Understanding Set Operations and Null Sets

This problem involves understanding set operations like intersection ($\cap$), union ($\cup$), and set difference (\ or $\setminus$), applied to sets defined by properties of numbers.

Let's first define the given sets A, B, and C based on the question:

• Set A: $\{ x \mid x \text{ is a multiple of } 3 \}$. This means A contains numbers like 3, 6, 9, 12, 15, 18, 21, 24, and so on.
• Set B: $\{ x \mid x \text{ is a multiple of } 4 \}$. This means B contains numbers like 4, 8, 12, 16, 20, 24, and so on.
• Set C: $\{ x \mid x \text{ is a multiple of } 12 \}$. This means C contains numbers like 12, 24, 36, 48, and so on.

We need to determine which of the given options represents a null set, also known as the empty set, denoted by $\emptyset$. The null set contains no elements.

Analyzing Each Option to Identify the Null Set

Let's evaluate each option one by one using our understanding of set operations.

Option 1: $(A \setminus B) \cup C$

• First, consider the set difference $A \setminus B$. This set contains elements that are in A but not in B. These are multiples of 3 that are NOT multiples of 4. For example, 3, 6, 9, 15, 18, 21 are in $A \setminus B$. Multiples of 12 (like 12, 24) are multiples of both 3 and 4, so they are not in $A \setminus B$. The set $A \setminus B$ is not empty.
• Next, consider the union $(A \setminus B) \cup C$. This is the set of elements that are in $A \setminus B$ or in C (or both). Since $A \setminus B$ is not empty (it contains numbers like 3, 6, 9) and C is not empty (it contains numbers like 12, 24), their union will contain elements and therefore is not a null set.

Option 2: $(A \setminus B) \setminus C$

• We already know that $A \setminus B$ is the set of multiples of 3 that are not multiples of 4. For example, $\{3, 6, 9, 15, 18, 21, ...\}$.
• Now, consider $(A \setminus B) \setminus C$. This set contains elements that are in $A \setminus B$ but not in C. Elements in C are multiples of 12. Multiples of 12 are also multiples of 3 and 4. Since $A \setminus B$ contains multiples of 3 that are *not* multiples of 4, none of its elements can be a multiple of 12. Therefore, removing elements of C from $A \setminus B$ doesn't remove any elements from $A \setminus B$. The set $(A \setminus B) \setminus C$ is the same as $A \setminus B$. Since $A \setminus B$ is not empty, $(A \setminus B) \setminus C$ is also not a null set.

Option 3: $(A \cap B) \cap C$

• First, consider the intersection $A \cap B$. This set contains elements that are in both A and B. These are numbers that are multiples of both 3 and 4. A number is a multiple of both 3 and 4 if and only if it is a multiple of their least common multiple (LCM). The LCM of 3 and 4 is 12. So, $A \cap B$ is the set of multiples of 12.
• From the problem definition, we know that set C is also the set of multiples of 12. Therefore, $A \cap B = C$.
• Next, consider the intersection $(A \cap B) \cap C$. Since $A \cap B = C$, this expression becomes $C \cap C$. The intersection of a set with itself is the set itself. So, $C \cap C = C$.
• Set C is the set of multiples of 12, which is $\{12, 24, 36, ...\}$. This set is not empty. Therefore, $(A \cap B) \cap C$ is not a null set.

Option 4: $(A \cap B) \setminus C$

• First, consider the intersection $A \cap B$. As we determined in Option 3, $A \cap B$ is the set of numbers that are multiples of both 3 and 4, which is the set of multiples of 12. So, $A \cap B = C$.
• Now, consider the set difference $(A \cap B) \setminus C$. This set contains elements that are in $A \cap B$ but not in C. Since $A \cap B$ is equal to C, this expression becomes $C \setminus C$.
• The set difference of a set with itself always results in the null set, $\emptyset$. This is because there are no elements in C that are not in C.
• Therefore, $(A \cap B) \setminus C = C \setminus C = \emptyset$.

Based on the analysis, the set $(A \cap B) \setminus C$ is the null set.

Revision Table: Key Set Concepts

Additional Information: Multiples and LCM

Understanding multiples and the Least Common Multiple (LCM) is crucial for this type of set theory problem.

• Multiples: A multiple of a number 'n' is any number that can be obtained by multiplying 'n' by an integer. For example, multiples of 3 are $3 \times 1, 3 \times 2, 3 \times 3, ...$ which are 3, 6, 9, ...
• Least Common Multiple (LCM): The LCM of two or more integers is the smallest positive integer that is a multiple of all the integers. For example, to find the LCM of 3 and 4:
  • Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, ...
  • Multiples of 4: 4, 8, 12, 16, 20, 24, 28, ...
  • The common multiples are 12, 24, ... The smallest common multiple is 12. So, LCM(3, 4) = 12.
• The set of numbers that are multiples of both 'a' and 'b' is exactly the set of numbers that are multiples of LCM(a, b). This is why the intersection of the set of multiples of 3 and the set of multiples of 4 is the set of multiples of 12 ($A \cap B = C$).