TITLE: Counting equivalence relations
QUESTION [5 upvotes]: I am aware that on a finite set the number of equivalence relations is the n-th Bell's Number. On the other hand, the only reference I could find on the web for infinite sets was this: On counting equivalence relations by D.J. Baylis, in which he proves that the set of equivalence relations on $\mathbb{N}$ is uncountable. I am interested in any sort of theorem that gives some rule about counting equivalence relations on an infinite set -something like $\beth_0 = \aleph_0$ and $\beth_n = 2^{\beth_{n-1}}$ - if there are any.

REPLY [5 votes]: Firstly, to address Beth numbers (as you noted, the math markup doesn't work with them. I'll use words instead)
$\beth$ (Beth) numbers are cardinal numbers defined as followed:

$\beth_0 = \aleph_0$,
$\beth_{\alpha+1} = 2^{\beth_n}$,
$\beth_\delta=\sup_{\gamma<\delta}\beth_\gamma$, for a limit ordinal $\delta$.

Now, given an infinite set $X$ and assuming the axiom of choice, we have that $|X\times X|=|X|$, therefore $|P(X\times X)| = |P(X)|$.
Define $R(X)$ as the set of all equivalence relations on $X$, then clearly $R(X) \subset P(X\times X)$ and therefore $|R(X)| \le |P(X)|$.
On the other hand there are $2^{|X|}$ partitions of $X$ into two sets (an easy exercise in cardinal arithmetic), so for every partition as such $\{X', X\setminus X'\}$ we can define an equivalence relation with two equivalence classes, namely $X'$ and its complement in $X$. Therefore $|R(X)| \ge |P(X)|$.
All in all we have that the number of equivalence relations on $X$ is $2^{|X|}$.