Q. 93.6( 5 Votes )

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 :

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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