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, think of the complement distributing to both sets, which "flips" the operation in between:
- Union (\(\cup\)) flips to Intersection (\(\cap\)).
- Intersection (\(\cap\)) flips to Union (\(\cup\)).
So, \((A \cup B)^C = A^C \cap B^C\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to identify the set equivalent to the complement of the union of two sets, \((A \cup B)^C\).
This is a standard theorem in set theory known as De Morgan's Law.
Key Formula or Approach:
De Morgan's Laws state:
1. \((A \cup B)^C = A^C \cap B^C\) (The complement of the union of two sets is the intersection of their complements).
2. \((A \cap B)^C = A^C \cup B^C\) (The complement of the intersection of two sets is the union of their complements).

Step 2: Detailed Explanation:

• Let us verify this identity logically using element membership:
Let \(x \in (A \cup B)^C\).
By definition of a complement:
\[ x \notin (A \cup B) \]

• Since \(x\) is not in the union of \(A\) and \(B\), \(x\) cannot belong to \(A\), nor can it belong to \(B\):
\[ x \notin A \quad \text{and} \quad x \notin B \]

• By definition of set complements:
\[ x \in A^C \quad \text{and} \quad x \in B^C \]

• Since \(x\) belongs to both \(A^C\) and \(B^C\), it must belong to their intersection:
\[ x \in (A^C \cap B^C) \]

• Therefore, any element in \((A \cup B)^C\) is also in \(A^C \cap B^C\). Thus, we can conclude:
\[ (A \cup B)^C = A^C \cap B^C \]

Step 3: Final Answer:
The expression \((A \cup B)^C\) is equal to \(A^C \cap B^C\).