Q.If $R_{1}$ and $R_{2}$ be two equivalence relations on a set $A$, then prove that $R_{1} \cap R_{2}$ be also an equivalence relation.

Intersection of Equivalence Relations

The Intuition First

Imagine two different ways of grouping people in a classroom.

• Relation $R_1$ — "same height group": related if in the same height bracket.
• Relation $R_2$ — "same eye colour": related if they share eye colour.

Each is an equivalence relation: reflexive (everyone is in their own group), symmetric (if A is in B's group, B is in A's), and transitive (same group as B, and B same as C, means same as C).

Now, what does it mean for two people to be related under both relations? They must be in the same height group and have the same eye colour. That "and" condition is exactly the intersection.

The Precise Statement

Theorem: The intersection of two equivalence relations on a set $S$ is itself an equivalence relation on $S$.

Let $R_1$ and $R_2$ be equivalence relations on $S$, and $R = R_1 \cap R_2$.

1. Reflexivity: For any $a \in S$, $a R_1 a$ and $a R_2 a$, so $(a,a) \in R_1 \cap R_2$. Thus $R$ is reflexive.

2. Symmetry: If $(a,b) \in R$, then $(a,b) \in R_1$ and $(a,b) \in R_2$. By symmetry of each, $(b,a) \in R_1$ and $(b,a) \in R_2$, so $(b,a) \in R$. Thus $R$ is symmetric.

3. Transitivity: If $(a,b) \in R$ and $(b,c) \in R$, then $(a,b),(b,c) \in R_1$ gives $(a,c) \in R_1$; the same reasoning gives $(a,c) \in R_2$. So $(a,c) \in R$. Thus $R$ is transitive.

What Does This Mean Visually?

The intersection $R_1 \cap R_2$ creates finer groups: each new group holds elements in the same $R_1$-class and the same $R_2$-class. In the height-and-eye-colour example, 3 height groups and 4 eye-colour groups give up to $3 \times 4 = 12$ combined groups (some may be empty). The intersection "refines" the partition into smaller, more specific classes.

A Quick Example

Let $S = \{1,2,3,4\}$, with $R_1$ having classes $\{1,2\}$ and $\{3,4\}$, and $R_2$ having classes $\{1,3\}$ and $\{2,4\}$.

$R_1 \cap R_2$ keeps only pairs where both relations agree:

• $(1,1), (2,2), (3,3), (4,4)$ are in both (reflexive). …