Question

Let S be the set of all sets and let R = {(A, B): (A ⊂ B)}, i.e. A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.

Answer 1

(i) Transitive

Consider A, B and C ∈ S, where (A, B) and (B, C) ∈ R

We get

(A, B) ∈ R => A ⊂ B ……. (1)

(B, C) ∈ R => B ⊂ C …….. (2)

Using both the equations we get

A ⊂ C => (A, C) ∈ R

Hence, R is a transitive relation S.

(ii) Non reflexive

We know that

A ⊄ A where (A, A) ∈ R

Hence, R is non reflexive.

(iii) Non symmetric

Consider A ⊂ B where (A, B) ∈ R

We know that B ⊄ A

So (B, A) ∉ R

We get (A, B) ∈ R and (B, A) ∉ R

Hence, R is non symmetric.