1.3. Functions and Partitions

Now in question number 2 again, you are given two equivalence relations over a non-empty set and we want to prove that $R_1 \cap R_2$ will be an equivalence relation if and only if the composition $R_1 \circ R_2$ is equal to the $R_1 \cup R_2$.

In this part, we are tasked with understanding under what conditions the intersection of two equivalence relations (denoted as $R_1 \cap R_2$) will still be an equivalence relation. An equivalence relation is defined by three properties: it must be reflexive, symmetric, and transitive. We need to prove that these properties hold for the intersection if the condition concerning the composition (denoted as $R_1 \circ R_2$) being equal to the union (denoted as $R_1 \cup R_2$) is satisfied. Specifically, we'll look at properties individually, showing that both relations maintain the required properties under these conditions.

Imagine two groups of friends at a party. If your two best friends each have their own friend groups (their equivalence relations), the intersection of those groups (friends they share) reflects mutual relationships. If every shared relationship can connect (like through common interests), it makes finding new connections easier. Thus, the way we can relate to their friends showcases the maintenance of the same properties, resulting in a more integrated social experience.

If I take $R_1$ and $R_2$ to be equivalence relations over the set $X$, then since $R_1$ and $R_2$ are individually reflexive relations... which shows that the intersection will be a symmetric relation.

This chunk discusses how to prove that the intersection of two equivalence relations retains the properties of reflexivity and symmetry. Reflexivity means that every element in set $X$ relates to itself. Since both equivalence relations possess this property, their intersection will also include these self-related pairs. For symmetry, if an element $(a, b)$ is in the intersection, it implies that both $R_1$ and $R_2$ have this pair which also implies the reverse pair, thus proving symmetry. This establishes that both reflexivity and symmetry hold true.

Consider two classes of students. If each class has several students, the intersection of students who belong to both classes will still contain students who relate to themselves. If you represent every student as a node, then connecting these nodes indicates mutual understanding and respect. Therefore, the intersection maintains the same qualities of reflexivity and symmetry.

Now consider the transitivity property for which I consider arbitrary ordered pairs $(a, b)$ and $(b, c)$ to be present in $R_1 \cap R_2$. ... this shows that the ordered pair $(a, c)$ will be present in the intersection as well.

This chunk explains the transitive property in the context of intersections. For a relation to be transitive, if two pairs $(a, b)$ and $(b, c)$ are related, then the pair $(a, c)$ must also be present. Since both $R_1$ and $R_2$ are equivalence relations, if $(a, b)$ and $(b, c)$ are in the intersection of these two relations, then we can conclude that $(a, c)$ must also be in the intersection. This is important because it solidifies the condition under which the intersection remains an equivalence relation.

Think of a friendship chain. If Alice is friends with Bob, and Bob is friends with Charlie, then Alice will naturally have a connection to Charlie as well. This chain of relationships exemplifies transitivity. Hence, in social structures, this property is essential for establishing solid networks of friendships.

So, now you can see that the argument that we have given here for $R_1 \cap R_2$ to be a transitive relation need not hold... this is precisely the reason due to which the union of two equivalence relations need not be a transitive relation.

The conclusion emphasizes that while the intersection of two equivalence relations will always be an equivalence relation (retaining reflexivity, symmetry, and transitivity), the union does not guarantee transitivity. This means that although both conditions can link together through shared elements in the intersection, combining all elements from both relations into a union may break the transitivity property, highlighting an essential distinction between the two types of operations on relations.

Imagine adding two different groups of friends together for a party. While individual friends may have connections (those in the intersection), creating a massive group may lead to unfamiliarity among some members, breaking the cohesion needed for everyone to feel included or connected (breaking transitivity). Therefore, while intersections help maintain relationships, unions may complicate them.

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