Question

Let A and B be two non-empty sets, then \((A \cup B)^c\) is equal to :

• A. \((A \cap B)^c\)
• B. \(A^c \cap B^c\)
• C. \(A^c \cup B^c\)
• D. \(A \cup B\)

Hint:

To remember De Morgan's laws easily:
When you distribute the complement symbol (\(^c\)) inside the parentheses, you must flip the operation:
- Union (\(\cup\)) flips to Intersection (\(\cap\)).
- Intersection (\(\cap\)) flips to Union (\(\cup\)).
Thus, \((A \cup B)^c\) becomes \(A^c \cap B^c\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question relates to set theory, specifically De Morgan's laws.
De Morgan's laws describe how mathematical complements of set unions and intersections are related to each other.

Step 2: Detailed Explanation:
Let us state De Morgan's two laws for sets:
1. The complement of the union of two sets is equal to the intersection of their individual complements:
\[ (A \cup B)^c = A^c \cap B^c \]
2. The complement of the intersection of two sets is equal to the union of their individual complements:
\[ (A \cap B)^c = A^c \cup B^c \]
Let us verify the first law logically:
- Suppose an element \(x\) belongs to \((A \cup B)^c\).
- This means \(x \notin (A \cup B)\).
- Therefore, \(x\) is not in \(A\) and \(x\) is not in \(B\).
- This means \(x \in A^c\) and \(x \in B^c\).
- Since \(x\) is in both \(A^c\) and \(B^c\), it must lie in their intersection: \(x \in (A^c \cap B^c)\).
- Conversely, if an element \(y\) belongs to \((A^c \cap B^c)\), it means \(y \in A^c\) and \(y \in B^c\).
- Therefore, \(y \notin A\) and \(y \notin B\).
- This implies \(y \notin (A \cup B)\), meaning \(y \in (A \cup B)^c\).
Since the subsets contain exactly the same elements, we have:
\[ (A \cup B)^c = A^c \cap B^c \]
This is exactly Option (B).

Step 3: Final Answer:
The correct option is (B).