LetX={1,2,3}P(X)=
Power set of X=Set of all subsets of X.={ϕ,{1}},{2},{3},{1,2},{2,3},{1,3},{1,2,3}}Since{1}⊂{1,2}{1}R{1,2}
∵ofABCD,allelementsofAareinB
ARB means A⊂B
here,relation is R={(A,B):AandBaresets,ACB}
Since every set is a subset of itself.
ACA∴(A,A)ϵR,R is reflexive.
To check whether symmetric or not,
If(A,B)ϵ,then(B,A)ϵR.
If(A,B)ϵR,A⊂BbutB⊂Aisnottrue.eg:−LetA={1}andB={1,2}
As all elements of A are in BA⊂B.
But all elements of B are not in A
∴B⊂A is not true.
∴R is not symmetric.
Since (A,B)ϵRand(B,C)ϵRofA⊂BandB⊂C
then A⊂B
⇒(A,C)ϵRSo,If(A,B)ϵRand(B,C)ϵR,then(A,C)ϵR
∴R is transitive.
hence R is reflexive and transitive but not symmetric.
hence, R is not an equivalence relation since it is not symmetric.