0
$\begingroup$

I was asked to show that there is some relation R, some set B and some subset A of B such that R is transitive on A but not transitive on B. The solution gave was A = the empty set, B = {0,1,2} and R = {(0,1),(1,2)}. Since A is the empty set, it is a subset of B and R is also transitive on it, but R is not transitive on B. Now I tried to find an example where A was not the empty set, but came up short. Is there general reasoning showing that, when A is not empty, if R is transitive on A and A a subset of B, then R is transitive on B? I felt like I nearly showed it but have been spinning my wheels.

$\endgroup$
3
  • 3
    $\begingroup$ Why would this be true? Suppose $A$ has only one element, which is equivalent to itself under $R$. That hardly implies that $R$ is transitive on the big set. $\endgroup$
    – lulu
    Commented Sep 7, 2022 at 18:28
  • 1
    $\begingroup$ Take $A=\{0\}$, $B=\{0,1,2,3\}$, $R=\{(0,0), (1,2),(2,3)\}$. $\endgroup$ Commented Sep 7, 2022 at 18:28
  • $\begingroup$ Take which-ever Set A [[having large number of elements including y & z]] you want & which-ever transitive relation you want. Then include new element x to make Set B, such that (x,y) & (y,z) but not (x,z) [[When we are given A, we can choose more such elements to include in B]] $\endgroup$
    – Prem
    Commented Sep 7, 2022 at 19:22

0

You must log in to answer this question.

Browse other questions tagged .