Q.If $R_1$ and $R_2$ are equivalence relations in a set $A$, show that $R_1 \cap R_2$ is also an equivalence relation.

Concept: Intersection of equivalence relations — the intersection of two equivalence relations inherits all three defining properties (reflexivity, symmetry, transitivity) from the original relations.

Reasoning:

1. Reflexivity: For any $a \in A$, $(a,a) \in R_1$ and $(a,a) \in R_2$ because both are reflexive. Hence $(a,a) \in R_1 \cap R_2$.

2. Symmetry: If $(a,b) \in R_1 \cap R_2$, then $(a,b) \in R_1$ and $(a,b) \in R_2$. Since $R_1$ and $R_2$ are symmetric, $(b,a) \in R_1$ and $(b,a) \in R_2$, so $(b,a) \in R_1 \cap R_2$.

3. Transitivity: If $(a,b) \in R_1 \cap R_2$ and $(b,c) \in R_1 \cap R_2$, then both pairs belong to $R_1$ and $R_2$ individually. By transitivity of each, $(a,c) \in R_1$ and $(a,c) \in R_2$, so $(a,c) \in R_1 \cap R_2$.

Since all three properties hold, $R_1 \cap R_2$ is an equivalence relation.

Why this works — the core idea

An equivalence relation is just a relation that satisfies three specific properties: reflexivity, symmetry, and transitivity. When we take the intersection $R_1 \cap R_2$, we're keeping only those pairs $(a,b)$ that are related in both $R_1$ and $R_2$.

The beautiful thing is: if a pair is in the intersection, it automatically satisfies the conditions of both relations. So to check any property for $R_1 \cap R_2$, we simply use the fact that the pair belongs to $R_1$ (so it obeys $R_1$'s rules) and to $R_2$ (so it obeys $R_2$'s rules). The properties then follow one after another.

Let's walk through each property carefully.

Step-by-step verification

1. Reflexivity — every element must be related to itself.

Take any element $a \in A$. Since $R_1$ is reflexive, $(a,a) \in R_1$. Since $R_2$ is reflexive, $(a,a) \in R_2$. Therefore $(a,a)$ belongs to both $R_1$ and $R_2$, which means $(a,a) \in R_1 \cap R_2$. So $R_1 \cap R_2$ is reflexive.

2. Symmetry — if $a$ is related to $b$, then $b$ must be related to $a$.

Assume $(a,b) \in R_1 \cap R_2$. By definition of intersection, this means $(a,b) \in R_1$ and $(a,b) \in R_2$.

Since $R_1$ is symmetric, $(a,b) \in R_1$ implies $(b,a) \in R_1$.

Since $R_2$ is symmetric, $(a,b) \in R_2$ implies $(b,a) \in R_2$.

Thus $(b,a)$ belongs to both $R_1$ and $R_2$, so $(b,a) \in R_1 \cap R_2$. Hence $R_1 \cap R_2$ is symmetric.

3. Transitivity — if $a$ is related to $b$ and $b$ is related to $c$, then $a$ must be related to $c$.

Assume $(a,b) \in R_1 \cap R_2$ and $(b,c) \in R_1 \cap R_2$. Then:

• From $(a,b) \in R_1 \cap R_2$: $(a,b) \in R_1$ and $(a,b) \in R_2$.
• From $(b,c) \in R_1 \cap R_2$: $(b,c) \in R_1$ and $(b,c) \in R_2$.

Now, since $R_1$ is transitive, $(a,b) \in R_1$ and $(b,c) \in R_1$ together imply $(a,c) \in R_1$.

Similarly, since $R_2$ is transitive, $(a,b) \in R_2$ and $(b,c) \in R_2$ together imply $(a,c) \in R_2$.

Therefore $(a,c)$ belongs to both $R_1$ and $R_2$, so $(a,c) \in R_1 \cap R_2$. Thus $R_1 \cap R_2$ is transitive.

A quick check with an example

Let $A = \{1,2,3\}$. Define:

• $R_1$: equivalence modulo 2 (i.e., numbers with same parity) — pairs: $(1,1),(2,2),(3,3),(1,3),(3,1)$
• $R_2$: equality relation — pairs: $(1,1),(2,2),(3,3)$

Then $R_1 \cap R_2 = \{(1,1),(2,2),(3,3)\}$, which is clearly an equivalence relation (it's just equality). The intersection "keeps" only what's common to both.

Method: Proving an Intersection Inherits the Equivalence Properties

This method handles any question of the form "$R_1$ and $R_2$ are equivalence relations, show $R_1 \cap R_2$ is too" — you never re-prove the relations from scratch, you only transfer the three properties through the intersection.

Steps

Step 1: Translate membership in the intersection into a pair of statements.

The single fact you use over and over is the definition of intersection: $(a,b) \in R_1 \cap R_2$ means $(a,b) \in R_1$ and $(a,b) \in R_2$. Every property below is proved by splitting a pair into these two memberships, applying the known property in each parent relation, then recombining.

Step 2: Reflexivity — use "holds in both."

For any element $a$, reflexivity of $R_1$ gives $(a,a) \in R_1$ and reflexivity of $R_2$ gives $(a,a) \in R_2$; hence $(a,a) \in R_1 \cap R_2$.

Step 3: Symmetry and transitivity — same recombination pattern.

Start from a pair (or chain of pairs) in the intersection, drop down into each parent relation, apply that relation's own symmetry/transitivity to get the required new pair in $R_1$ and separately in $R_2$, then conclude the new pair lies in $R_1 \cap R_2$. Because each property is a "for all" condition that must succeed in both relations simultaneously, the conclusion always lands back in the intersection.

Step 4: Recognise the generalisation.

The identical argument works for the intersection of any family of equivalence relations — and it does not work for unions (a union can break transitivity), so never assume the union has the same behaviour.

Common Mistakes

Mistake 1: Assuming the result holds for the union too.

Why it's wrong: students often write "similarly $R_1 \cup R_2$ is also an equivalence relation." The union can fail transitivity — a pair may reach a middle element through $R_1$ and the middle element reach a third through $R_2$, with no single relation joining the ends. Correct approach: prove only what is asked; if union comes up, remember it is generally not an equivalence relation.

Mistake 2: Proving reflexivity/symmetry for only one parent relation.

Why it's wrong: to place a pair in $R_1 \cap R_2$ you must show it lies in $R_1$ and in $R_2$; showing it in just one is not enough. Correct approach: for each property, apply it in both $R_1$ and $R_2$ separately, then combine to conclude membership in the intersection.

Mistake 3: Re-proving that $R_1$ and $R_2$ are equivalence relations.

Why it's wrong: they are given as equivalence relations, so re-deriving their properties wastes time and misses the point. Correct approach: use their properties as established facts and only show the intersection inherits them.