Equivalence relation proofs: general or specific?

Question

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!

Answer 1

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).

Answer 2

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!