start with a finite R and use this property to "grow" R. $\endgroup$
    – Gowtham K
    Commented May 25, 2013 at 1:36

1 Answer 1

1
$\begingroup$

(This is more of a comment than an answer, but too long for the comment box.)

Your property is saying precisely that $[(R\circ R')\upharpoonright\text{dom}(R)]\subseteq R$.

I'm not sure I like your "closure" terminology, since that suggests that you start with a relation, and then close it. But there are in general many relations $R$ that satisfy your property with a given relation $R'$. For example,

  • the empty relation $R$ has your property with respect to any $R'$.
  • similarly, the full relation $R$ also has this property.
  • Also, if $R$ is the transitive closure of $R'$, then this property holds.

So you do not seem to be starting with something and then taking a "closure", but rather asserting that the given relation $R$ is itself already closed under this kind of application with $R'$.

$\endgroup$
3
  • $\begingroup$ Is $\textrm{dom}(R) = \{a : \exists b\ R(a,b)\}$? $\endgroup$
    – Jason Rute
    Commented May 25, 2013 at 5:28
  • $\begingroup$ The intriguing aspect of this is that he requires R(a,b) before "augmenting" R; just having R(a',b) is not enough. This makes it different (and more restrictive) than many constructs that augment relations similarly. It is barely possible that Tarski might have considered something like this, and that Steve Givant would know of a reference. Perhaps Joel could bring this to Steve's attention? Gerhard "Thinks Tolerances May Also Relate" Paseman, 2013.05.24 $\endgroup$ Commented May 25, 2013 at 5:54
  • $\begingroup$ Jason, yes, that's right. Gerhard, yes, that aspect makes it unusual; I don't know any reference. $\endgroup$ Commented May 25, 2013 at 11:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .