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.
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.