Question 1:  Find the smallest set $A$ such that $A \cup \{ 1, 2 \} = \{ 1, 2, 3, 5, 9 \}$

Smallest set $A = \{ 1, 2, 3, 5, 9 \} - \{ 1, 2 \} = \{ 3, 5, 9 \}$

$\\$

Question 2: Let $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$. Verify the following identities:

(i) $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$   (ii) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

(iii) $A \cap (B -C) =(A \cap B) -(A \cap C)$   (iv) $A - (B \cup C) =(A -B) \cap (A-C)$

(v) $A -(B \cap C) =(A -B) \cup (A-C)$   (vi) $A \cap (B \triangle C) =(A \cap B) \triangle (A \cap C)$

(i)  Given $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$

$A \cup (B \cap C) = \{ 1,2,4,5 \} \cup (\{ 2,3,5,6 \} \cap \{ 4,5,6,7 \}) = \{ 1,2,4,5 \} \cup \{ 5,6 \}$ $= \{ 1, 2, 4, 5, 6 \}$

$(A \cup B) \cap (A \cup C) = (\{ 1,2,4,5 \} \cup \{ 2,3,5,6 \}) \cap ( \{ 1,2,4,5 \} \cup \{ 4,5,6,7 \})$ $= \{ 1,2,3, 4,5,6 \} \cap \{ 1, 2, 4, 5, 6, 7 \} = \{ 1, 2, 4, 5, 6 \}$

Hence $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

(ii)  Given $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$

$A \cap (B \cup C) = \{ 1,2,4,5 \} \cap (\{ 2,3,5,6 \} \cup \{ 4,5,6,7 \} )$ $= \{ 1,2,4,5 \} \cap \{ 2,3 ,4,5,6,7 \} = \{ 2, 4, 5 \}$

$(A \cap B) \cup (A \cap C) = (\{ 1,2,4,5 \} \cap \{ 2,3,5,6 \}) \cup (\{ 1,2,4,5 \} \cap \{ 4,5,6,7 \})$ $= \{ 2, 5 \} \cup \{ 4, 5 \} = \{ 2, 4, 5 \}$

Hence $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

(iii) Given $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$

$B - C = \{ 2,3,5,6 \} - \{ 4,5,6,7 \} = \{ 2, 3 \}$

$A \cap (B -C) = \{ 1,2,4,5 \} \cap \{ 2, 3 \} = \{ 2 \}$

$A \cap B = \{ 1,2,4,5 \} \cap \{ 2,3,5,6 \} = \{ 2, 5 \}$

$A \cap C = \{ 1,2,4,5 \} \cap \{ 4,5,6,7 \} = \{ 4, 5 \}$

$A \cap B - A \cap C = \{ 2 \}$

Hence $A \cap (B -C) =(A \cap B) -(A \cap C)$

(iv) $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$

$B \cup C = \{ 2,3,5,6 \} \cup \{ 4,5,6,7 \} = \{2, 3, 4, 5, 6, 7 \}$

$A - (B \cup C) = \{ 1,2,4,5 \} - \{2, 3, 4, 5, 6, 7 \} = \{ 1 \}$

$A - B = \{ 1,2,4,5 \} - \{ 2,3,5,6 \} = \{1, 4 \}$

$A - C = \{ 1,2,4,5 \} - \{ 4,5,6,7 \} = \{ 1, 2 \}$

$(A -B) \cap (A-C) = \{1, 4 \} - \{ 1, 2 \} = \{1 \}$

Hence $A - (B \cup C) =(A -B) \cap (A-C)$

(v)   Given $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$

$B \cap C = \{ 2,3,5,6 \} \cap \{ 4,5,6,7 \} = \{5, 6 \}$

$A -(B \cap C) = \{ 1,2,4,5 \} - \{5, 6 \} = \{1, 2, 4 \}$

$A-C = \{ 1,2,4,5 \} - \{ 4,5,6,7 \} = \{ 1, 2 \}$

$A - B = \{ 1,2,4,5 \} - \{ 2,3,5,6 \} = \{1, 4 \}$

$(A -B) \cup (A-C) = \{1, 4 \} - \{ 1, 2 \} = \{1, 2, 4 \}$

Hence $A -(B \cap C) =(A -B) \cup (A-C)$

(vi) Given $A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}$

$A \cap C = \{ 1,2,4,5 \} \cap \{ 4 ,5,6,7 \} = \{ 4, 5 \} = X$

$A \cap B = \{ 1,2,4,5 \} \cap \{ 2,3,5,6 \} = \{ 2, 5 \} = Y$

$Y - X = \{ 2, 5 \} - \{ 4, 5 \} =\{ 2 \}$

$X - Y = \{ 4, 5 \} - \{ 2, 5 \} = \{ 4 \}$

$(A \cap B) \triangle (A \cap C) = Y \triangle X = (Y-X) \cup (X-Y) = \{ 2 \} \cup \{ 4 \} = \{ 2, 4 \}$

