0
$\begingroup$

I have a similar question that is posted here:

finding different equivalence relations

But I don't understand the answers that people gave there.

I have seen some where that when you have the set S = {x,y,z} that the equivalence relation could be the following pairs: {(x,x), (y,y), (z,z)}

I don't see how to show that this is symmetric and transitive.

In the link one person stated that it is "vacuously" because we essentially can't test for symmetric and transitive. Well, if we need all of the 3 conditions to be met, and we can't apply 2 of them, then I am thinking that we hence can't call it an equivalence relation!

SO not sure how the logic of this goes! Hope someone can shed some light on this.

$\endgroup$
8
  • 1
    $\begingroup$ The point is that we can "apply" all three conditions, and those conditions are satisfied. $\endgroup$ Oct 12 '18 at 4:46
  • $\begingroup$ I still don't understand. I don't see how the other 2 conditions are applied. This is the problem with the Abstract Algebras, there is some subtle logic that is not really explained well in general. $\endgroup$
    – Palu
    Oct 12 '18 at 4:47
  • 2
    $\begingroup$ There is no subtle logic whatsoever: for symmetry, just consider all pairs $(a,b)\in R$ and check that $(b,a)\in R$ for all of them. I mean, there are only three of them.... $\endgroup$ Oct 12 '18 at 4:50
  • $\begingroup$ So for symmetry case : (a,b) relation to (b,a), can one show this by using similar elements, I assume that in this definition that a != b (means 'a' not equal to 'b') $\endgroup$
    – Palu
    Oct 12 '18 at 4:56
  • 1
    $\begingroup$ If what i wrote is now correct, then it means that these work as "positional arguments". $\endgroup$
    – Palu
    Oct 12 '18 at 5:00

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.