Cardinality of all equivalence relations on N

Question

I haven't been able to prove that the cardinality of the set of all equivalence relations on N is equal to 2^N. I think that there's probably a bijection between the set of all equivalence classe on N and P(N) but I can't find it. Would apreciate some help.

Answer 1

It is often easier to show inequalities both ways. To show there are at least $2^{\Bbb N}$, break $\Bbb N$ into pairs $(n,n+1)$. Each pair can be separated or together, showing that many equivalence relations.

To show there are at most $2^{\Bbb N}$ note that each equivalence class is a subset of $\Bbb N$. There are only $2^{\Bbb N}$ of those and the relation is a collection of at most $|\Bbb N|$ of them (which have to be disjoint), so there are only $(2^{\Bbb N})^{\Bbb N}=2^{\Bbb N}$ collections of subsets.