1.1.2. Question 1: Equivalence Relations - Part A

This question you are given two equivalence relations on a non-empty set X. You are asked to prove or disprove whether R1 ∪ R2 is an equivalence relation or not? So, it turns out at a union of two equivalence relations need not be an equivalence relation.

In this section, we are introduced to the idea of equivalence relations and their union. An equivalence relation is a relation that is reflexive, symmetric, and transitive. In this specific problem, we are examining two equivalence relations, denoted as R1 and R2, defined over a set X. The task is to determine whether the union of these two relations, written as R1 ∪ R2, also forms an equivalence relation. Interestingly, it has been concluded that this is not necessarily the case, meaning the union may lack one or more properties of equivalence relations.

Think of equivalence relations like a group of friends who are connected by a unique set of common interests (like hobbies). If Friend A is connected to Friend B (R1) and Friend B is connected to Friend C (R2), you would expect them to all know each other (R1 ∪ R2). However, if Friend A doesn't share any interests with Friend C directly, their connection might not exist in the combined interests, just like how R1 ∪ R2 might not be an equivalence relation.

And this is demonstrated by this counter example. So, I consider my set X={a,b,c}. And let me have equivalence relations R1 and R2. So, R1 = {(a,a),(b,b),(c,c),(a,b),(b,a)} and R2 = {(a,a),(b,b),(c,c),(b,c),(c,b)}.

Here, a counterexample is introduced using a specific set X, which contains the elements {a, b, c}. Two equivalence relations, R1 and R2, are specified. R1 consists of the pairs emphasizing reflexivity and symmetry (a is related to b, and b is related to a). Similarly, R2 also captures these properties. This specific construction allows us to explore the union of R1 and R2 to see if it retains the properties of equivalence relations.

Imagine you have three friends: Alice (a), Bob (b), and Charlie (c). If Alice thinks of Bob as a friend (R1), sensing a connection, and Bob thinks of Charlie the same way (R2), you might think Alice and Charlie are connected. However, if Alice doesn’t know Charlie directly, they don’t share a friendship (R1 ∪ R2), illustrating that knowing one friend does not always mean knowing their friends.

Now, if you take the union of these two relations, R1 ∪ R2, you will have these ordered pairs. And it is easy to verify that R1 ∪ R2 is a reflexive relation over the set X, because you have {(a,a),(b,b),(c,c)} present in the relation. The union is also symmetric because if you have any (a,b) in the union present and ordered pair (b,a) is also present in the union.

In this chunk, after forming the union of relations R1 and R2, we can easily examine the properties of reflexivity and symmetry. Reflexivity is confirmed as all elements {a, b, c} have themselves as pairs in the relation, meaning every element relates to itself. The symmetry condition is also satisfied because for every connection (a, b), there’s a corresponding (b, a) present, maintaining the symmetric property expected from equivalence relations.

Think of it like a mutual friendship network. If Alice is friends with Bob, and Bob is friends with Alice, their relationship is both mutual and connective. This reflects reflexivity because everyone believes they are at least their own friend, and symmetry shows that if one person is friends with another, the reverse is also true.

But it is easy to see that union here is not a transitive relation, specifically you have (a,b) and an ordered pair (b,c) present in R1 ∪ R2, but (a,c) is not present in the union and hence the transitivity properties violated.

In examining the union of R1 and R2, we find that while reflexivity and symmetry are satisfied, transitivity fails. If we have (a, b) from R1 and (b, c) from R2, we would expect (a, c) to also be included for transitivity to hold. However, since (a, c) is missing from the union, we see that the union does not maintain the transitive property required for an equivalence relation.

Imagine organized friendships again: if Alice likes Bob and Bob likes Charlie, one might assume Alice also likes Charlie. However, if Alice doesn’t know Charlie at all, this assumption crumbles, mirroring how missing connections ruin the logical flow—like missing links in a chain.

So, in general the union of two equivalence relations need not be transitive. We can prove that they will be always reflexive and symmetric. If I take the union it will be always reflexive and symmetric, but the union need not be a transitive relation.

To conclude this section, it’s highlighted that while unions of equivalence relations will consistently exhibit reflexivity and symmetry, they do not guarantee transitivity. Therefore, while examining unions of equivalence relations, one must be prepared to check these properties individually to establish if the resulting relation is indeed an equivalence relation.

This is like asserting that while many friendships are either beneficial (reflexive) or mutually acknowledged (symmetric), a lack of common connections can make certain relationships weak (failing transitivity), underscoring the need for personal connections to truly form a united front.

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