Sets – Complement of a union in a fixed universe: Given the universe U = {a, b, c, d, e, f} and A = {a, b, c}. Compute the complement (U ∪ A)′ with respect to U.

Introduction / Context:
In elementary set theory, complements are taken with respect to a fixed universe U. The complement S′ means all elements of U that are not in S. Here, U and subset A are given; we must find (U ∪ A)′.

Given Data / Assumptions:

• Universe U = {a, b, c, d, e, f}
• Set A = {a, b, c}
• Notation X′ denotes complement with respect to U

Concept / Approach:
The union U ∪ A contains every element that is in U or in A (or both). Because A ⊆ U, their union equals U itself. The complement of U inside U is the empty set Φ, since there is no element in U that is outside U.

Step-by-Step Solution:
Observe A ⊆ U ⇒ U ∪ A = UComplement: (U ∪ A)′ = U′ = Φ

Verification / Alternative check:
List-wise: U has 6 elements. U ∪ A also has exactly these 6. Nothing remains in U to complement, so the complement is empty.

Why Other Options Are Wrong:
U is not a complement; A is not a complement of U; “None of these” is unnecessary because Φ is the exact result.

Common Pitfalls:
Confusing universal complement (relative to an absolute “universe of discourse”) with complement inside a larger, unspecified universe. Always confirm the reference universe.

Final Answer:
Φ