$B \triangle C = ( B - C ) \cup ( C - B ) = ( \{ 2,3,5,6 \} - \{ 4,5,6,7 \} ) \cup ( \{ 4,5,6,7 \} - \{ 2,3,5,6 \} ) = \{ 2,3 \} \cup \{ 4, 7 \} = \{ 2,3,4,7 \}$

$A \cap B \triangle C = \{ 1, 2, 4, 5 \} \cap \{ 2, 3, 4, 7 \} = \{ 2, 4 \}$

Hence $A \cap (B \triangle C) = (A \cap B) \triangle (A \cap C)$

$\\$

Question 3: If $U= \{ 2,3,5,7,9 \}$ is the universal set and $A= \{3,7\}$, $B= \{2,5,7,9 \}$, then prove that: (i) $(A \cup B)'= A' \cap B'$    (ii) $(A \cap B)' = A' \cup B'$

(i)   Given $U= \{ 2,3,5,7,9 \}$,$A= \{3,7\}$ and $B= \{2,5,7,9 \}$

$A \cup B = \{3,7\} \cup\{2,5,7,9 \} = \{ 2, 3, 5, 7, 9 \}$

$(A \cup B)' = \phi$

$A' = \{ 2, 5, 9 \}$

$B' = \{ 3 \}$

$A' \cap B' = \phi$

Hence $(A \cup B)'= A' \cap B'$

(ii)  Given $U= \{ 2,3,5,7,9 \}$,$A= \{3,7\}$ and $B= \{2,5,7,9 \}$

$A' = \{ 2, 5, 9 \}$

$B' = \{ 3 \}$

$A' \cup B' = \{ 2, 3, 5, 9 \}$

$A \cap B = \{3,7\} \cap\{2,5,7,9 \} = \{ 7 \}$

$(A \cap B)' = \{ 2, 3, 5, 9 \}$

Hence $(A \cap B)' = A' \cup B'$

$\\$

Question 4: For any two sets $A$ and $B$, prove that

(i) $B \subset A \cup B$ (ii) $A \cap B \subset A$    (iii) $A \subset B \Rightarrow A \cap B=A$

(i)   $B \subset A \cup B$

For all $x \in B$

$\Rightarrow x \in A$ or $x \in B$

$\Rightarrow x \in A \cup B$ (Definition of union of sets)

$\Rightarrow B \subset A \cup B$

(ii) $A \cap B \subset A$

Fora all $x \in A \cap B$

$\Rightarrow x \in A$ and $x \in B$ (definition of intersection of sets)

$\Rightarrow x \in A$

(iii) $A \subset B \Rightarrow A \cap B=A$

Let $x \in A \subset B$

$\Rightarrow x \in B$

Let $x \in A \cap B$

$\Rightarrow x \in A$ and $x \in B$

$\Rightarrow x \in A$ and $x \in B$ (as $A \subset B$)

$\therefore A \cap B = A$

$\\$

Question 5: For any two sets $A$ and $B$, show that the following statements are equivalent:

(i) $A \subset B (ii) A-B = \phi$    (iii) $A \cup B = B (iv) A \cap B=A$

In order to show that the four statements are equivalent, we need to show that $i) \Rightarrow ii)$ ,   $ii) \Rightarrow iii)$$iii) \Rightarrow iv)$ and  $iv) \Rightarrow i)$

i)    Let us assume $A \subset B$

Let $x \in A$, since $A \subset B, x \in B$

Now $A - B \Rightarrow x \in B$ and $x \notin B$

Since $A \subset B$, each element of $A$ is an element of $B$

$\Rightarrow A - B = \phi$

ii)   Let $A - B = \phi$

$\Rightarrow x \in A, x \in B$

Therefore every element of $A$ is also an element of $B$.

Therefore $A \cup B = B$

iii)  Let us assume $A \cup B = B$

$\Rightarrow x \in B$ and $x \in B$

Therefore $A \cap B = A$ since every element of $A$ is in $B$

iv)   Let $A \cap B = A$

$\Rightarrow$ if $x \in A, x \in B$ also

Therefore $A \subset B$. All elements of $A$ are elements of $B$ also.

