\displaystyle \textbf{Question 1: Find the smallest set }A\text{ such that }A\cup\{1,2\}=\{1,2,3,5,9\}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Since }\{1,2\}\text{ already contains }1\text{ and }2,
\displaystyle \text{the smallest set }A\text{ must contain only the remaining elements.}
\displaystyle A=\{1,2,3,5,9\}-\{1,2\}=\{3,5,9\}.
\displaystyle \\

\displaystyle \textbf{Question 2: Let }A=\{1,2,4,5\},\;B=\{2,3,5,6\}\text{ and }C=\{4,5,6,7\}.
\displaystyle \text{Verify the following identities:}
\displaystyle \text{(i) }A\cup(B\cap C)=(A\cup B)\cap(A\cup C)
\displaystyle \text{(ii) }A\cap(B\cup C)=(A\cap B)\cup(A\cap C)
\displaystyle \text{(iii) }A\cap(B-C)=(A\cap B)-(A\cap C)
\displaystyle \text{(iv) }A-(B\cup C)=(A-B)\cap(A-C)
\displaystyle \text{(v) }A-(B\cap C)=(A-B)\cup(A-C)
\displaystyle \text{(vi) }A\cap(B\triangle C)=(A\cap B)\triangle(A\cap C)
\displaystyle \textbf{Answer:}
\displaystyle \text{Given, }A=\{1,2,4,5\},\quad B=\{2,3,5,6\},\quad C=\{4,5,6,7\}.

\displaystyle \text{(i) }B\cap C=\{5,6\}.
\displaystyle A\cup(B\cap C)=\{1,2,4,5\}\cup\{5,6\}=\{1,2,4,5,6\}.
\displaystyle A\cup B=\{1,2,3,4,5,6\}.
\displaystyle A\cup C=\{1,2,4,5,6,7\}.
\displaystyle (A\cup B)\cap(A\cup C)=\{1,2,4,5,6\}.
\displaystyle \therefore A\cup(B\cap C)=(A\cup B)\cap(A\cup C).

\displaystyle \text{(ii) }B\cup C=\{2,3,4,5,6,7\}.
\displaystyle A\cap(B\cup C)=\{1,2,4,5\}\cap\{2,3,4,5,6,7\}=\{2,4,5\}.
\displaystyle A\cap B=\{2,5\},\qquad A\cap C=\{4,5\}.
\displaystyle (A\cap B)\cup(A\cap C)=\{2,5\}\cup\{4,5\}=\{2,4,5\}.
\displaystyle \therefore A\cap(B\cup C)=(A\cap B)\cup(A\cap C).

\displaystyle \text{(iii) }B-C=\{2,3\}.
\displaystyle A\cap(B-C)=\{1,2,4,5\}\cap\{2,3\}=\{2\}.
\displaystyle A\cap B=\{2,5\},\qquad A\cap C=\{4,5\}.
\displaystyle (A\cap B)-(A\cap C)=\{2,5\}-\{4,5\}=\{2\}.
\displaystyle \therefore A\cap(B-C)=(A\cap B)-(A\cap C).

\displaystyle \text{(iv) }B\cup C=\{2,3,4,5,6,7\}.
\displaystyle A-(B\cup C)=\{1,2,4,5\}-\{2,3,4,5,6,7\}=\{1\}.
\displaystyle A-B=\{1,4\},\qquad A-C=\{1,2\}.
\displaystyle (A-B)\cap(A-C)=\{1,4\}\cap\{1,2\}=\{1\}.
\displaystyle \therefore A-(B\cup C)=(A-B)\cap(A-C).

\displaystyle \text{(v) }B\cap C=\{5,6\}.
\displaystyle A-(B\cap C)=\{1,2,4,5\}-\{5,6\}=\{1,2,4\}.
\displaystyle A-B=\{1,4\},\qquad A-C=\{1,2\}.
\displaystyle (A-B)\cup(A-C)=\{1,4\}\cup\{1,2\}=\{1,2,4\}.
\displaystyle \therefore A-(B\cap C)=(A-B)\cup(A-C).

