# Equivalence relation criteria

I'm studying set theory/equivalence relation.

For an equivalence relation, $$3$$ conditions are to be met:

$$1)$$ reflexivity

$$2)$$ symmetry

$$3)$$ transitivity

Does the following set and the relation on it fulfill these criteria, even though $$(2)$$ and $$(3)$$ are missing? They are not disproven, they're just absent.

$$\{0,1,2\}$$

$$R=\{(0, 0), (1, 1),(2, 2)\}$$

thanks ralph

• What makes you think that symmetry and transitivity are 'absent'? – Clive Newstead Aug 10 '20 at 11:35
• Actually, this is just the equality relation on the given set, and equality is the main prototype of equivalence relations. – Berci Aug 10 '20 at 11:55
• Be careful when you see something like $\forall x, y \in R, \dots$. It means "for all $x$ and for all $y$", not "for all distinct $x$ and $y$". So you don't even have to worry about the empty set stuff in some of the answers, because indeed for the $x$ and $y$ that are both $0$, $R$ is symmetric, for example. – Izaak van Dongen Aug 10 '20 at 14:10

## 2 Answers

1. Since $$(\forall x \in A): (x,x) \in R$$, then $$R$$ is clearly reflexive.

I think I’m understanding what you meant by “absent”. But remember the definitions:

1. For symmetry: $$R$$ is symmetric iff $$xRy$$ implies $$yRx$$ for all $$x, y \in A$$. Note that in the relation $$R$$ you can shift the numbers and the ordered pairs resulting from this will still be in $$R$$. Hence $$R$$ is symmetric.

2. For transitivity: $$R$$ is transitive iff $$xRy$$ and $$yRz$$ implies $$xRz$$. How do we check this? Start by picking the first element of $$R$$, $$(0,0)$$. What are the elements in $$R$$ that have $$0$$ in the first place? It is $$(0,0)$$. Now check $$(0,0)$$ (this $$(0,0)$$ would be the result from combining both the elements mentioned). You note that $$(0,0)$$ is still in $$R$$. Doing this for every element, tou deduce that $$R$$ is transitive.

The point is that in these definitions $$x, y$$ and $$z$$ can be equal to each other. As long they appear in the “right places”, the relation will agree with these properties.

If there is no counter-example for a statement, the statement is necessarily true. This also applies to your example.

To give another example, let $$S(A)$$ be the statement $$\forall a \in A: a>0$$. Then $$S(\varnothing)$$ is a true statement, because there exists no counter-example to it.