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
 A: *

*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:


*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.


*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.
A: 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.
