Define $\sim_\mathrm{bi}$ by $$\sim_\mathrm{bi} = \{(S_1,S_2)\mid \text{there is a bijection } f:S_1 \to S_2\}$$ for $S_1,S_2 \subseteq \mathbb{N}$
My proof comes as: In order to prove that $\sim_\mathrm{bi}$ is an equivalence relation on $\mathcal{P}(N)$, we should prove that the relation is reflexive, symmetric, and transitive.
Reflexive: For all $S_1 \subseteq \mathcal{P}(N)$, then the identity function provides a bijection from $S_1$ to $S_1 \subseteq \mathcal{P}(N)$, so the relation is reflexive.
Symmetric: For $S_1 \subseteq \mathcal{P}(N)$, if $S_1 \sim_\mathrm{bi} S_2$, then there is a bijection $f: S_1 \to \S_2$, so there is a bijection $f^{-1} : S_2 \to S_1$, so $S_2 \sim_\mathrm{bi} S_1$, so the relation is symmetric.
Transitive: Assume that $S_1 \sim_\mathrm{bi} S_2$ and $S_2 \sim_\mathrm{bi} S_3$, then we know that there are bijections $f: S_1 \to S_2$ and $g: S_2 \to S_3$. The composition $g \circ f$ is then a bijection from $S_1$ to $S_3$. Thus, we know that $S_1 \sim_\mathrm{bi} S_3$, so the relation is transitive.
By combining the three mini-proofs above, we can conclude that the relation is an equivalence relation.
Also, I was thinking of examples of two distinct equivalence classes. May I know if:
$$ \begin{align} & [\{1,2\}] =\{\{a,b\} \mid a \in \mathbb{N} \text{ and } b \in \mathbb{N} \setminus a\}, \\[6pt] & [\{1,2,3\}] = \{\{a,b,c\} \mid a \in \mathbb{N} \mbox{ and } b \in \mathbb{N} \setminus a \text{ and } c \in \mathbb{N} \setminus a \setminus b\} \end{align} $$
would suffice?
Hope some one can help me to improve my proof. Thank you!