\displaystyle \text{(vi) }B-C=\{2,3\}\text{ and }C-B=\{4,7\}.
\displaystyle B\triangle C=(B-C)\cup(C-B)=\{2,3,4,7\}.
\displaystyle A\cap(B\triangle C)=\{1,2,4,5\}\cap\{2,3,4,7\}=\{2,4\}.
\displaystyle A\cap B=\{2,5\},\qquad A\cap C=\{4,5\}.
\displaystyle (A\cap B)-(A\cap C)=\{2\}.
\displaystyle (A\cap C)-(A\cap B)=\{4\}.
\displaystyle (A\cap B)\triangle(A\cap C)=\{2\}\cup\{4\}=\{2,4\}.
\displaystyle \therefore A\cap(B\triangle C)=(A\cap B)\triangle(A\cap C).
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{If }U=\{2,3,5,7,9\}\text{ is the universal set and }A=\{3,7\},
\displaystyle \text{ }B=\{2,5,7,9\},\text{ then prove that: (i) }(A\cup B)'=A'\cap B'
\displaystyle \text{(ii) }(A\cap B)'=A'\cup B'.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }U=\{2,3,5,7,9\},\ A=\{3,7\}\text{ and }B=\{2,5,7,9\}.
\displaystyle A\cup B=\{3,7\}\cup\{2,5,7,9\}=\{2,3,5,7,9\}=U.
\displaystyle (A\cup B)'=\emptyset.
\displaystyle A'=U-A=\{2,5,9\}.
\displaystyle B'=U-B=\{3\}.
\displaystyle A'\cap B'=\{2,5,9\}\cap\{3\}=\emptyset.
\displaystyle \therefore (A\cup B)'=A'\cap B'.
\displaystyle \\

\displaystyle \text{(ii) Given }U=\{2,3,5,7,9\},\ A=\{3,7\}\text{ and }B=\{2,5,7,9\}.
\displaystyle A\cap B=\{3,7\}\cap\{2,5,7,9\}=\{7\}.
\displaystyle (A\cap B)'=U-\{7\}=\{2,3,5,9\}.
\displaystyle A'=U-A=\{2,5,9\}.
\displaystyle B'=U-B=\{3\}.
\displaystyle A'\cup B'=\{2,5,9\}\cup\{3\}=\{2,3,5,9\}.
\displaystyle \therefore (A\cap B)'=A'\cup B'.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{For any two sets }A\text{ and }B,\text{ prove that}
\displaystyle \text{(i) }B\subset A\cup B\qquad\text{(ii) }A\cap B\subset A\qquad\text{(iii) }A\subset B\Rightarrow A\cap B=A.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }x\in B.
\displaystyle \therefore x\in A\text{ or }x\in B.
\displaystyle \therefore x\in A\cup B\text{ (Definition of union of sets).}
\displaystyle \therefore B\subset A\cup B.
\displaystyle \\

\displaystyle \text{(ii) Let }x\in A\cap B.
\displaystyle \therefore x\in A\text{ and }x\in B\text{ (Definition of intersection of sets).}
\displaystyle \therefore x\in A.
\displaystyle \therefore A\cap B\subset A.
\displaystyle \\

