I'm confused about whether a specific example must exist to prove an aspect of an equivalence relation.

For example: if a set, $A$, only contains one element, $A = \{1\}$, and a relation, $R$, on that set is defined by: "for all $x$ and $y$ in $R$, $xRy$ if and only if $x \neq y$".

Is that relation symmetric?

My reasoning is that if $x \neq y,$ then $y \neq x$. However, there are no $x, y \in A$ such that $x \neq y.$ The same could be true for transitivity. Reflexivity as I understand it fails by definition of R.

My text defines transitivity and symmetry as a conditional statement. So if the antecedent is always false, the statement is vacuously true. So is the relation on A symmetric, transitive, but not reflexive?

*I do intend to pick up the MathJaX code as I go along. Thanks!

  • $\begingroup$ As a general rule, of $\forall x,y: \lnot R(x,y)$ then $\forall x,y: R(x,y)\implies R(y,x)$. Same for transitivity. $\endgroup$ – Thomas Andrews Apr 22 '13 at 20:57

Exactly. Your relation is merely non-reflexive. It is vacuously true that it is symmetric and transitive.

That is, we could go even further and say that any relation defined on the empty set is vacuously an equivalence relation, because it cannot NOT be reflexive, symmetric, transitive. Such a relation, would hardly be interesting or useful, with essentially no domain.

If there is no way that a relation can fail to be reflexive, it is reflexive. If a relation cannot fail to be symmetric, it is symmetric, and ditto for transitivity. These properties are held to be satisfied, unless there exists any counterexample(s).

  • $\begingroup$ This had been bothering me for some time now, thank you! $\endgroup$ – GotterdammerunG Apr 22 '13 at 21:09
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    $\begingroup$ You're very welcome. You seem to have a firm grasp as to why the property holds (many students struggle to understand the logic component of conditional statements, and the conditional nature of the definitions of symmetry and transitivity). But it does leave us wanting to add "but...only vacuously"...that I understand. $\endgroup$ – Namaste Apr 22 '13 at 21:12
  • $\begingroup$ I'll find out tomorrow if manage to come up with a counter-example to my understanding. :) Feeling strong though, thank you again. $\endgroup$ – GotterdammerunG Apr 22 '13 at 21:22
  • $\begingroup$ Good luck! You seem to have a handle on the properties, and their satisfaction (or lack-there-of!). Understanding what you posted is the "hardest" part. $\endgroup$ – Namaste Apr 22 '13 at 21:26
  • $\begingroup$ @amWhy: That is the best feedback! +1 $\endgroup$ – Amzoti Apr 23 '13 at 0:13

Yes, it's as you say for a set of one element. The logic is unambiguous - it is true that "whenever condition, result". This shows the importance of considering the set/domain/etc. for definitions like these. I strongly suspect this particular example isn't very interesting though!

  • $\begingroup$ Merely a specific example regarding general vs specific criteria to address a general question of mine. Thanks, I'll be sure to note the domain during my final exam tomorrow. :) $\endgroup$ – GotterdammerunG Apr 22 '13 at 21:12

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