Set Equivalence behaves like Equivalence Relation/Transitive

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Theorem

Set equivalence behaves like a transitive relation:

$S_1 \sim S_2 \land S_2 \sim S_3 \implies S_1 \sim S_3$


Proof

\(\ds \) \(\) \(\ds S_1 \sim S_2 \land S_2 \sim S_3\)
\(\ds \) \(\leadsto\) \(\ds \exists \phi_1: S_1 \to S_2 \land \exists \phi_2: S_2 \to S_3\) Definition of Set Equivalence: $\phi_1$ and $\phi_2$ are bijections
\(\ds \) \(\leadsto\) \(\ds \exists \phi_2 \circ \phi_1: S_1 \to S_3\) Composite of Bijections is Bijection: $\phi_2 \circ \phi_1$ is a bijection
\(\ds \) \(\leadsto\) \(\ds S_1 \sim S_3\) Definition of Set Equivalence

$\blacksquare$


Sources