\displaystyle \text{(iii) Given }A\subset B.
\displaystyle \text{Let }x\in A\cap B.
\displaystyle \therefore x\in A\text{ and }x\in B.
\displaystyle \therefore x\in A.
\displaystyle \therefore A\cap B\subset A.
\displaystyle \text{Now let }x\in A.
\displaystyle \text{Since }A\subset B,\ x\in B.
\displaystyle \therefore x\in A\cap B.
\displaystyle \therefore A\subset A\cap B.
\displaystyle \therefore A\cap B=A.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{For any two sets }A\text{ and }B,\text{ show that the following statements}
\displaystyle \text{are equivalent:}
\displaystyle \text{(i) }A\subset B\qquad\text{(ii) }A-B=\emptyset\qquad\text{(iii) }A\cup B=B
\displaystyle \text{(iv) }A\cap B=A.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that the statements are equivalent, it is sufficient to prove}
\displaystyle \text{(i)}\Rightarrow\text{(ii)},\quad\text{(ii)}\Rightarrow\text{(iii)},\quad\text{(iii)}\Rightarrow\text{(iv)}\quad\text{and}\quad\text{(iv)}\Rightarrow\text{(i)}.
\displaystyle \text{(i)}\Rightarrow\text{(ii):}\text{ Let }A\subset B.
\displaystyle \text{Suppose }x\in A-B.
\displaystyle \therefore x\in A\text{ and }x\notin B.
\displaystyle \text{But }A\subset B\text{ implies that }x\in A\Rightarrow x\in B,
\displaystyle \text{which is a contradiction.}
\displaystyle \therefore \text{No element belongs to }A-B.
\displaystyle \therefore A-B=\emptyset.
\displaystyle \\

\displaystyle \text{(ii)}\Rightarrow\text{(iii):}\text{ Let }A-B=\emptyset.
\displaystyle \text{Let }x\in A.
\displaystyle \text{Since }A-B=\emptyset,\ x\notin A-B.
\displaystyle \therefore x\in B.
\displaystyle \therefore A\subset B.
\displaystyle \text{Now }A\subset A\cup B\text{ and }B\subset A\cup B.
\displaystyle \text{Since }A\subset B,\text{ every element of }A\cup B\text{ belongs to }B.
\displaystyle \therefore A\cup B\subset B.
\displaystyle \text{Also, }B\subset A\cup B.
\displaystyle \therefore A\cup B=B.
\displaystyle \\

\displaystyle \text{(iii)}\Rightarrow\text{(iv):}\text{ Let }A\cup B=B.
\displaystyle \text{Let }x\in A.
\displaystyle \therefore x\in A\cup B.
\displaystyle \text{Since }A\cup B=B,\ x\in B.
\displaystyle \therefore x\in A\cap B.
\displaystyle \therefore A\subset A\cap B.
\displaystyle \text{Also, }A\cap B\subset A.
\displaystyle \therefore A\cap B=A.
\displaystyle \\

\displaystyle \text{(iv)}\Rightarrow\text{(i):}\text{ Let }A\cap B=A.
\displaystyle \text{Let }x\in A.
\displaystyle \text{Since }A=A\cap B,\ x\in A\cap B.
\displaystyle \therefore x\in A\text{ and }x\in B.
\displaystyle \therefore x\in B.
\displaystyle \therefore A\subset B.
\displaystyle \therefore \text{The four statements are equivalent.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{For three sets }A,\ B\text{ and }C,\text{ show that}
\displaystyle \text{(i) }A\cap B=A\cap C\text{ need not imply }B=C
\displaystyle \text{(ii) }A\subset B\Rightarrow C-B\subset C-A.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }A=\{1,2,3\},\ B=\{2,4,6\}\text{ and }C=\{2,5,7\}.
\displaystyle A\cap B=\{1,2,3\}\cap\{2,4,6\}=\{2\}.
\displaystyle A\cap C=\{1,2,3\}\cap\{2,5,7\}=\{2\}.
\displaystyle \therefore A\cap B=A\cap C.
\displaystyle \text{However, }B=\{2,4,6\}\neq\{2,5,7\}=C.
\displaystyle \therefore A\cap B=A\cap C\text{ need not imply }B=C.
\displaystyle \\

