Set Equivalence is Equivalence Relation

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Set equivalence is an equivalence relation.


For two sets to be equivalent, there needs to exist a bijection between them.

In the following, let $\phi$, $\phi_1$, $\phi_2$ etc. be understood to be bijections.


From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ is the required bijection.

Thus there exists a bijection from $S$ to itself and $S$ is therefore equivalent to itself.

Therefore set equivalence is reflexive.



\(\ds \) \(\) \(\ds S \sim T\)
\(\ds \) \(\leadsto\) \(\ds \exists \phi: S \to T\) Definition of Set Equivalence
\(\ds \) \(\leadsto\) \(\ds \exists \phi^{-1}: T \to S\) Bijection iff Inverse is Bijection
\(\ds \) \(\leadsto\) \(\ds T \sim S\) Definition of Set Equivalence: $\phi^{-1}$ is also a bijection

Therefore set equivalence is symmetric.



\(\ds \) \(\) \(\ds S_1 \sim S_2 \land S_2 \sim S_3\)
\(\ds \) \(\leadsto\) \(\ds \exists \phi_1: S_1 \to S_2 \land \exists \phi_2: S_2 \to S_3\) Definition of Set Equivalence
\(\ds \) \(\leadsto\) \(\ds \exists \phi_2 \circ \phi_1: S_1 \to S_3\) 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

Therefore set equivalence is transitive.


Also see

The definition of a cardinal of a set as the equivalence class of that set under set equivalence.