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Relations and Functions
Question
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If R1 and R2 are equivalence relations in a set A, show that R1 ∩ R2 is also an equivalence relation.

Answer

Suppose that R1 and R2 are two equivalence relations on a non-empty set X.
First we prove that R1 ∩ R2 in an equivalence relation on X.
(i) R2 ∩ R2 is reflexive :
Let a ∈ X arbitrarily.
Then (a, a) ∈ R1 and (a, a) ∈ R2 , since R1, R2 both being equivalence relations are reflexive.
So. (a, a) ∈ R1 ∩ R2
⇒ R1 ∩ R2 is reflexive.
(ii) R1 ∩ R2 is symmetric :
Let a, b ∈ X such that (a, b) ∈ R1 ∩ R2 ∴ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1and (b, a) ∈ R2, since R1 and R2 being equivalence relations are also symmetric.
(b, a) ∈ R1∩ R2
(a, b) ∈ R1 ∩ R2 implies that (b, a) ∈ R1 ∩ R2.
∴ R1 ∩ R2 is a symmetric relation.
(iii) R1 ∩ R2 is transitive :
Let a, b, c ∈ X such that (a, b) ∈ R1 ∩ R2 and (b, c) ∈ R1 ∩ R2.
(a, b) ∈ R1 ∩ R2 ⇒ (a, b) ∈ R1 and (a, b) ∈ R2    ...(i)
(b, c) ∈ R1 ∪ R2 ⇒ (b, c) ∈ R1 and (b, c) ∈ R2    ...(ii)
(i) and (ii) ⇒ (a, b) and (b, c) ∈ R1
⇒ (a. c) ∈ R1, since R1 being an equivalence relation is also transitive.
Similarly, we can prove that (a, c) ∈ R2 ∴ (a, c) ∈ R1 ∩ R2 So, R1 ∩ R2 is transitive.
Thus R1 ∩ R2 is reflexive, symmetric and also transitive. Thus R1 ∩ R2 is an equivalence relation.

Sponsor Area

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Give an example of a relation which is

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(v) Symmetric and transitive but not reflexive.

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 Determine whether each of the following relations are reflexive, symmetric and transitive :

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(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.