# Does defining all equivalence classes within a set define an equivalence relation on that set?

I am just starting to learn about equivalence relations and have a question that I think will help solidify my understanding.

Say you want to define an equivalence relation on set $$A$$. Is it sufficient to define $$S_1, S_2,...,$$ and $$S_k$$ such that $$\bigcup\limits_{i=1}^k S_i = A$$ and all $$S_i$$ are pairwise disjoint, setting each $$S_i$$ to be an equivalence class?

EDIT: I found a proof that there is a one-to-one correspondence between the equivalence relations on a set $$A$$ and the partitions of $$A$$, so the answer to the above question is "yes" -- thank you also to the commenters who pointed this out.

Then, I presume, the equivalence relation $$p=\bigcup\limits_{i=1}^k (S_i \times S_i)$$. Is this correct?

Also, if this is true, then the number of distinct equivalence relations on a set of cardinality $$a$$ would be $$\sum_{i=1}^{a}\frac{a!}{(a-i)!}i^{a-i}$$, where $$i$$ represents the number of equivalence classes (at least 1 and at most $$a$$). My thinking is that, since you have to put at least 1 element in each group, you can do this in $$\frac{a!}{(a-i)!}$$ ways. Then, for the rest of the $$a-i$$ elements, you have $$i$$ choices. But this is not correct... the correct way of counting equivalence relations is described here (as pointed out in the comments). So where did I go wrong in my counting?

EDIT: the above counting method overcounts by distinguishing by order of groups and order of elements within groups. See the link posted in the comments for the correct solution.

I think the right formula is $$\sum_{i=1}^{a}S(a,i)$$ where $$S$$ denotes a Stirling number of the second kind.

• The answer to the question in your 2nd paragraph is "Yes". Jun 23, 2019 at 22:49
• yes, a partition corresponds to an equivalence relation Jun 23, 2019 at 22:49