Equivalence relations on $\{1,2,3\} \times \{1,2,3\}$

Question

Well, I am having problem with understanding why the minimum elements of an equivalence relation on the Cartesian Product $\{1,2,3\} \times \{1,2,3\}$ is $3$. In our lecture the explanation was, that we have defined transitivity for three elements (see bellow, the definition), thus we need a minimum of $3$ different elements.

But in this post Equivalence relation on set $\{0,1,2,3\}$

They argue, that the definition:

$$aRb ∧ bRc ⇒ aRc$$

does not state that $a, b, c$ have to be distinct. Thus since we have the pairs $\{\{1,1\}, \{2,2\}, \{3,3\}\}$ which are the minimum amount of elements of an equivalence relation.

Which is now the right explanation?

I would argue, that since an equivalence relation must be defined on all elements of a set, thus the most trivial one is $\{\{1,1\}, \{2,2\}, \{3,3\}\}$ which is the smallest as well. To make it clear: we can define an equivalence relation on a set with one element, right?

Answer 1

The reflexivity condition states that a binary relation $R \subseteq A \times A$ is reflexive if we have that $x R x$ for all $x \in A$.

But the transitivity condition is implicative. It says that for all $x,y,z \in A$ we have that if $x R y$ and $y R z$ then $x R z$. Therefore we do not require that the elements $x,y,z$ must be distinct.

Answer 2

Symmetric means if $(1,2) \in R$, then $(2,1) \in R$. In your example, all elements are of the form $(1,1)$ so it is true.

A set X is transitive if whenever an element 1 is related to an element 2, and 2 is in turn related to an element 3, then 1 is also related to 3. Its like they have similarity.

In your case 1 = 2 = 3.

And yes we can define an equivalence relation on one element set also.