TITLE: Equivalence relations on natural numbers
QUESTION [5 upvotes]: How many equivalence relations are there on $\mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers? How can one compute it?

REPLY [3 votes]: There are $2^{\aleph_0}$ many. As mentioned above, equivalence relation corresponds to partition of $\mathbb{N}$. For each $f \in 2^{\mathbb{N}}$, i.e. a function $f : \mathbb{N} \rightarrow \{0,1\}$. Define a partition, $P_f$, such that, $P_f$ has a subset of size $n$ if and only if $f(n) = 1$. Then if $f \neq g$, then $P_f \neq P_g$. Hence the number of equivalence relations on $\mathbb{N}$ is at least as many as $2^{\aleph_0}$. It can not be bigger because the power set $\mathcal{P}(\mathbb{N})$ has cardinality $2^{\aleph_0}$. 

Explicitly, you can have construct $P_f$ as follows: If $f(i) = 1$, then put into $P_f$, the subset $(\sum_{k < i \ st \ f(k) = 1} k + 1, \sum_{k < i \ st \ f(k) = 1} k + 1 + i)$.