22.5.1. Reflexivity

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.

A partition of a set means dividing that set into smaller groups (subsets) where each group is distinct and does not overlap with any other group. To visualize this, think of a pie: if you cut the pie into slices, each slice represents a subset, and together they represent the whole pie (the original set). Importantly, every slice must have some content (no empty slices) and they cannot overlap (no slice half in one and half in another).

Imagine organizing a group of students into teams for a project. Each team (subset) must have at least one member and must be unique (no two teams can share members). All students must belong to a team, and when you combine all the teams together, you get everyone back — that's a partition.

More formally the requirements here are the following. Each subset ≠ ∅ that means each subset should have at least one element. They should be pairwise disjoint. That means if I take any two subsets, then their intersection must be empty, and their union should equal the set C. One trivial partition of the set C is the set C itself.

A valid partition must meet three main criteria: each subset must not be empty (at least one item each), the subsets cannot overlap (no common elements), and when combined, all subsets must recreate the original set. For example, the set of all fruits can be partitioned into apples, oranges, and bananas — where each fruit type must only belong to its respective subset, ensuring no overlap.

Think of a bookcase organized by genre. Each shelf (subset) must contain at least one book (non-empty), sci-fi books are on one shelf and romance books on another (no overlap), and all books on the shelves must represent the entire collection (recreating the original set).

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, 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 C1, ..., Cm. Then my claim here is that the equivalence classes C1, ..., Cm constitute a partition of the set C.

This highlights a deep connection: every equivalence relation (a specific way of grouping items based on their relations) naturally creates subsets known as equivalence classes. These classes form a partition because they fulfill the requirements: each class has at least one element, they don't overlap (elements belong to only one class), and they together contain every element in the original set.

Consider social groups formed by friendships. Each friendship can be seen as a relation, and all friends from a group form an 'equivalence class'. Grouping all friends this way means you've created subsets (friend groups) that cover every friend without overlap, thus forming a partition of your entire social circle.

So, just 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, because my relation R is an equivalence relation, it will be a reflexive relation that means the element will be related to itself. That means none of these equivalence classes will be an empty set.

Reflexivity states that every element must relate to itself, ensuring that all equivalence classes indeed have at least one member. This clause states that any defined group of similar things must contain something from that group; otherwise, it wouldn't be a valid grouping.

Think about a neighborhood: if each house has at least one person living in it (non-empty) and every person has a home they belong to (reflexivity), then no household is left without occupants, helping ensure every group of neighbors defined by similar characteristics exists.

The second requirement from the partition is that the union of the various subsets should give me back the original set. So, my claim here is that if I take the union of all these m equivalence classes, I will definitely get back my original set C. And this is because you take any element x ∈ Ci, it is bound to be present in at least one equivalence class.

This property emphasizes the importance of coverage: by insisting that every element must fall into a class, the union of all the classes must collectively equal the entire original set C. This means every item must be represented by these classes — a complete circle around the original set.

Imagine a sports league where every player belongs to one team. If you combine these teams (equivalence classes), you'll end up with every player accounted for, thus recreating the whole league (the original set).

Third requirement from the partition was that the various subsets in the partition should be pairwise disjoint. So, in this specific case, I have to show that you take any two equivalence classes, they should be pairwise disjoint and that is easy because in the last lecture we proved that two equivalence classes are either same or they are disjoint.

This ensures precision in classification: no item can belong to more than one equivalence class at the same time. Two classes being identical or completely distinct guarantees that you maintain clear boundaries and avoid overlaps in categorization.

Consider a library: books can be categorized as fiction or non-fiction, but each book fits uniquely into one category. Therefore, a book cannot be both fiction and non-fiction at once – it’s either one or the other (pairwise disjoint).

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 means you can start with groups (partitions) and define relationships (equivalence relations) such that each partition corresponds to a unique equivalence class. This further reinforces the interconnectedness between these two conceptual frameworks.

If you have different sports teams, thinking of each group as a separate class allows you to determine relationships like 'members of the same team'. Thus, the partition (teams) informs us of the equivalence relation (membership), showing how they're connected.

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