\displaystyle \text{(ii) Given }A\subset B.
\displaystyle \text{Let }x\in C-B.
\displaystyle \therefore x\in C\text{ and }x\notin B\text{ (Definition of difference of sets).}
\displaystyle \text{Since }A\subset B,\ x\notin B\Rightarrow x\notin A.
\displaystyle \therefore x\in C\text{ and }x\notin A.
\displaystyle \therefore x\in C-A.
\displaystyle \therefore C-B\subset C-A.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{For any two sets }A\text{ and }B,\text{ prove that:}
\displaystyle \text{(i) }A\cup(A\cap B)=A\qquad\text{(ii) }A\cap(A\cup B)=A.
\displaystyle \text{Answer:}
\displaystyle \text{(i) To prove }A\cup(A\cap B)=A,\text{ we prove both inclusions.}
\displaystyle \text{Let }x\in A\cup(A\cap B).
\displaystyle \therefore x\in A\text{ or }x\in A\cap B.
\displaystyle \text{If }x\in A\cap B,\text{ then }x\in A.
\displaystyle \therefore x\in A.
\displaystyle \therefore A\cup(A\cap B)\subset A.
\displaystyle \text{Also, if }x\in A,\text{ then }x\in A\cup(A\cap B).
\displaystyle \therefore A\subset A\cup(A\cap B).
\displaystyle \therefore A\cup(A\cap B)=A.
\displaystyle \\

\displaystyle \text{(ii) To prove }A\cap(A\cup B)=A,\text{ we prove both inclusions.}
\displaystyle \text{Let }x\in A\cap(A\cup B).
\displaystyle \therefore x\in A\text{ and }x\in A\cup B.
\displaystyle \therefore x\in A.
\displaystyle \therefore A\cap(A\cup B)\subset A.
\displaystyle \text{Also, let }x\in A.
\displaystyle \therefore x\in A\cup B.
\displaystyle \therefore x\in A\text{ and }x\in A\cup B.
\displaystyle \therefore x\in A\cap(A\cup B).
\displaystyle \therefore A\subset A\cap(A\cup B).
\displaystyle \therefore A\cap(A\cup B)=A.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find sets }A,\ B\text{ and }C\text{ such that }A\cap B,\ A\cap C\text{ and}
\displaystyle \text{ }B\cap C\text{ are non-empty sets, but }A\cap B\cap C=\emptyset.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\{5,6,10\},\ B=\{6,8,9\}\text{ and }C=\{9,10,11\}.
\displaystyle A\cap B=\{6\}\neq\emptyset.
\displaystyle A\cap C=\{10\}\neq\emptyset.
\displaystyle B\cap C=\{9\}\neq\emptyset.
\displaystyle A\cap B\cap C=\emptyset.
\displaystyle \therefore A,\ B\text{ and }C\text{ satisfy all the required conditions.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{For any two sets }A\text{ and }B,\text{ prove that}
\displaystyle A\cap B=\emptyset\Rightarrow A\subseteq B'.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A\cap B=\emptyset.
\displaystyle \text{Let }x\in A.
\displaystyle \text{Since }A\cap B=\emptyset,\ x\notin B.
\displaystyle \therefore x\in B'.
\displaystyle \therefore x\in A\Rightarrow x\in B'.
\displaystyle \therefore A\subseteq B'.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }A\text{ and }B\text{ are sets, prove that }A-B,\ A\cap B\text{ and}
\displaystyle B-A\text{ are pairwise disjoint.}
\displaystyle \text{Answer:}
\displaystyle \text{We need to prove that}
\displaystyle (A-B)\cap(A\cap B)=\emptyset,
\displaystyle (A\cap B)\cap(B-A)=\emptyset
\displaystyle \text{and }(A-B)\cap(B-A)=\emptyset.
\displaystyle \text{First, suppose }x\in(A-B)\cap(A\cap B).
\displaystyle \therefore x\in A-B\text{ and }x\in A\cap B.
\displaystyle x\in A-B\Rightarrow x\in A\text{ and }x\notin B.
\displaystyle x\in A\cap B\Rightarrow x\in A\text{ and }x\in B.
\displaystyle \text{This is a contradiction.}
\displaystyle \therefore (A-B)\cap(A\cap B)=\emptyset.
\displaystyle \\

