Let f: `X rarr Y` and A , B are non void subsets of y then where the symbols have their usual interpretation

Options

A. `f^(-1)(A)-f^(-1)(B)supset f^(-1) (A-B)` but the opposite does not hold
B. `f^(-1)(A)-f^(-1)(B)subset f^(-1) (A-B)` but the opposite does not hold
C. `f^(-1)(A-B)=f^(-1)(A)- f^(-1) (B)`
D. `f^(-1)(A-B)=f^(-1)(A)cup f^(-1) (B)`

Answer

To solve the problem step by step, we need to analyze the given function \( f: X \to Y \) and the non-void subsets \( A \) and \( B \) of \( Y \). We will explore the relationships between the pre-images of these sets under the function \( f \).

### Step 1: Understand the Definitions
- **Function**: A function \( f \) maps elements from set \( X \) to set \( Y \).
- **Pre-image**: The pre-image of a subset \( A \subseteq Y \) under \( f \) is defined as:
\[ f^{-1}(A) = \{ x \in X \mid f(x) \in A \} \]
- **Set Difference**: The set difference \( A - B \) is defined as:
\[ A - B = \{ y \in A \mid y \notin B \} \]

### Step 2: Analyze the Given Options
We need to evaluate the following statements regarding the pre-images:

1. \( f^{-1}(A) - f^{-1}(B) \supseteq f^{-1}(A - B) \)
2. \( f^{-1}(A) - f^{-1}(B) \subseteq f^{-1}(A - B) \)
3. \( f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B) \)
4. \( f^{-1}(A - B) = f^{-1}(A) \cup f^{-1}(B) \)

### Step 3: Evaluate Each Option
#### Option 1: \( f^{-1}(A) - f^{-1}(B) \supseteq f^{-1}(A - B) \)
- This statement suggests that the pre-image of \( A - B \) is contained within the difference of the pre-images of \( A \) and \( B \).
- This is not necessarily true because elements in \( f^{-1}(B) \) could also map to elements in \( A \) that are not in \( B \).

#### Option 2: \( f^{-1}(A) - f^{-1}(B) \subseteq f^{-1}(A - B) \)
- This statement suggests that the difference of the pre-images is a subset of the pre-image of the difference.
- This is also not true, as there may be elements in \( f^{-1}(A) \) that are not in \( f^{-1}(B) \) but still map to elements in \( B \).

#### Option 3: \( f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B) \)
- This statement suggests that the pre-image of the difference \( A - B \) is equal to the difference of the pre-images.
- This is true because if \( x \in f^{-1}(A) \) and \( x \notin f^{-1}(B) \), then \( f(x) \in A \) and \( f(x) \notin B \), which means \( f(x) \in A - B \).

#### Option 4: \( f^{-1}(A - B) = f^{-1}(A) \cup f^{-1}(B) \)
- This statement suggests that the pre-image of the difference is equal to the union of the pre-images.
- This is false because the union would include elements that are in both pre-images, which does not correspond to the elements in \( A - B \).

### Conclusion
The correct option is **Option 3**:
\[ f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B) \]