1.1.4. Question 2: Union and Composition of Equivalence Relations

Let us prove the implication in the direction where I assume $R_1 \circ R_2 = R_1 \cup R_2$. Under that assumption I will be showing that $R_1 \cup R_2$ is an equivalence relation. I will focus on proving that $R_1 \cup R_2$ is a transitive relation. Because we can always show that $R_1 \cup R_2$ will satisfy the reflexive property and symmetric property.

In this section, we are discussing the properties of union and composition of equivalence relations. First, we have the equivalence relations $R_1$ and $R_2$ over a non-empty set. We claim that the union of these two equivalence relations is also an equivalence relation under certain conditions. To do so, we only need to verify its transitivity property because reflexivity and symmetry can be established through the inclusion of elements in each relation. We take two pairs of elements $(a, b)$ and $(b, c)$ from the union and look into three different scenarios to establish transitivity. If both pairs are from one relation, transitivity follows directly from the properties of equivalence relations. If one is from each relation, we must show how they lead to a related pair in the union.

Consider a class of students where some are friends. Let $R_1$ represent 'is friends with' for one group and $R_2$ for another. If Alice and Bob are friends (in $R_1$) and Bob and Charlie are friends (in $R_2$), we can infer a transitive relationship. If Alice is friends with Bob and Bob is a friend with Charlie through both relations, we conclude that Alice and Charlie have a connection through friendship as well.

Let us prove the implication in the other direction. So, I am going to prove that if $R_1 \cup R_2$ is an equivalence relation, then $R_1 \circ R_2 = R_1 \cup R_2$. So, I have to prove the equality of two sets. I have to prove that $R_1 \circ R_2 \subseteq R_1 \cup R_2$ and I have to prove that $R_1 \cup R_2 \subseteq R_1 \circ R_2$.

In this chunk, we switch our approach and test the converse: if the union of two equivalence relations is itself an equivalence relation, does it mean that their composition is the same as their union? To establish this, we check the subset relationships: if we take an element from the composition, it must appear in the union, and vice versa. We prove each side of the subset relationship. For the first side, if a pair can be formed through a third element from either relation, it shows that the composition indeed falls within the union. Then we scrutinize the union's pairs to illustrate that they would necessarily align with the composition if the union is transitive.

Imagine two car rental companies that have their networks. Let $R_1$ be the set of rental agreements from company A and $R_2$ from company B. If utilizing both companies implies you can join two contracts (composition), establishing that all agreements from A and B shared reflect the total agreements (union), then we've created a greater network. The equality signifies that every shared rental agreement provides a consistent outing experience across both organizations from the perspective of transitive connections.

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