22.1.2. Equivalence Relations and Partitions

So, let us start with the definition of a partition of a set. So, imagine you are given a set C which may be finite or it may be infinite. Now, what is the partition of this set C? The partition here is basically a collection of pairwise disjoint, non-empty subsets say m subsets of C which should be pairwise disjoint such that if you take their union, you should get back the original set C.

So intuitively, say for example, you have the map of India you can say that the various states of India partition the entire country India into various subsets such that there is no intersection among the states here. So, in that sense, I am just trying to find out some subsets of the set C such that there should not be any overlap among those subsets and if I take the union of all those subsets I should get back the original set C, there should not be any element of C which is missing.

So, more formally the requirements here are the following. Each subset $A_i \neq \emptyset$ that means each subset should have at least one element. They should be pairwise disjoint. That means if I take any $A_i, A_j$ then $A_i \cap A_j = \emptyset$ and (A_1 \cup A_2 \cup \ldots \cup A_m = C.

A partition of a set is a way of dividing the set into non-overlapping subsets that cover the entire set without leaving out any elements. For example, if you have a set of students in a class, you could partition them into groups based on their hobbies. Each group would contain students with the same hobby, and no student would belong to more than one group. The key characteristics of these groups (the subsets) are that they must not overlap (no student belongs to two groups at the same time) and together they must include every student (no student is left out).

Consider a box of assorted candies divided into different jars based on flavors: chocolates, sour candies, gummies, etc. Each jar contains candies of only one flavor, fulfilling the requirement of non-empty, pairwise disjoint subsets that collectively represent all candies in the box.

What we now want to establish here is a very interesting relationship between the equivalence from an equivalence relation to the partition of a set. So, we want to establish a relationship between equivalence relation and the partition of a set. So, imagine you are given a set C consisting of $n$ elements. Now what I can prove here is that if R is an equivalence relation over the set C and if the equivalence classes which I can form with respect to the relation R are $A_1, A_2, \ldots, A_m$. Then my claim here is that the equivalence classes $A_1, A_2, \ldots, A_m$ constitute a partition of the set C.

Equivalence classes are formed from an equivalence relation, which is a relationship that groups elements in a set based on certain criteria (like being equal, related, etc.). When we say that these equivalence classes form a partition, we mean that each class is a distinct, non-overlapping group, and together they cover the entire set. For instance, if we have a set of integers and we define an equivalence relation based on divisibility, the equivalence classes would be groups of numbers that share the same remainder when divided by a given number. This aligns with the idea of partitioning the set into non-overlapping subsets.

Think of a library that categorizes books by genre (like fiction, non-fiction, science, and history). Each genre represents an equivalence class, and together they partition the entire collection of books. No book belongs to more than one genre at a time, yet collectively, every book in the library is classified into one of these genres.

So, to recall, the definition of partition demands me to prove three properties, the first property is that each of this subset should be non-empty. And that is trivial because I know that each of these equivalence classes is non-empty because each of these equivalence classes is bound to have at least one element, $x \in [A_i]$ since my relation R is an equivalence relation, it will be a reflexive relation that means the element x will be related to itself. That means none of these equivalence classes will be an empty set. So, the first requirement is satisfied.

In proving that equivalence classes form a partition, we start by confirming each class is non-empty: since any element must relate to itself through reflexivity, every equivalence class must contain at least one element. Then, we verify that the union of all equivalence classes equals the original set (meaning every element is included), and finally, we show that no element belongs to more than one equivalence class (meaning they are distinct and non-overlapping).

Imagine a school organized by grades. Each grade (like grade 1, grade 2, etc.) represents an equivalence class. Each class must have at least one student (non-empty), if you combine all students from each grade, you'll account for every student in the school (the union is the entire school population), and no student is counted in multiple grades (distinct without overlap).

Now, I can prove the property in the reverse direction as well. What do I mean by that? I claim here that you give me any partition of a set C, say you give me a collection of m subsets which constitute a partition of the set C. Then I can give you an equivalence relation R whose equivalence classes will be the subsets, which you have given me in the partition.

This segment discusses the reverse relationship: starting from a partition and forming an equivalence relation. Given distinct subsets that totally cover a set, you can create a relation where elements from the same subset are considered equivalent. The process is systematic; iterate over each subset and link all its elements in the equivalence relation. This proves that any valid partition implies a corresponding equivalence relation.

Think of a group of friends who form separate clubs based on interests: cooking, sports, and music. Each club forms a partition of the total friendship group. You can create a relationship where members from the same club are 'friends' with each other, establishing an equivalence relation based on club membership.

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