\displaystyle \text{Next, suppose }x\in(A\cap B)\cap(B-A).
\displaystyle \therefore x\in A\cap B\text{ and }x\in B-A.
\displaystyle x\in A\cap B\Rightarrow x\in A\text{ and }x\in B.
\displaystyle x\in B-A\Rightarrow x\in B\text{ and }x\notin A.
\displaystyle \text{This is a contradiction.}
\displaystyle \therefore (A\cap B)\cap(B-A)=\emptyset.
\displaystyle \\

\displaystyle \text{Finally, suppose }x\in(A-B)\cap(B-A).
\displaystyle \therefore x\in A-B\text{ and }x\in B-A.
\displaystyle x\in A-B\Rightarrow x\in A\text{ and }x\notin B.
\displaystyle x\in B-A\Rightarrow x\in B\text{ and }x\notin A.
\displaystyle \text{This is a contradiction.}
\displaystyle \therefore (A-B)\cap(B-A)=\emptyset.
\displaystyle \therefore A-B,\ A\cap B\text{ and }B-A\text{ are pairwise disjoint.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Using properties of sets, show that for any two sets }A\text{ and }B,
\displaystyle (A\cup B)\cap(A\cup B')=A.
\displaystyle \text{Answer:}
\displaystyle (A\cup B)\cap(A\cup B')=A\cup(B\cap B')
\displaystyle \text{(By the distributive law).}
\displaystyle =A\cup\emptyset\qquad\text{(Since }B\cap B'=\emptyset\text{).}
\displaystyle =A\qquad\text{(Since }A\cup\emptyset=A\text{).}
\displaystyle \therefore (A\cup B)\cap(A\cup B')=A.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{For any two sets }A\text{ and }B,\text{ prove that:}
\displaystyle \text{(i) }A'\cup B=U\Rightarrow A\subset B\qquad\text{(ii) }B'\subset A'\Rightarrow A\subset B.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }A'\cup B=U.
\displaystyle \text{Let }x\in A.
\displaystyle \therefore x\notin A'.
\displaystyle \text{Since }x\in U\text{ and }U=A'\cup B,\ x\in A'\cup B.
\displaystyle \therefore x\in A'\text{ or }x\in B.
\displaystyle \text{But }x\notin A'.
\displaystyle \therefore x\in B.
\displaystyle \therefore x\in A\Rightarrow x\in B.
\displaystyle \therefore A\subset B.
\displaystyle \\

