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184 Relations
Note that in this directed graph there is an edge from ∅ to itself, because
(∅, ∅) ∈ S. Edges such as this one that go from a vertex to itself are called
loops. In fact, in Figure 3 there is a loop at every vertex, because S has the
property that ∀x ∈ B((x, x) ∈ S). We describe this situation by saying that S
is reflexive.

Definition 4.3.2. Suppose R is a relation on A.

1. R is said to be reflexive on A (or just reflexive,if A is clear from context) if
∀x ∈ A(xRx), or in other words ∀x ∈ A((x, x) ∈ R).
2. R is symmetric if ∀x ∈ A∀y ∈ A(xRy → yRx).
3. R is transitive if ∀x ∈ A∀y ∈ A∀z ∈ A((xRy ∧ yRz) → xRz).

As we saw in Example 4.3.1, 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.

Example 4.3.3. Is the relation G from part 2 of Example 4.2.2 reflexive? Is it
symmetric?Transitive?AretherelationsinExample4.3.1reflexive,symmetric,
or transitive?

Solution

Recall that the relation G from Example 4.2.2 is a relation on R and that for
any real numbers x and y, xGy means x > y. Thus, to say that G is reflexive
would mean that ∀x ∈ R(xGx), or in other words ∀x ∈ R(x > x), and this
is clearly false. To say that G is symmetric would mean that ∀x ∈ R∀y ∈
R(x > y → y > x), and this is also clearly false. Finally, to say that G is
transitive would mean that ∀x ∈ R∀y ∈ R∀z ∈ R((x > y ∧ y > z) → x > z),
and this is true. Thus, G is transitive, but not reflexive or symmetric.
The analysis of the relations in Example 4.3.1 is similar. For the relation S in
part 1 we use the fact that for any x and y in B, xSy means x ⊆ y.Aswehave
already observed, S is reflexive, since ∀x ∈ B(x ⊆ x), but it is not true that
∀x ∈ B∀y ∈ B(x ⊆ y → y ⊆ x). For example, {1}⊆{1, 2},but {1, 2} ⊆{1}.
You can see this in Figure 3 by noting that there is an edge from {1} to
{1, 2} but not from {1, 2} to {1}. Thus, S is not symmetric. S is transitive,