补集bǔjí

Core Concept

The complement of a set A, written as $\complement_U A$ or $\overline{A}$ or $A^c$, is the set of all elements in the universal set U that are NOT in A.

Mathematical Definition

$\complement_U A = \{x | x \in U \text{ and } x \notin A\}$

The complement contains exactly those elements that belong to the universal set but not to A.

Notation Variants

• $\complement_U A$ - Standard notation emphasizing universal set
• $\overline{A}$ - Bar notation
• $A^c$ or $A'$ - Prime/superscript notation
• $U - A$ - Set difference notation

Visual Representation

In a Venn diagram, the complement is the region outside set A but inside the universal set.

  U: [#############]
     [####]  A  [  ]

The shaded region [####] represents $\complement_U A$.

Key Properties

1. Complement of Complement

$\complement_U(\complement_U A) = A$

2. Complement of Universal Set

$\complement_U U = \emptyset$

3. Complement of Empty Set

$\complement_U \emptyset = U$

4. Union with Complement

$A \cup \complement_U A = U$

5. Intersection with Complement

$A \cap \complement_U A = \emptyset$

6. De Morgan's Laws

$\complement_U(A \cup B) = \complement_U A \cap \complement_U B$ $\complement_U(A \cap B) = \complement_U A \cup \complement_U B$

Worked Examples

Example 1: Finite Sets

Given: U = {1, 2, 3, 4, 5, 6}, A = {2, 4, 6}

Find: $\complement_U A$

Solution: Elements in U but not in A: 1, 3, 5

Answer: $\complement_U A$ = {1, 3, 5}

Example 2: Real Number Sets

Given: U = ℝ, A = {x | x ≥ 2}

Find: $\complement_U A$

Solution: Real numbers NOT ≥ 2, meaning < 2

Answer: $\complement_U A$ = {x | x < 2} = (-∞, 2)

Example 3: Interval Complement

Given: U = ℝ, A = (-1, 3]

Find: $\complement_U A$

Solution: All reals except those in (-1, 3]

Answer: $\complement_U A$ = (-∞, -1] ∪ (3, +∞)

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus.

Example 1: Basic (Difficulty ★★☆☆☆)

If U = {1, 2, 3, 4, 5} and A = {1, 3, 5}, find $\complement_U A$.

Options:

• A. {1, 3, 5}
• B. {2, 4}
• C. {1, 2, 3, 4, 5}
• D. ∅

Solution: Elements in U but not in A: 2, 4

Answer: B

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Example 2: Intermediate (Difficulty ★★★☆☆)

Given U = ℝ, A = {x | x² - 4 ≤ 0}, find $\complement_U A$.

Solution:

First, solve the inequality: $x² - 4 ≤ 0$ $(x-2)(x+2) ≤ 0$ $A = [-2, 2]$

Complement is all reals outside this interval: $\complement_U A = (-\infty, -2) \cup (2, +\infty)$

Answer: $(-\infty, -2) \cup (2, +\infty)$

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Example 3: Advanced (Difficulty ★★★★☆)

If U = ℝ, A = {x | x > 1}, B = {x | x > 2}, find $\complement_U A \cup B$.

Solution:

$\complement_U A$ = {x | x ≤ 1} = (-∞, 1] B = {x | x > 2} = (2, +∞)

$\complement_U A \cup B = (-\infty, 1] \cup (2, +\infty)$

Answer: $(-\infty, 1] \cup (2, +\infty)$

De Morgan's Laws in Detail

Law 1: Complement of Union

$\complement_U(A \cup B) = \complement_U A \cap \complement_U B$

Example: If A = {1, 2}, B = {2, 3}, U = {1, 2, 3, 4}

• A ∪ B = {1, 2, 3}
• $\complement_U(A \cup B)$ = {4}
• $\complement_U A$ = {3, 4}, $\complement_U B$ = {1, 4}
• $\complement_U A \cap \complement_U B$ = {4} ✓

Law 2: Complement of Intersection

$\complement_U(A \cap B) = \complement_U A \cup \complement_U B$

Common Mistakes

❌ Mistake 1: Forgetting the Universal Set

Wrong: $\complement A$ = {all elements not in A} ✗

Correct: $\complement_U A$ = {elements in U but not in A} ✓

❌ Mistake 2: Wrong Interval Boundary

Wrong: If A = [1, 3], then $\complement_\mathbb{R} A$ = (-∞, 1] ∪ [3, +∞) ✗

Correct: $\complement_\mathbb{R} A$ = (-∞, 1) ∪ (3, +∞) ✓

❌ Mistake 3: De Morgan Sign Error

Wrong: $\complement(A \cup B)$ = $\complement A \cup \complement B$ ✗

Correct: $\complement(A \cup B)$ = $\complement A \cap \complement B$ ✓

Study Tips

1. ✅ Always identify U first: The universal set determines the complement
2. ✅ Flip boundaries for intervals: Open ↔ closed when taking complement
3. ✅ Master De Morgan's Laws: "Break the bar, change the sign"
4. ✅ Double complement returns original: $\complement(\complement A) = A$

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💡 Exam Tip: When taking complements of intervals, remember: closed boundary becomes open, and open becomes closed!