$\\$

Question 6: For three sets $A, B$ and $C$, show that

(i) $A \cap B = A \cap C$ need not imply $B=C$    (ii) $A \subset B \Rightarrow C - B \subset C-A$

(i)   $A \cap B = A \cap C$ need not imply $B=C$

Let $A = \{ 1, 2, 3 \}, B = \{ 2, 4, 6 \}$ and $C = \{ 2, 5, 7 \}$

Therefore $A \cap B = \{ 2 \}$

$A \cap C = \{ 2 \}$

Hence $A \cap B = A \cap C$

But we can clearly see that $B \neq C$

(ii)  $A \subset B \Rightarrow C - B \subset C-A$

Given $A \subset B$

Let $x \in C - B$

$\Rightarrow x \in C$ but $x \notin B$ (By definition  of $C - B$ )

$\Rightarrow x \in C$ and $x \notin A$ (Since $A \subset B$ )

Thus $x \in C-B \Rightarrow x \in C - A$. This is true for all $x \in C - B$

Therefore  $C - B \subset C-A$

$\\$

Question 7: For any two sets, prove that: (i) $A \cup (A \cap B) = A$    (ii) $A \cap (A \cup B) = A$

(i)   $A \cup (A \cap B) = A$

$A \cup ( A \cap B) = (A \cup A ) \cap (A \cup B)$

[Since union is distributive over intersection ]

$= A \cap (A \cup B)$ [ Since $A \cup A = A$ ]

$= A$

[ Since $A \subset (A \cup B)$ as union of two sets is biggest than each of the individual sets ]

Hence $A \cup (A \cap B) = A$. Hence Proved.

(ii)  $A \cap (A \cup B) = A$

$A \cap (A \cup B) = (A \cap A ) \cup (A \cap B)$

[ Since intersection is distributive on union ]

$= A \cup ( A \cap B)$ [Since $A \cap A = A$ ]

$= A$

$\\$

Question 8: Find sets $A, \ B$ and $C$ such that $A \cap B, A \cap C$ and $B \cap C$ are non-empty sets and $A \cap B \cap C = \phi$

Consider $A = \{ 5, 6, 10 \}, B = \{ 6, 8, 9 \}$ and $C = \{ 9, 10, 11 \}$

Therefore $A \cap B = \{ 6 \} \Rightarrow A \cap B \neq \phi$

$B \cap C = \{ 10 \} \Rightarrow B \cap C \neq \phi$

Also $A \cap B \cap C = \phi$

Hence there are no common elements between all three sets.

$\\$

Question 9: For any two sets $A$ and $B$, prove that $A \cap B = \phi + A \subseteq B'$.

Given $A \cap B = \phi$

$\Rightarrow$ for $x \in A, x \notin B$

There are no common elements between $A$ and $B$

$\Rightarrow x \in B'$

Thus $x \in A$ and $x \in B' \Rightarrow A \subseteq B'$

$\\$

Question 10: If $A$ and $B$ are sets, then prove that $A - B, A \cap B$ and $B - A$ are pair wise disjoint.

We need to show:

$(A-B) \cap (A \cap B) = \phi$

$(A \cap B) \cap (B-A) = \phi$

$(A-B) \cap (B-A) = \phi$

Let us first show $(A-B) \cap (A \cap B) = \phi$

Let $x \in (A - B)$ $\Rightarrow x \in A$ but $x \notin B$

$\Rightarrow x \notin A \cap B$

This is true for all $x \in (A - B)$

Hence $(A- B) \cap (A \cap B) = \phi$

Similarly, let us prove $(A \cap B) \cap (B-A) = \phi$

$x \in (B-A)$ $\Rightarrow x \in B$ but $x \notin A$

$\Rightarrow x \notin A \cap B$

Therefore $(A \cap B ) \cap (B - A ) = \phi$

Now prove $(A-B) \cap (B-A) = \phi$

If $x \in (A - B) \Rightarrow x \in A$ but $x \notin B$

If $y \in (B - A) \Rightarrow y \in B$ but $y \notin A$

We can see that there is no intersection of $x$ and $y$. Hence $(A-B) \cap (B-A) = \phi$

$\\$

Question 11: Using properties of sets, show that for any two sets $A$ and $B$, $(A \cup B) \cap (A \cup B') = A$.

