Cross-Relation is Equivalence Relation
Contents
Theorem
Let $\left({S, \circ}\right)$ be a commutative semigroup with cancellable elements.
Let $\left({C, \circ {\restriction_C}}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.
Let $\left({S_1, \circ {\restriction_1}}\right) \subseteq \left({S, \circ}\right)$ be a subsemigroup of $S$.
Let $\left({S_2, \circ {\restriction_2}}\right) \subseteq \left({C, \circ {\restriction_C}}\right)$ be a subsemigroup of $C$.
Let $\left({S_1 \times S_2, \oplus}\right)$ be the (external) direct product of $\left({S_1, \circ {\restriction_1}}\right)$ and $\left({S_2, \circ {\restriction_2}}\right)$, where $\oplus$ is the operation on $S_1 \times S_2$ induced by $\circ {\restriction_1}$ on $S_1$ and $\circ {\restriction_2}$ on $S_2$.
Let $\boxtimes$ be the cross-relation on $S_1 \times S_2$, defined as:
- $\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$
Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$.
Proof
Reflexivity
- $\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$
So $\boxtimes$ is a reflexive relation.
$\Box$
Symmetry
| \(\displaystyle \tuple {x_1, y_1}\) | \(\boxtimes\) | \(\displaystyle \tuple {x_2, y_2}\) | |||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle x_1 \circ y_2\) | \(=\) | \(\displaystyle x_2 \circ y_1\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle x_2 \circ y_1\) | \(=\) | \(\displaystyle x_1 \circ y_2\) | $\circ$ is commutative | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \tuple {x_2, y_2}\) | \(\boxtimes\) | \(\displaystyle \tuple {x_1, y_1}\) |
So $\boxtimes$ is a symmetric relation.
$\Box$
Transitivity
| \(\displaystyle \tuple {x_1, y_1}\) | \(\boxtimes\) | \(\displaystyle \tuple {x_2, y_2}\) | |||||||||||
| \(\, \displaystyle \land \, \) | \(\displaystyle \tuple {x_2, y_2}\) | \(\boxtimes\) | \(\displaystyle \tuple {x_3, y_3}\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle x_1 \circ y_2\) | \(=\) | \(\displaystyle x_2 \circ y_1\) | ||||||||||
| \(\, \displaystyle \land \, \) | \(\displaystyle x_2 \circ y_3\) | \(=\) | \(\displaystyle x_3 \circ y_2\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle x_1 \circ y_3 \circ y_2\) | \(=\) | \(\displaystyle x_1 \circ y_2 \circ y_3\) | $\circ$ is commutative | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x_2 \circ y_1 \circ y_3\) | substituting $x_2 \circ y_1$ for $x_1 \circ y_2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x_2 \circ y_3 \circ y_1\) | $\circ$ is commutative | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x_3 \circ y_2 \circ y_1\) | substituting $x_3 \circ y_2$ for $x_2 \circ y_3$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x_3 \circ y_1 \circ y_2\) | $\circ$ is commutative | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle x_1 \circ y_3\) | \(=\) | \(\displaystyle x_3 \circ y_1\) | as $y_2 \in S_2$, therefore $y_2 \in C$, and so is cancellable for $\circ$ | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \tuple {x_1, y_1}\) | \(\boxtimes\) | \(\displaystyle \tuple {x_3, y_3}\) |
So $\boxtimes$ is a transitive relation.
$\Box$
All the criteria are therefore seen to hold for $\boxtimes$ to be an equivalence relation.
$\blacksquare$
Also see
Examples
- Cross-Relation on Natural Numbers is Equivalence Relation
- Cross-Relation on Real Numbers is Equivalence Relation
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 20$
