Relation Isomorphism Preserves Transitivity

Theorem

Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.

Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.

Let $\phi: S \to T$ be a relation isomorphism.

Without loss of generality, it is therefore sufficient to prove only that if $\RR_1$ is transitive, then $\RR_2$ is also transitive.

So, suppose $\RR_1$ is a transitive relation.

Let $y_1, y_2, y_3 \in T$ such that $y_1 \mathrel {\RR_2} y_2$ and $y_2 \mathrel {\RR_2} y_3$.

Let $x_1 = \map {\phi^{-1} } {y_1}$, $x_2 = \map {\phi^{-1} } {y_2}$ and $x_3 = \map {\phi^{-1} } {y_3}$.

As $\phi$ is a bijection it follows from Inverse Element of Bijection that:

$y_1 = \map \phi {x_1}$

$y_2 = \map \phi {x_2}$

$y_3 = \map \phi {x_3}$

As $\phi^{-1}$ is a relation isomorphism it follows that:

$x_1 = \map {\phi^{-1} } {y_1} \mathrel {\RR_1} \map {\phi^{-1} } {y_2} = x_2$

$x_2 = \map {\phi^{-1} } {y_2} \mathrel {\RR_1} \map {\phi^{-1} } {y_3} = x_3$

As $\RR_1$ is a transitive relation it follows that:

$x_1 \mathrel {\RR_1} x_3$

As $\phi$ is a relation isomorphism it follows that:

$y_1 = \map \phi {x_1} \mathrel {\RR_2} \map \phi {x_3} = y_3$

Hence if $y_1 \mathrel {\RR_2} y_2$ and $y_2 \mathrel {\RR_2} y_3$, then also:

$y_1 \mathrel {\RR_2} y_3$

Hence, $\RR_2$ is a transitive relation, by definition.

Hence the result.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$