Answer :

Let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B, be a relation defined on S.

Now,

Any set is a subset of itself, but not a proper subset.

⇒ (A,A) ∉ R ∀ A ∈ S

⇒ R is not reflexive.

Let (A,B) ∈ R ∀ A, B ∈ S

⇒ A is a proper subset of B

⇒ all elements of A are in B, but B contains at least one element that is not in A.

⇒ B cannot be a proper subset of A

⇒ (B,A) ∉ R

For e.g. , if B = {1,2,5} then A = {1,5} is a proper subset of B . we observe that B is not a proper subset of A.

⇒ R is not symmetric

Let (A,B) ∈ R and (B,C) ∈ R ∀ A, B,C ∈ S

⇒ A is a proper subset of B and B is a proper subset of C

⇒ A is a proper subset of C

⇒ (A,C) ∈ R

For e.g. , if B = {1,2,5} then A = {1,5} is a proper subset of B .

And if C = {1,2,5,7} then B = {1,2,5} is a proper subset of C.

We observe that A = {1,5} is a proper subset of C also.

⇒ R is transitive.

Thus, R is transitive but not reflexive and not symmetric.

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