Equivalence relations on natural numbers

Question

How many equivalence relations are there on $\mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers? How can one compute it?

Answer 1

An equivalence relation on any set partitions that set.

So we can approach your question by asking, equivalently: How many ways are there to partition $\mathbb{N}$?

• A partition of $X$ with is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.

• $|\mathbb{N}| = \aleph_0$. There are uncountably many possible distinct partitions of $\mathbb{N}: 2^{\aleph_0}$ distinct partitions.
  (See this post for a nicely detailed proof of this - not just a sketch of a proof.)

• Since there is a natural bijection from the set of all possible equivalence relations on a set $\mathbb{N}$ and the set of all partitions of $\mathbb{N}$, there are $2^{\aleph_0}$ equivalence relations on $\mathbb{N}$.

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• For counting the number of partitions on a finite sets, see Counting Partitions.

• For an elaboration on the relationship between an equivalence relation on a set and the partition of the set induced by that relation, vice versa, see this previous post

Answer 2

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 $2^{\aleph_0}$.

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