Proof of Properties of Equivalence Relation

Let's begin with the definition of a partition of a set. A partition of a set, say C, is a collection of non-empty, pairwise disjoint subsets of C, such that their union returns the original set C.

Can you explain what pairwise disjoint means?

Great question! Pairwise disjoint means that any two subsets do not have any elements in common. For example, if A and B are two subsets of C, then A ∩ B equals the empty set.

Could you give an example of a partition?

Certainly! If C = {1, 2, 3, 4} then one possible partition could be {{1, 2}, {3, 4}}.

What if the subsets overlapped?

If subsets overlap, it wouldn't be a proper partition. Partitions require that the subsets are completely disjoint!

To summarize, a partition includes subsets that are non-empty, collectively exhaustive (their union forms the original set), and pairwise disjoint.

Now let's delve into how equivalence relations relate to partitions. Can anyone remind me what an equivalence relation is?

It's a relation that is reflexive, symmetric, and transitive.

Exactly! When we have an equivalence relation R defined on a set C, it allows us to group elements into equivalence classes. I claim that these classes form a partition of C. Can anyone identify one of the properties we need to prove that?

They must be non-empty, right?

Correct! Each equivalence class must contain at least one element because each element is related to itself due to the reflexity of R. What’s our next proof requirement?

We need to show that the union of all classes equals C.

Exactly! Since every element belongs to some equivalence class, the union gives back the original set. Lastly, do we remember the final proof condition?

They need to be pairwise disjoint!

Yes! That follows from our earlier classes, where we discussed that two different equivalence classes cannot share elements. For our final recap: equivalence classes must be non-empty, their union must yield C, and they must be pairwise disjoint.

Suppose you have a partition of a set. How could we create an equivalence relation from it?

Do we link every element in the subclass?

Precisely! For each subset in the partition, we form pairs by considering all elements in that subset.

Can you give an example of that?

Certainly! Let's say we have a partition of {1, 2, 3, 4} into {{1, 2}, {3, 4}}. For the equivalence relation, we would form pairs like (1, 1), (1, 2), (2, 1), (2, 2) for the first subset, and similar pairs for the second subset.

And each of those subsets will form their own equivalence class?

Exactly! Hence the equivalence classes correspond directly to the subsets of the partition.

Let's summarize: the process of constructing an equivalence relation creates classes that mirror the subsets in the partition.

Now that we have constructed our equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.

For reflexivity, each element should relate to itself.

Exactly! Each element in a subset generates a pair to itself, demonstrating reflexivity. What about symmetry?

If (a, b) is in the relation, then (b, a) must also be in it!

Correct! We add pairs (x, y) from any subset, ensuring symmetry naturally follows. Now, who can tell me about transitivity?

If we have (a, b) and (b, c), then (a, c) should also be part of the relation!

Well done! By ensuring all elements are in the same subset, transitivity holds. So, we established that this constructed relation is indeed an equivalence relation.

To recap: all three properties—reflexivity, symmetry, and transitivity—are satisfied in our relation defined from the partition.