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I came across with this topic. It looks straight forward for $5$ elements, but what if I want to find how many different equivalence relations with exactly two different equivalence classes are there on a set with $n$ elements?

It says that the set of equivalence relations on a set are in direct bijection with the set of partitions on a set. This made me read about Bell numbers but I'm not sure how help us here. How to solve this problem?

vesii
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  • If you have any subset $A\in S$ then we define $x\sim y$ iff either $x,y$ are both in $A$ or if neither $x,y$ are in $A$. Note that we get the same relation if we replace $A$ by it's complement $A^c$. – lulu Jan 14 '21 at 21:02

2 Answers2

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Since equivalence relations bijectively correspond to partitions on the same set (via their equivalence classes) we see that the number of equivalence relations with exactly two different equivalence classes is the number of partitions of a set with $n$ elements into two non-empty sets. There are in total $2^n$ different subsets of which $2^n-2$ are non-empty and not the whole set. Therefore the number of these partitions is $\frac{2^n-2}{2}=2^{n-1}-1$ (we need to divide by $2$ since otherwise we would count each partition twice)

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Pick an element $a\in S$. Then we have an obvious bijection between $$\{\text{equivalence relations on }S\text{ with exactly two equivalence classes}\}$$ and $$\{\text{proper subsets of }S\setminus\{a\}\,\}$$ given by $$ {\sim}\mapsto \{\,x\in S\setminus\{a\}\mid x\sim a\,\}$$

Hagen von Eitzen
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