If U= {1,2,3,4,5,6},A = {1,2},B={3,4,5}, find `A^(c),B^(c),A^(c)capB^(c)andAcapB`.

Answer

To solve the problem, we need to find the complements and intersections of the given sets. Let's break it down step by step.

### Given:
- Universal set \( U = \{1, 2, 3, 4, 5, 6\} \)
- Set \( A = \{1, 2\} \)
- Set \( B = \{3, 4, 5\} \)

### Step 1: Find \( A^c \) (Complement of A)
The complement of set \( A \) consists of all the elements in the universal set \( U \) that are not in \( A \).

\[ A^c = U - A = \{1, 2, 3, 4, 5, 6\} - \{1, 2\} \]

Thus, we remove \( 1 \) and \( 2 \) from \( U \):

\[ A^c = \{3, 4, 5, 6\} \]

### Step 2: Find \( B^c \) (Complement of B)
Similarly, the complement of set \( B \) consists of all the elements in \( U \) that are not in \( B \).

\[ B^c = U - B = \{1, 2, 3, 4, 5, 6\} - \{3, 4, 5\} \]

Thus, we remove \( 3, 4, \) and \( 5 \) from \( U \):

\[ B^c = \{1, 2, 6\} \]

### Step 3: Find \( A^c \cap B^c \) (Intersection of A complement and B complement)
The intersection \( A^c \cap B^c \) consists of elements that are common to both \( A^c \) and \( B^c \).

\[ A^c = \{3, 4, 5, 6\} \]
\[ B^c = \{1, 2, 6\} \]

The common element between \( A^c \) and \( B^c \) is \( 6 \):

\[ A^c \cap B^c = \{6\} \]

### Step 4: Find \( A \cap B \) (Intersection of A and B)
The intersection \( A \cap B \) consists of elements that are common to both \( A \) and \( B \).

\[ A = \{1, 2\} \]
\[ B = \{3, 4, 5\} \]

Since there are no common elements between \( A \) and \( B \):

\[ A \cap B = \emptyset \]

### Final Results:
- \( A^c = \{3, 4, 5, 6\} \)
- \( B^c = \{1, 2, 6\} \)
- \( A^c \cap B^c = \{6\} \)
- \( A \cap B = \emptyset \)