I know that the relation is symmetric if $\forall x \forall y \ xRy \implies yRx $.

Consider the set $ A = \{{a, b, c, d}\}$ and $R = \{{(a, a),(a, b),(a, d),(b, a),(b, b),(c, c),(d, a),(d, d)}\}.$

My textbook claims that this relation is symmetric. But what about $(c,d)$ and $(d,c)$ that are not part of the $R$ set? In the definition, it said for all $x$ and $y$, so shouldn't this violate the symmetricity?

  • 2
    $\begingroup$ There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.) $\endgroup$ – Stefan Mesken Nov 21 '17 at 9:29
  • $\begingroup$ @StefanMesken Thanks a lot! I edited it. It is still confusing me though. $\endgroup$ – Avocado Nov 21 '17 at 9:37

If $(c,d)$ is in the relation, then we check whether $(d,c)$ is in the set.

However, in this case, $(c,d)$ is not in $R$, hence we should not expect $(d,c)$ to be in the relation for it to be symmetric.

  • $\begingroup$ Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A? $\endgroup$ – Avocado Nov 21 '17 at 9:41
  • 1
    $\begingroup$ yes, it is not required. $\endgroup$ – Siong Thye Goh Nov 21 '17 at 9:42
  • $\begingroup$ Thank you! I am so amazed how quickly the community here helps you out. This is great :) $\endgroup$ – Avocado Nov 21 '17 at 9:43

Definition 1: A relation R over set A is symmetric if for all x, y from A the following is true: (x,y) is in R implies (y,x) is in R.

Definition 2: A relation R over set A is symmetric if for all x, y from A the following is true: if (x,y) is in R, then (y,x) is in R.

These definitions do not require that every (x,y) has to be in R. They only require that for those (x,y) that are in R, (y,x) has to be in R.

Suppose that no single (x,y), where x is different from y, is in R. In this case R is symmetric.

This is so because proposition

  if (x,y) is in R, then  (y,x) is in R

is automatically true when proposition

  (x,y) is in R

is false (i.e. when (x,y) is not in R).

  • $\begingroup$ Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines). $\endgroup$ – Xander Henderson Jan 10 '18 at 4:43

So $(a,c)$ is not included in $\mathbb{R}$. So we can not expect $(c,a)$ in $\mathbb{R}$.

If $(a,c)$ is included in $\mathbb{R}$. So we can expect $(c,a)$ in $\mathbb{R}$

And so both are symmetric


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.