For any two sets A and B, A' – B'=

Options

A. `A-B`
B. `B-A`
C. `A cup B`
D. `A cap B`

Answer

To solve the problem \( A' - B' \), we will use the properties of set complements and the universal set. Here’s a step-by-step solution:

### Step 1: Understand the notation
- \( A' \) (or \( A^c \)) represents the complement of set \( A \), which includes all elements not in \( A \).
- \( B' \) (or \( B^c \)) represents the complement of set \( B \), which includes all elements not in \( B \).

**Hint:** Remember that the complement of a set contains all elements from the universal set that are not in the given set.

### Step 2: Write the expression in terms of the universal set
Using the definition of the complement, we can express \( A' \) and \( B' \) in terms of the universal set \( U \):
- \( A' = U - A \)
- \( B' = U - B \)

**Hint:** The universal set \( U \) contains all possible elements under consideration.

### Step 3: Substitute the complements into the expression
Now, substitute the expressions for \( A' \) and \( B' \) into the original expression:
\[ A' - B' = (U - A) - (U - B) \]

**Hint:** When subtracting sets, think of it as removing elements of one set from another.

### Step 4: Simplify the expression
Using the property of set difference, we can simplify:
\[ A' - B' = (U - A) \cap (B) \]
This means we take all elements in \( U - A \) and intersect it with \( B \).

**Hint:** The intersection of two sets contains only the elements that are present in both sets.

### Step 5: Reinterpret the expression
The expression \( (U - A) \cap B \) can be interpreted as all elements in \( B \) that are not in \( A \), which is the definition of the set difference \( B - A \).

**Hint:** The set difference \( B - A \) includes all elements in \( B \) excluding those that are also in \( A \).

### Final Result
Thus, we conclude that:
\[ A' - B' = B - A \]

### Summary
The final answer is \( B - A \).