To prove: $(A \cup B) \cap (A \cup B') = A$

Now $(A \cup B) \cap (A \cap B') = [ (A \cup B )\cap A ] \cap B'$  (using associative property)

$= \Big( (A \cap A) \cup (B \cap A ) \Big) \cap B'$ [ Since $A \cap A = A$ and $B \cap A = A \cap B$ ]

$= A \cap B'$ [Since $A \cup (A \cap B ) = A$ ]

$= A$

$\\$

Question 12: For any two sets of $A$ and $B$, prove that:   (i) $A' \cup B = U \Rightarrow A \subset B$    (ii) $B' \subset A' \Rightarrow A \subset B$

(i) To prove: $A' \cup B = U \Rightarrow A \subset B$

Let $x \in A \Rightarrow x \notin A'$ [ Since $A \cap A' = \phi$ ]

Therefore $x \in A$ and $A \subset U$

$\Rightarrow x \in U$

$\Rightarrow x \in (A' \cup B)$ [Since $U = A' \cup B$ ]

$\Rightarrow x \in A'$ or $x \in B$

But $x \notin A' \Rightarrow x \in B$

Thus $x \in A \Rightarrow x \in B$

This is true for all $x \in A$

Therefore $A \subset B$

(ii)  To prove $B' \subset A' \Rightarrow A \subset B$

Let $x \in A \Rightarrow x \notin A'$ [ Since $A \cap A' = \phi$]

$\Rightarrow x \notin B'$ [Since $B' \subset A'$]

$\Rightarrow x \in B$ [Since $B \cap B' = \phi$ ]

Thus $x \in A \Rightarrow x \in B$

This is true for all $x \in A$. Therefore $A \subset B$

$\\$

Question 13: Is it true that for any sets $A$ and $B$, $P (A) \cup P (B) = P (A \cup B)$ Justify your answer.

False

Let $X \in P(A) \cup P(B)$

$\Rightarrow X \in P(A)$ or $X \in P(B)$

$\Rightarrow X \subset A$ or $X \subset B$

$\Rightarrow X \subset A \cup B$

$\Rightarrow X \in P(A \cap B)$

Therefore $P(A) \cup P(B) \subset P(A \cup B)$

Again let $X \in P(A \cup B)$ but $X \notin P(A)$ or $X \notin P(B)$

Therefore $X \notin P(A) \cup P(B)$

Thus $P(A \cup B)$ is not necessarily a subset of $P(A) \cup P(B)$

$\\$

Question 14: Show, that for any sets $A$ and $B$,   (i) $A=(A \cap B) \cap (A-B)$     (ii) $A \cup (B-A) =A \cup B$

(i)   We know that $(A \cap B) \subset A$ and $(A - B) \subset A$

$\Rightarrow (A \cap B) \cap (A - B)$ … … … … … i)

Let $x \in (A \cap B) \cap (A - B)$

$\Rightarrow x \in (A \cap B) \Rightarrow x \in A$ and $x \in B$

and $x \in (A - B) \Rightarrow x \in A$ but $x \notin B$

$\Rightarrow x \in A$ [ Since $x \in B$ and $x \notin B$ are not possible simultaneously]

Therefore $(A \cap B) \cap (A - B) \subset A$    … … … … … ii)

From i) and ii) we get

$A = (A \cap B) \cap (A - B)$

(ii)  Let $x \in A \cup (B-A)$

$\Rightarrow x \in A$ or $x \in (B-A) \Rightarrow x \in B$ or $x \notin A$

$\Rightarrow x \in A$ or $x \in B$

$\Rightarrow x \in ( A \cup B)$

Therefore $A \cup (B - A) \subset (A \cup B)$   … … … … … i)

Let $x \in (A \cup B) \Rightarrow x \in A$ or $x \in B$

$\Rightarrow x \in A$ or $x \in B$ and $x \notin A$

$\Rightarrow x \in A$ or $x \in (B-A)$

$\Rightarrow x \in A \cup ( B - A)$

Therefore $(A \cup B) \subset A \cup (B - A)$   … … … … … ii)

From i) and ii) we get $A \cup (B-A) =A \cup B$

$\\$

Question 15: Each set $X$, contains $5$ elements and each set $Y$, contains $2$ elements and $\underset{r=1}{\overset{20}{U}} X_r = S = \underset{r=1}{\overset{n}{U}} Y_r$. If each element of $S$ belongs to exactly $10$ of the ${X'}_r^s$ and to exactly $4$ of ${Y'}_r^s$, find the value of $n$.

It is given that each set $X$ contains $5$ elements $\underset{r=1}{\overset{20}{U}} X_r = S$

Therefore $n(S) = 20 \times 5 = 100$

But it is given that each element of $S$ belongs to exactly $10$ of that $X_r's$

Number of distinct elements in $S =$ $\frac{100}{10}$ $= 10$  … … … … … i)

It is also given that each set $Y$ contains $2$ elements and $\underset{r=1}{\overset{n}{U}} Y_r = S$

Therefore $n(S) = n \times 2 = 2n$

Also each element of $S$ belongs to exactly $4$ of $Y_r's =$ $\frac{1}{4}$ $(2n) =$ $\frac{n}{2}$  … … … … … ii)

Therefore from i) and ii) we get

$10 =$ $\frac{n}{2}$ $\Rightarrow n = 20$