Composite of Transitive Relations is not necessarily Transitive

Theorem

Let $A$ be a set.

Let $\RR$ and $\SS$ be transitive relations on $A$.

Then their composite $\RR \circ \SS$ is not necessarily also transitive.

Proof

Recall the definition of composition of relations:

Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.

Then the composite of $\RR_1$ and $\RR_2$ is defined and denoted as:

$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$

Proof by Counterexample

Let $A = \set {a, b, c}$.

Let $\RR$ be defined as:

$\RR = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {b, c} }$

Let $\SS$ be defined as:

$\SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b} }$

Both $\RR$ and $\SS$ can be seen to be transitive.

Then we have:

$\RR \circ \SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, c} }$

We note that:

$\tuple {a, b}, \tuple {b, c} \in \RR \circ \SS$

but:

$\tuple {a, c} \notin \RR \circ \SS$

Hence, by definition, $\RR \circ \SS$ is not transitive.

$\blacksquare$