I have been reading Velleman's How to prove book and have come across following section while reading "Relations" chapter:
Suppose R is a relation on A.
- R is said to be reflexive on A (or just reflexive, if A is clear from context) if ∀x ∈ A(x Rx), or in other words ∀x ∈ A((x, x) ∈ R).
- R is symmetric if ∀x ∈ A∀y ∈ A(x Ry → y Rx).
- R is transitive if ∀x ∈ A∀y ∈ A∀z ∈ A((x Ry ∧ y Rz) → x Rz).
As we saw if R is reflexive on A, then the directed graph representing R will have loops at all vertices. If R is symmetric, then whenever there is an edge from x to y, there will also be an edge from y to x. If x and y are distinct, it follows that there will be two edges connecting x and y, one pointing in each direction. Thus, if R is symmetric, then all edges except loops will come in such pairs. If R is transitive, then whenever there is an edge from x to y and y to z, there is also an edge from x to z.
Now it is stated above that if $R$ is symmetric then all the edges except loops will come in such pairs. But why isn't the loops also included ? Imagine a relation like this: $\{(1,1),(1,2),(2,1)\}$
Now even the element $(1,1)$ satisfies $xRy \implies yRx$. So, why aren't they part of symmetric ?
Your relation is symmetric. Velleman’s point is that we draw one directed edge for each ordered pair in the relation. Thus, edges between distinct vertices come in pairs: if we have an edge $1\longrightarrow 2$, then we also have an edge $1\longleftarrow 2$, because if $\langle 1,2\rangle\in R$, then $\langle 2,1\rangle\in R$. Loops don’t come in pairs, however, because there’s only one loop from $1$ to $1$, say: the ordered pair $\langle 1,1\rangle$ is identical to its reversal.