# Equivalence relation gives unique partition?

I read lecture notes from mit ocw

https://ocw.mit.edu/courses/mathematics/18-703-modern-algebra-spring-2013/lecture-notes/MIT18_703S13_pra_l_3.pdf

I came across the lemma

Let $X$ be a set. Given an equivalence relation ~ on $X$ there is a 'unique partition ' on $X$

But in the proof there is nothing about uniqueness of partition.I want to know in which sense the partition is unique ( what does it meant by 'unique partition ' and why not 'just partition').

Thanks

• Page 2 on linked lecture notes – Cloud JR Sep 18 '18 at 17:53

The word "unique" is misleading here. What is meant is that there is a one to one correspondence between partitions of a set and equivalence relations on it. Each partition induces an equivalence relation, and each equivalence relation partitions a set. It looks like he leaves part of this as an exercise for the reader.

Sketch of Proof:

An equivalence relation partitions a set:

The equivalence classes form the disjoint "pieces" of the partition. Note that

a.) every element is in some equivalence class (namingly its own).

b.) The equivalence classes are disjoint.

A partition induces an equivalence relation:

Define two elements to be "equivalent" if they reside in the same "piece" of the partition. Show that this relation is reflexive, transitive, and symmetric.

• Give an equivalence relation ,is it possible to partition a set in two different ways? How can we claim there is one to one correspondence between them – Cloud JR Sep 18 '18 at 18:07
• @CloudJR I'm sorry I don't quite know how to explain it any differently than I have above. The Wikipedia section : en.wikipedia.org/wiki/… may make it clearer. – David Reed Sep 18 '18 at 18:16
• This wikipedia article helps a lot thanks – Cloud JR Sep 18 '18 at 18:20
• Thanks for your help – Cloud JR Sep 18 '18 at 18:21
• @CloudJR Sure. I actually remember struggling with this concept when I initially came across it as well. It is important though for making combinatorial ("counting") arguments in certain fundamental theorems --like Lagrange's Theorem from Algebra:en.wikipedia.org/wiki/… – David Reed Sep 18 '18 at 18:25