Relation Isomorphism Preserves Antisymmetry

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 therefore suffices to prove only that if $\RR_1$ is antisymmetric, then also $\RR_2$ is antisymmetric.

So, suppose $\RR_1$ is an antisymmetric relation.

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

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

As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \map \phi {x_1}$ and $y_2 = \map \phi {x_2}$.

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

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

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

As $\RR_1$ is an antisymmetric relation it follows that $x_1 = x_2$.

As $\phi$ is a bijection it follows that:

$y_1 = \map \phi {x_1} = \map \phi {x_2} = y_2$

Hence $y_1 = y_2$ and so by definition, $\RR_2$ is an antisymmetric relation.

Hence the result.

$\blacksquare$

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