Analyzing Set Operations and Disjoint Sets

The question asks us to evaluate two statements regarding set operations based on the given conditions about four sets A, B, C, and D. The given conditions are that the intersection of set A and set B is empty ($A \cap B = \phi$), and the intersection of set C and set D is empty ($C \cap D = \phi$). When the intersection of two sets is empty, we say the sets are disjoint.

• Given: $A \cap B = \phi$ (A and B are disjoint)
• Given: $C \cap D = \phi$ (C and D are disjoint)

Let's analyze each statement.

Statement 1: A ∪ C and B ∪ D are always disjoint.

This statement claims that $(A \cup C) \cap (B \cup D) = \phi$ must always be true under the given conditions. Let's examine this intersection using properties of set operations.

The intersection of unions can be expanded:

$(A \cup C) \cap (B \cup D) = (A \cap B) \cup (A \cap D) \cup (C \cap B) \cup (C \cap D)$

We are given that $A \cap B = \phi$ and $C \cap D = \phi$. Substituting these into the expanded expression:

$(A \cap B) \cup (A \cap D) \cup (C \cap B) \cup (C \cap D) = \phi \cup (A \cap D) \cup (C \cap B) \cup \phi$

$= (A \cap D) \cup (C \cap B)$

For $(A \cup C) \cap (B \cup D)$ to be equal to $\phi$, the result of the expansion $(A \cap D) \cup (C \cap B)$ must be $\phi$. This means both $A \cap D$ and $C \cap B$ must be empty sets.

However, the given conditions ($A \cap B = \phi$ and $C \cap D = \phi$) do not provide any information about the intersections $A \cap D$ or $C \cap B$. It is possible for $A \cap D$ or $C \cap B$ (or both) to be non-empty while satisfying the initial conditions.

Let's consider a counterexample to show that Statement 1 is not always true:

• Let $A = \{1\}$
• Let $B = \{2\}$
• Let $C = \{3\}$
• Let $D = \{1\}$

Let's check the given conditions:

• $A \cap B = \{1\} \cap \{2\} = \phi$. (Condition met)
• $C \cap D = \{3\} \cap \{1\} = \phi$. (Condition met)

Now let's check Statement 1:

• $A \cup C = \{1\} \cup \{3\} = \{1, 3\}$
• $B \cup D = \{2\} \cup \{1\} = \{1, 2\}$
• $(A \cup C) \cap (B \cup D) = \{1, 3\} \cap \{1, 2\} = \{1\}$

Since $\{1\}$ is not an empty set ($\phi$), the statement $(A \cup C) \cap (B \cup D) = \phi$ is false in this case. Therefore, Statement 1 is not always correct.

Statement 2: A ∩ C and B ∩ D are always disjoint.

This statement claims that $(A \cap C) \cap (B \cap D) = \phi$ must always be true under the given conditions. Let's examine this intersection.

The intersection of four sets $A, C, B, D$ means we are looking for elements that are present in all four sets simultaneously. An element $x$ is in $(A \cap C) \cap (B \cap D)$ if and only if $x \in A$ AND $x \in C$ AND $x \in B$ AND $x \in D$.

Using the associative property of intersection, we can group the sets differently:

$(A \cap C) \cap (B \cap D) = A \cap C \cap B \cap D = (A \cap B) \cap (C \cap D)$

We are given that $A \cap B = \phi$ and $C \cap D = \phi$. Substituting these values:

$(A \cap B) \cap (C \cap D) = \phi \cap \phi$

The intersection of any set with the empty set is always the empty set. So,

$\phi \cap \phi = \phi$

Thus, $(A \cap C) \cap (B \cap D) = \phi$ is always true, regardless of what sets A, B, C, and D are, as long as $A \cap B = \phi$ and $C \cap D = \phi$. Statement 2 is always correct.

Conclusion

Based on our analysis:

• Statement 1: $(A \cup C) \cap (B \cup D) = \phi$ is not always true.
• Statement 2: $(A \cap C) \cap (B \cap D) = \phi$ is always true.

Therefore, only Statement 2 is correct.

Revision Table: Set Theory Concepts

Let's review the key set operations used in this problem.

Additional Information: Properties of Set Operations

Understanding the properties of set operations is crucial for solving problems like this. Some relevant properties include:

• Commutative Properties:
  • $A \cup B = B \cup A$
  • $A \cap B = B \cap A$
• Associative Properties:
  • $(A \cup B) \cup C = A \cup (B \cup C)$
  • $(A \cap B) \cap C = A \cap (B \cap C)$ (This was used in Statement 2 analysis)
• Distributive Properties:
  • $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
  • $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
  • The expansion of $(A \cup C) \cap (B \cup D)$ in Statement 1 analysis uses the distributive property twice or can be seen as a form of expansion.
• Properties involving the Empty Set ($\phi$):
  • $A \cap \phi = \phi$ (The intersection of any set with the empty set is the empty set)
  • $A \cup \phi = A$ (The union of any set with the empty set is the set itself)

These properties allow us to manipulate set expressions and simplify them, which is often necessary to determine if a statement about sets is always true under given conditions.