Power set of the set $A=\{1,2\}$ is ........... .
$$\therefore A=\{1,2\}$$
So, the subsets of $A$ are $\phi,\{1\},\{2\}$ and $\{1,2\}$.
$$\therefore \quad P(A)=\{\phi,\{1\},\{2\},\{1,2\}\}$$
If the sets $A=\{1,3,5\}, B=\{2,4,6\}$ and $C=\{0,2,4,6,8\}$. Then, the universal set of all the three sets $A, B$ and $C$ can be .......... .
Universal set for $A, B$ and $C$ is given by $U=\{0,1,2,3,4,5,6,8\}$
If $U=\{1,2,3,4,5,6,7,8,9,10\}, A=\{1,2,3,5\}, B=\{2,4,6,7\}$ and $C=\{2,3,4,8\}$. Then,
(i) $(B \cup C)^{\prime}$ is ............ .
(ii) $(C-A)^{\prime}$ is .......... .
If $\begin{aligned} & U=\{1,2,3,4,5, \ldots, 10\} \\ & A=\{1,2,3,5\}, B=\{2,4,6,7\} \text { and } C=\{2,3,4,8\}\end{aligned}$
$$\therefore \quad B \cup C=\{2,3,4,6,7,8\}$$
(i) $(B \cup C)^{\prime}=U-(B \cup C)=\{1,5,9,10\}$
(ii) $C-A=\{4,8\}$
$\therefore \quad(C-A)^{\prime}=U-(C-A)=\{1,2,3,5,6,7,9,10\}$
For all sets $A$ and $B, A-(A \cap B)$ is equal to .......... .
$A-(A \cap B)=A-B=A \cap B^{\prime}$
Match the following sets for all sets $A, B$ and $C$
| Column I | Column II | ||
|---|---|---|---|
| (i) | $\left(\left(A^{\prime} \cup B^{\prime}\right)-A\right)^{\prime}$ | (a) | $A-B$ |
| (ii) | $\left[\left(B^{\prime} \cup\left(B^{\prime}-A\right)\right]^{\prime}\right.$ | (b) | $A$ |
| (iii) | $(A-B)-(B-C)$ | (c) | $B$ |
| (iv) | $(A-B) \cap(C-B)$ | (d) | $(A \times B) \cap(A \times C)$ |
| (v) | $A \times(B \cap C)$ | (e) | $(A \times B) \cup(A \times C)$ |
| (vi) | $A \times(B \cup C)$ | (f) | $(A \cap C)-B$ |
(i) $\left[\left(A^{\prime} \cup B^{\prime}\right)-A\right]^{\prime}=\left[\left(A^{\prime} \cup B^{\prime}\right) \cap A^{\prime}\right]^{\prime} \quad\left[\because A-B=A \cap B^{\prime}\right]$
$=\left[(A \cap B)^{\prime} \cap A^{\prime}\right]^{\prime} \quad\left[\because(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\right]$
$=\left[(A \cap B)^{\prime}\right]^{\prime} \cup\left(A^{\prime}\right)^{\prime}=(A \cap B) \cup A \quad\left[\because\left(A^{\prime}\right)^{\prime}=A\right]$
$=A$
(ii) $\left[B^{\prime} \cup\left(B^{\prime}-A\right)\right]^{\prime}=\left[B^{\prime} \cup\left(B^{\prime} \cap A^{\prime}\right)\right]^{\prime} \quad\left[\because A-B=A \cap B^{\prime}\right]$
$=\left[B^{\prime} \cup(B \cup A)^{\prime}\right]^{\prime} \quad\left[\because A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}\right]$
$=\left(B^{\prime}\right)^{\prime} \cap\left[(B \cup A)^{\prime}\right]^{\prime} \quad\left[\because(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\right]$
$=B \cap(B \cup A) \quad\left[\because\left(A^{\prime}\right)^{\prime}=A\right]$
$=B$
(iii) $(A-B)-(B-C)=\left(A \cap B^{\prime}\right)-\left(B \cap C^{\prime}\right) \quad\left[\because A-B=A \cap B^{\prime}\right]$
$$\begin{aligned} & =\left(A \cap B^{\prime}\right) \cap\left(B \cap C^{\prime}\right)^{\prime} \\ & =\left(A \cap B^{\prime}\right) \cap\left[B^{\prime} \cup\left(C^{\prime}\right)^{\prime}\right] \\ & =\left(A \cap B^{\prime}\right) \cap\left(B^{\prime} \cup C\right)\quad \left[\because\left(A^{\prime}\right)^{\prime}=A\right] \\ & =\left[A \cap\left(B^{\prime} \cup C\right)\right] \cap\left[B^{\prime} \cap\left(B^{\prime} \cup C\right)\right] \\ & =\left[A \cap\left(B^{\prime} \cup C\right)\right] \cap B^{\prime} \\ & =\left(A \cap B^{\prime}\right) \cap\left[\left(B^{\prime} \cup C\right) \cap B^{\prime}\right] \\ & =\left(A \cap B^{\prime}\right) \cap B^{\prime}=A \cap B^{\prime}=A-B \end{aligned}$$
Alternate Method
It is clear from the diagram, $(A-B)-(B-C)=A-B$.
$\begin{array}{lll}\text{(iv)}\quad (A-B) \cap(C-B) & & \\ \Rightarrow & \left(A \cap B^{\prime}\right) \cap\left(C \cap B^{\prime}\right) & {\left[\because A-B=A \cap B^{\prime}\right]} \\ \Rightarrow & (A \cap C) \cap B^{\prime} & \\ \Rightarrow & (A \cap C)-B & {\left[\because A \cap B^{\prime}=A-B\right]}\end{array}$
(v) $A \times B \cap C=(A \times B) \cap(A \times C)$
(vi) $A \times(B \cup C)=(A \times B) \cup(A \times C)$
Hence, the correct matches are
(i) $\leftrightarrow$ (b),
(ii) $\leftrightarrow$ (c),
(iii) $\leftrightarrow$ (a),
(iv) $\leftrightarrow(f)$,
$(\mathrm{v}) \leftrightarrow$ (d),
$(\mathrm{vi}) \leftrightarrow(\mathrm{e})$