集合的定义
CSCA 集合的定义 study guide aligned to the official syllabus. Practice 数学 questions on aicsca.com.
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This study guide follows the official CSCA exam syllabus for international undergraduate applicants.
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International students preparing for CSCA Math, Physics, Chemistry, or Chinese exams.
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Tutorial Content
Definition of a Set
1. Core Concepts
In mathematics, a **Set** is a collection of **well-defined, distinct** objects. Each object in a set is called an **Element**.

**Three Key Properties (Exam Focus)**:
1. **Well-definedness**:
* For any object, logic must clearly determine whether it belongs to the set. Ambiguity is not allowed.
* *Example*: "Tall people" implies ambiguity; "People taller than 180cm" is a set.
2. **Distinctness (Mutual Exclusivity) —— 🚩 Common Pitfall**:
* Every element in a set must be unique. **If a calculation results in duplicate elements, that solution must be rejected.**
* *Example*: $\{1, 2, 2\}$ is incorrect notation; it should be written as $\{1, 2\}$.
3. **Unorderedness**:
* The order of elements does not matter.
* *Example*: $\{1, 2, 3\}$ is the same set as $\{3, 1, 2\}$.
2. Notation & Number Sets
* **Relationship between Element and Set**:
* $a \in A$: $a$ belongs to set $A$ (is an element of).
* $a \notin A$: $a$ does not belong to set $A$.
* **Standard Number Set Symbols (Must Memorize)**:

* **$\mathbb{N}$**: Natural Numbers, includes $0, 1, 2, ...$
* **$\mathbb{Z}$**: Integers, includes $..., -1, 0, 1, ...$
* **$\mathbb{Q}$**: Rational Numbers (fractions, terminating/repeating decimals).
* **$\mathbb{R}$**: Real Numbers.
* *Note*: Sometimes $\mathbb{N}^*$ or $\mathbb{N}_+$ represents **Positive Integers** (excluding 0).
3. Worked Examples
**Example 1: Determining Sets**
(1) Numbers very close to 0.
(2) The solution to $x^2 - 2x + 1 = 0$.
**Solution**:
(1) **No**. "Very close" is not well-defined.
(2) **Yes**. The solution is $x=1$. The set is $\{1\}$.
**Example 2: Testing Distinctness (Important)**
Given set $A = \{a, a^2, 2\}$ and $1 \in A$, find the value of real number $a$.
**Solution**:
Since $1 \in A$, $1$ must be equal to either $a$ or $a^2$ (obviously $1 \neq 2$).
**Case 1**: If $a = 1$.
Substitute into $A$: $A = \{1, 1^2, 2\} = \{1, 1, 2\}$. This creates duplicate elements, which **violates the property of Distinctness**. Therefore, **$a \neq 1$**.
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**Case 2**: If $a^2 = 1$.
Solving gives $a = 1$ or $a = -1$.
From Case 1, we know $a \neq 1$.
Check $a = -1$: Substitute into $A$: $A = \{-1, (-1)^2, 2\} = \{-1, 1, 2\}$. All elements are distinct. This is valid.
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**Final Answer**: $a = -1$.