\displaystyle \text{(ii) Given }B'\subset A'.
\displaystyle \text{Let }x\in A.
\displaystyle \therefore x\notin A'.
\displaystyle \text{Suppose }x\notin B.
\displaystyle \therefore x\in B'.
\displaystyle \text{Since }B'\subset A',\ x\in A'.
\displaystyle \text{This contradicts }x\notin A'.
\displaystyle \therefore x\in B.
\displaystyle \therefore x\in A\Rightarrow x\in B.
\displaystyle \therefore A\subset B.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Is it true that for any sets }A\text{ and }B,
\displaystyle P(A)\cup P(B)=P(A\cup B)?\text{ Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{The statement is false.}
\displaystyle \text{Consider }A=\{1\}\text{ and }B=\{2\}.
\displaystyle P(A)=\{\emptyset,\{1\}\},\qquad P(B)=\{\emptyset,\{2\}\}.
\displaystyle \therefore P(A)\cup P(B)=\{\emptyset,\{1\},\{2\}\}.
\displaystyle A\cup B=\{1,2\}.
\displaystyle P(A\cup B)=\{\emptyset,\{1\},\{2\},\{1,2\}\}.
\displaystyle \text{Since }\{1,2\}\in P(A\cup B)\text{ but }\{1,2\}\notin P(A)\cup P(B),
\displaystyle P(A)\cup P(B)\neq P(A\cup B).
\displaystyle \therefore \text{The given statement is false.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Show that for any sets }A\text{ and }B,
\displaystyle \text{(i) }A=(A\cap B)\cup(A-B)\qquad\text{(ii) }A\cup(B-A)=A\cup B.
\displaystyle \text{Answer:}
\displaystyle \text{(i) To prove }A=(A\cap B)\cup(A-B),\text{ we prove both inclusions.}
\displaystyle \text{Let }x\in A.
\displaystyle \text{Either }x\in B\text{ or }x\notin B.
\displaystyle \text{If }x\in B,\text{ then }x\in A\cap B.
\displaystyle \text{If }x\notin B,\text{ then }x\in A-B.
\displaystyle \therefore x\in(A\cap B)\cup(A-B).
\displaystyle \therefore A\subseteq(A\cap B)\cup(A-B).
\displaystyle \text{Now let }x\in(A\cap B)\cup(A-B).
\displaystyle \therefore x\in A\cap B\text{ or }x\in A-B.
\displaystyle \text{In either case, }x\in A.
\displaystyle \therefore (A\cap B)\cup(A-B)\subseteq A.
\displaystyle \therefore A=(A\cap B)\cup(A-B).
\displaystyle \\

\displaystyle \text{(ii) To prove }A\cup(B-A)=A\cup B,\text{ we prove both inclusions.}
\displaystyle \text{Let }x\in A\cup(B-A).
\displaystyle \therefore x\in A\text{ or }x\in B-A.
\displaystyle \text{If }x\in B-A,\text{ then }x\in B.
\displaystyle \therefore x\in A\text{ or }x\in B.
\displaystyle \therefore x\in A\cup B.
\displaystyle \therefore A\cup(B-A)\subseteq A\cup B.
\displaystyle \text{Now let }x\in A\cup B.
\displaystyle \therefore x\in A\text{ or }x\in B.
\displaystyle \text{If }x\in A,\text{ then }x\in A\cup(B-A).
\displaystyle \text{If }x\in B\text{ and }x\notin A,\text{ then }x\in B-A.
\displaystyle \text{If }x\in B\text{ and }x\in A,\text{ then }x\in A.
\displaystyle \therefore x\in A\cup(B-A).
\displaystyle \therefore A\cup B\subseteq A\cup(B-A).
\displaystyle \therefore A\cup(B-A)=A\cup B.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Each set }X_r\text{ contains }5\text{ elements and each set }Y_r\text{ contains}
\displaystyle 2\text{ elements, where }\bigcup_{r=1}^{20}X_r=S=\bigcup_{r=1}^{n}Y_r.
\displaystyle \text{If each element of }S\text{ belongs to exactly }10\text{ of the sets }X_r\text{ and exactly}
\displaystyle 4\text{ of the sets }Y_r,\text{ find the value of }n.
\displaystyle \text{Answer:}
\displaystyle \text{Each of the }20\text{ sets }X_r\text{ contains }5\text{ elements.}
\displaystyle \therefore \text{Total number of element-set incidences among the sets }X_r=20\times5=100.
\displaystyle \text{Each element of }S\text{ belongs to exactly }10\text{ of the sets }X_r.
\displaystyle \therefore 10\,n(S)=100.
\displaystyle \therefore n(S)=10.
\displaystyle \text{Each of the }n\text{ sets }Y_r\text{ contains }2\text{ elements.}
\displaystyle \therefore \text{Total number of element-set incidences among the sets }Y_r=2n.
\displaystyle \text{Each element of }S\text{ belongs to exactly }4\text{ of the sets }Y_r.
\displaystyle \therefore 4\,n(S)=2n.
\displaystyle 4\times10=2n.
\displaystyle 40=2n.
\displaystyle \therefore n=20.
\displaystyle \\

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