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Question: For any sets A, B and C, prove that \(\begin{aligned} & [i]\left( A\bigcup B \right)-C=\left( A-...

For any sets A, B and C, prove that
[i](A⋃B)−C=(A−C)⋃(B−C) [ii](A⋂B)−C=(A−C)⋂(B−C) \begin{aligned} & [i]\left( A\bigcup B \right)-C=\left( A-C \right)\bigcup \left( B-C \right) \\\ & [ii]\left( A\bigcap B \right)-C=\left( A-C \right)\bigcap \left( B-C \right) \\\ \end{aligned}

Explanation

Solution

Hint: Prove the R.H.S. of each statement is equal to the L.H.S. of the statement. Use the fact that A−B=A⋂BcA-B=A\bigcap {{B}^{c}}. Use distributive laws and associative laws of union and intersection to simplify R.H.S. and hence prove L.H.S. is equal to R.H.S.

Complete step-by-step answer:
[i] We have R.H.S. =(A−C)⋃(B−C)=\left( A-C \right)\bigcup \left( B-C \right)
We know that A−B=A⋂BcA-B=A\bigcap {{B}^{c}}
Hence, we have R.H.S. =(A⋂Cc)⋃(B⋂Cc)=\left( A\bigcap {{C}^{c}} \right)\bigcup \left( B\bigcap {{C}^{c}} \right)
We know that the intersection of two sets distributes over the union, i.e. A⋂(B⋃C)=(A⋂B)⋃(A⋂C)A\bigcap \left( B\bigcup C \right)=\left( A\bigcap B \right)\bigcup \left( A\bigcap C \right)
Hence, we have
R.H.S. =Cc⋂(A⋃B)={{C}^{c}}\bigcap \left( A\bigcup B \right)
Now, we know that A−B=A⋂BcA-B=A\bigcap {{B}^{c}}
Hence, we have R.H.S. =A⋃B−C=A\bigcup B-C
Hence, RHS = LHS.

[ii] RHS =(A−C)⋂(B−C)=\left( A-C \right)\bigcap \left( B-C \right)
We know that A−B=A⋂BcA-B=A\bigcap {{B}^{c}}
Hence, we have R.H.S. =(A⋂Cc)⋂(B⋂Cc)=\left( A\bigcap {{C}^{c}} \right)\bigcap \left( B\bigcap {{C}^{c}} \right)
We know that the intersection of two sets is associative, i.e. A⋂(B⋂C)=(A⋂B)⋂CA\bigcap \left( B\bigcap C \right)=\left( A\bigcap B \right)\bigcap C
Hence, we have R.H.S. =A⋂(CC⋂(B⋂Cc))=A\bigcap \left( {{C}^{C}}\bigcap \left( B\bigcap {{C}^{c}} \right) \right)
We know that the intersection of two sets is associative, i.e. A⋂(B⋂C)=(A⋂B)⋂CA\bigcap \left( B\bigcap C \right)=\left( A\bigcap B \right)\bigcap C
Hence, we have R.H.S. =A⋂((Cc⋂Cc)⋂B)=A\bigcap \left( \left( {{C}^{c}}\bigcap {{C}^{c}} \right)\bigcap B \right)
We know that A⋂A=AA\bigcap A=A
Hence, we have
R.H.S. =A⋂(Cc⋂B)=A⋂(B⋂Cc)=A\bigcap \left( {{C}^{c}}\bigcap B \right)=A\bigcap \left( B\bigcap {{C}^{c}} \right)
We know that the intersection of two sets is associative, i.e. A⋂(B⋂C)=(A⋂B)⋂CA\bigcap \left( B\bigcap C \right)=\left( A\bigcap B \right)\bigcap C
Hence, we have
R.H.S. =(A⋂B)⋂Cc=\left( A\bigcap B \right)\bigcap {{C}^{c}}
Now, we know that A−B=A⋂BcA-B=A\bigcap {{B}^{c}}
Hence, we have
R.H.S. =A⋂B−C=A\bigcap B-C
Hence, LHS = RHS

Note: Verification using Venn diagrams:
[i] Diagram of A⋃BA\bigcup B

Diagram of A⋃B−CA\bigcup B-C

Diagram of A−CA-C

Diagram of B−CB-C

Diagram of (A−C)⋃(B−C)\left( A-C \right)\bigcup \left( B-C \right)

Hence from Venn diagrams, it is verified that (A⋃B)−C=(A−C)⋃(B−C)\left( A\bigcup B \right)-C=\left( A-C \right)\bigcup \left( B-C \right)
Similarly, it can be verified from Venn diagrams that (A⋂B)−C=(A−C)⋂(B−C)\left( A\bigcap B \right)-C=\left( A-C \right)\bigcap \left( B-C \right)