Cartesian Product of Equivalence Relations

Theorem
Let $A$ and $B$ be sets.

Let $\RR$ and $\SS$ be equivalence relations on $A$ and $B$ respectively.

Let $\TT$ be the relation on $A \times B$ defined as:
 * $\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tuple {x, y} \iff u \mathrel \RR x \land v \mathrel \SS y$

Then $\TT$ is an equivalence relation on $A \times B$.

The equivalence class $\eqclass {\tuple {a, b} } \TT$ of an element $\tuple {a, b}$ of $\TT$ is:
 * $\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$

Proof
We have that $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.

Thus they are both reflexive, symmetric and transitive.

Checking in turn each of the criteria for equivalence:

Reflexivity
Thus $\TT$ is seen to be reflexive.

Symmetry
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in A \times B$ such that:
 * $\tuple {x_1, y_1} \mathrel \TT \tuple {x_2, y_2}$

Then:

Thus $\TT$ is seen to be symmetric.

Transitivity
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2}, \tuple {x_3, y_3} \in A \times B$ such that:

Then:

Thus $\TT$ is seen to be transitive.

$\TT$ has been shown to be reflexive, symmetric and transitive.

Hence by definition $\TT$ is an equivalence relation.

The equivalence class $\eqclass {\tuple {a, b} } \TT$ follows directly:
 * $\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$