If \[{{R}_{1}}\] and \[{{R}_{2}}\] be two equivalence relations on a set A, prove that \[{{R}_{1}}\cap {{R}_{2}}\] is also an equivalence relation on A. |
OR |
Let X be a non-empty set and P(X) be its power set. Let '*' be an operation defined on elements of P(X) by \[A*B=A\cap B,\] \[\forall A,\] \[B\in P(X).\] then, |
(i) Prove that ?*? is a biliary operation in P(X), |
(ii) Is * commutative? |
(iii) Is * associative? |
(iv) Find the identity element in P(X) w.r.t. ?*?. |
(v) If o is another binary operation defined on P(X) as \[AoB=A\cup B,\] then verify that O distributes itself over ?*?. |
Answer:
Let \[{{R}_{1}}\] and \[{{R}_{2}}\] be two equivalence relations on a set A. Then, \[{{R}_{1}}\subseteq A\times A,\] \[{{R}_{2}}\subseteq A\times A\] \[\Rightarrow \] \[({{R}_{1}}\cap {{R}_{2}})\subseteq A\times A.\] So, \[({{R}_{1}}\cap {{R}_{2}})\] is a relation on A. This relation on A satisfies the following properties. (i) Reflexivity \[{{R}_{1}}\] is reflexive and \[{{R}_{2}}\] is reflexive \[\Rightarrow \] \[(a,\,\,a)\in {{R}_{1}}\] and \[(a,\,\,a)\in {{R}_{2}}\] for all \[a\in A\] \[\Rightarrow \] \[(a,\,\,a)\in {{R}_{1}}\cap {{R}_{2}}\] for all \[a\in A\] \[\Rightarrow \] \[{{R}_{1}}\cap {{R}_{2}}\] is reflexive. (ii) Symmetry Let (a, b) be an arbitrary element of \[{{R}_{1}}\cap {{R}_{2}}.\]then, \[(a,\,\,b)\in {{R}_{1}}\cap {{R}_{2}}\] \[\Rightarrow \] \[(a,\,\,b)\in R,\] and \[(a,\,\,b)\in {{R}_{2}}\] \[\Rightarrow \] \[(b,\,\,a)\in {{R}_{1}}\] and \[(b,\,\,a)\in {{R}_{2}}\] \[[\because \,\,{{R}_{1}}\]is symmetric and \[{{R}_{2}}\] is symmetric] \[\Rightarrow \] \[(b,\,\,a)\in {{R}_{1}}\cap {{R}_{2}}\] This shows that \[{{R}_{1}}\cap {{R}_{2}}\] is symmetric (iii) Transitivity \[(a,\,\,b)\in {{R}_{1}}\cap {{R}_{2}}\,\,and\,\,(b,\,\,c)\in {{R}_{1}}\cap {{R}_{2}}\] \[\Rightarrow \] \[(a,\,\,b)\in {{R}_{1}},\] \[(a,b)\in {{R}_{2}}\] and \[\,(b,\,\,c)\in {{R}_{1}},\] \[\,(b,\,\,c)\in {{R}_{2}}\] \[\Rightarrow \] \[\{(a,\,\,b)\in {{R}_{1}},(b,\,\,c)\in {{R}_{1}}\},\] and \[\{(a,\,\,b)\in {{R}_{2}},(b,\,\,c)\in {{R}_{2}}\] \[\Rightarrow \] \[\{a,\,\,c\}\in {{R}_{1}}\,\,and\,\,(a,\,\,c)\in {{R}_{2}}\] \[[\because \,\,{{R}_{1}}\]is transitive and \[{{R}_{2}}\] is transitive] \[\Rightarrow \] \[(a,\,\,c)\in {{R}_{1}}\cap {{R}_{2}}\] This shows that \[({{R}_{1}}\cap {{R}_{2}})\] is transitive. Thus, \[{{R}_{1}}\cap {{R}_{2}}\] is reflexive, symmetric and transitive. Hence, \[{{R}_{1}}\cap {{R}_{2}}\] is an equivalence relation. OR We have, \[A*B=A\cap B,\] \[\forall A,\] \[B\in P(X)\] (i) Let \[A,\,\,B\in P(X)\] be any arbitrary elements. Then, \[A\subset X\,\,\text{and}\,\,B\subset X\] \[\Rightarrow \] \[A\cap B\subset X\] \[\Rightarrow \] \[A\cap B\in P(X)\] \[\Rightarrow \] \[A*B\in P(X)\] \[\therefore \] \[*\] is a binary operation in P(X) (ii) Let \[A,\,\,B\in P(X)\] be any arbitrary elements. Then, \[A*B=A\cap B=B\cap A=B*A\] Thus, \[A*B=B*A\,\,\forall \,\,A,\] \[B\in P(X)\] Hence, \[*\] is commutative. (iii) Let \[A,\,\,B,\,\,C\in P(X)\] be any arbitrary elements. Then, \[(A*B)*C=(A\cap B)\cap C\] \[=A\cap B\cap C=A\cap (B\cap C)=A*(B*C)\] Thus, \[=(A*B)*C=A*(B*C),\] \[\forall A,\,\,B,\,\,C\in P(X)\] Hence, \[*\] is commutative. (iv) Let, if possible, E be an identity element in P(X). Then, \[A*E=A=E*A,\] \[\forall \,\,A\in P(X)\] \[\Rightarrow \] \[4x+3y<110,\] \[\forall \,\,A\in P(X)\] \[\Rightarrow \] \[A\subset E\,\,\forall \,\,A\in P(X)\] This is possible only when E = X. Also, \[X\in P(X)\] Thus, X is the identity element. (v) Let A be an invertible element of P(X). Then, there exist an element B in P(X) such that \[A\cap B=X=B\cap A\] This is possible only when A = B = X Thus, X is the only invertible element in P(X) and \[{{X}^{-1}}=X\] (vi) Let \[A,\,\,B,\,\,C\in P(X)\] be any arbitrary elements. To verify \[Ao(B*C)\text{ }=\text{ (}AoB\text{)}*(AoC)\] Consider \[LHS=Ao(B*C)\text{ }=\text{ A}\cup \text{(}B\cap C\text{)}\] \[=(A\cup B)\cap (A\cup C)\] \[=(AoB)*(AoC)=RHS\] Hence, 'o' distributes itself over *.
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