Question

Let \( X \) and \( Y \) be two non-empty sets such that \( X \cap A = Y \cap A = \emptyset \) and \( X \cup A = Y \cup A \) for some non-empty set \( A \). Then

• A. \( X \) is a proper subset of \( Y \)
• B. \( Y \) is a proper subset of \( X \)
• C. \( X = Y \)
• D. \( X \) and \( Y \) are disjoint sets
• E. \( X \setminus A = \emptyset \)

Hint:

When sets are disjoint from a common set, equality of unions directly implies equality of the original sets.

Answer 1

The Correct Option is C

Concept: We use properties of sets:
• If two sets are disjoint with a third set, they lie entirely outside it
• Equality of unions with a common set implies equality of remaining parts

Step 1: Given conditions
\[ X \cap A = \emptyset, \quad Y \cap A = \emptyset \] So, \[ X \subseteq A^c, \quad Y \subseteq A^c \]

Step 2: Use union condition
\[ X \cup A = Y \cup A \]

Step 3: Interpret the union
Since neither \(X\) nor \(Y\) contains elements of \(A\), adding \(A\) to both just appends the same set. Thus, any difference between \(X\) and \(Y\) must appear outside \(A\).

Step 4: Logical deduction
If \(X \neq Y\), then there exists an element belonging to one but not the other. That difference would still appear after union with \(A\), contradicting: \[ X \cup A = Y \cup A \]

Step 5: Conclusion
Therefore, \[ X = Y \] \[ \boxed{X = Y} \]