Write the empty and singleton set from the following sets :
(i) `A={x:x+2=5}`
(ii) `B={x:x^(2)-8x+16=0}`
(iii) `C={x:x+2=3 and 2x=1}`
(iv) `D={x:xltx}`.

Answer

To solve the problem of identifying empty and singleton sets from the given sets A, B, C, and D, we will analyze each set one by one.

### Step-by-Step Solution:

**(i) Set A:**
- The set is defined as \( A = \{ x : x + 2 = 5 \} \).
- To find the value of \( x \), we solve the equation:
\[ x + 2 = 5 \implies x = 5 - 2 \implies x = 3 \]
- Therefore, the set A can be expressed as \( A = \{ 3 \} \).
- Since it contains exactly one element, A is a **singleton set**.

**(ii) Set B:**
- The set is defined as \( B = \{ x : x^2 - 8x + 16 = 0 \} \).
- We can factor the quadratic equation:
\[ x^2 - 8x + 16 = (x - 4)^2 = 0 \]
- This gives us the solution:
\[ x - 4 = 0 \implies x = 4 \]
- Therefore, the set B can be expressed as \( B = \{ 4 \} \).
- Since it contains exactly one element, B is also a **singleton set**.

**(iii) Set C:**
- The set is defined as \( C = \{ x : x + 2 = 3 \text{ and } 2x = 1 \} \).
- We solve each equation:
1. From \( x + 2 = 3 \):
\[ x = 3 - 2 \implies x = 1 \]
2. From \( 2x = 1 \):
\[ x = \frac{1}{2} \]
- The solutions \( x = 1 \) and \( x = \frac{1}{2} \) are different, which means there are two distinct values for \( x \).
- Therefore, the set C can be expressed as \( C = \{ 1, \frac{1}{2} \} \).
- Since it contains more than one element, C is neither an empty set nor a singleton set.

**(iv) Set D:**
- The set is defined as \( D = \{ x : x < x \} \).
- This statement is always false because no number is less than itself.
- Therefore, the set D is an **empty set**, denoted as \( D = \emptyset \).

### Summary of Results:
- **Set A**: Singleton set \( A = \{ 3 \} \)
- **Set B**: Singleton set \( B = \{ 4 \} \)
- **Set C**: Neither empty nor singleton \( C = \{ 1, \frac{1}{2} \} \)
- **Set D**: Empty set \( D = \emptyset \)