simple clarification of equivalence relation and order relation notation meaning

Question

So by definition an equivalence relation is a binary relation on a subset of the cartesian product of our set A, i.e $A\times A$ satisfying the 3 conditions. But I'm a little confused about interpreting the notation. So we express an equivalence relation as $(x,y) \in A\times A$ so is this conisdered a point for example say in the $x \times y$ i.e the usual 2-D plane we see in lower level math courses or is this just an arbitrary collection of two elements from our set so it isn't talking about a specific coordinate? Because the condition of symmetry $(x,y) = (y,x)$ throws me right off if these are coordinates.

Answer 1

An equivalence relation on $A$ starts life as a subset of $A \times A$.

If I hand you a subset $S$ from $A \times A$ and tell you it is an equivalence relation, then the symmetric property says that for any $x \in A$ and for any $y \in A$, if $(x,y) \in S$ then $(y,x) \in S$.

This does not hold for arbitrary subsets of $A \times A$, but any subset of $A \times A$ that is an equivalence relation on $A$ must satisfy this property.

Let's take your example of points in the $xy$-plane. Then those points look like $(x,y)$ and are elements of $\mathbb{R} \times \mathbb{R}$. You observed that because of the symmetric property, any subset $S$ of $\mathbb{R} \times \mathbb{R}$ where $S$ is an equivalence relation on $\mathbb{R}$ must have $(y,x) \in S$ whenever $(x,y) \in S$. Set $S=\{(x,y)\ |\ x=y \}$. You can verify that $S$ is an equivalence relation on $\mathbb{R}$.

You can't just pick an arbitrary point from the plane and expect to have it be an element of an equivalence relation on $\mathbb{R}$.

Answer 2

Equivalence relation on a subset of the $A\times A$ does not mean it is an element on the set $A \times A$. It is a binary relation between the elements of $A \times A$ satisfying the 3 conditions.

For example, if you say "$(x,y) = (y,x)$ for all $(x,y)\in A \times A$" then the relation '$=$' is symmetric.