Prove that with arbitrary method If A and B are two subsets of a universal set, then

(1) Α-Α - Φ

(ii) A-B=

(iii) A = U-A'

(iv) A - Φ= A

(v) A-B=B-A, if and only if A = B (vi) A - B = Φ if and only if  A ⊆ B
(Vii ) A- B = A  if and only if A∩B = Φ

To prove these set-theoretic properties, let's carefully analyze each statement:

### **(i) A - A = Φ**
- **Proof:**
  - The difference \( A - A \) means all elements in \( A \) that are not in \( A \). Since every element of \( A \) is in \( A \), there are no elements left, hence \( A - A = \emptyset \).

### **(ii) A - B**
- **Clarification Required:**
  - This seems incomplete. Did you mean to prove something specific about \( A - B \)? For example, \( A - B = A \setminus B \)?

### **(iii) A = U - A'**
- **Proof:**
  - Here, \( U \) is the universal set, and \( A' \) denotes the complement of \( A \) in \( U \). The difference \( U - A' \) consists of all elements in \( U \) that are not in \( A' \). By the definition of a complement, \( A' \) consists of all elements not in \( A \). Thus, \( U - A' \) is precisely \( A \), so \( A = U - A' \).

### **(iv) A - Φ = A**
- **Proof:**
  - The difference \( A - \emptyset \) means all elements in \( A \) that are not in \( \emptyset \). Since \( \emptyset \) contains no elements, \( A - \emptyset = A \).

### **(v) A - B = B - A \text{ if and only if } A = B**
- **Proof:**
  - **(⇒ direction):** Suppose \( A - B = B - A \). Consider any element \( x \in A \). Since \( A - B = B - A \), \( x \) must also belong to \( B \), otherwise it would be in \( A - B \) but not in \( B - A \). Similarly, any element \( x \in B \) must belong to \( A \). Thus, \( A = B \).
  - **(⇐ direction):** If \( A = B \), then clearly \( A - B = B - A = \emptyset \).

### **(vi) A - B = Φ \text{ if and only if } A \subseteq B**
- **Proof:**
  - **(⇒ direction):** Suppose \( A - B = \emptyset \). This means there are no elements in \( A \) that are not in \( B \), implying \( A \subseteq B \).
  - **(⇐ direction):** If \( A \subseteq B \), then all elements of \( A \) are in \( B \), so there are no elements left in \( A - B \), hence \( A - B = \emptyset \).

### **(vii) A - B = A \text{ if and only if } A \cap B = Φ**
- **Proof:**
  - **(⇒ direction):** Suppose \( A - B = A \). This implies no elements from \( A \) are removed when subtracting \( B \), which means \( A \) and \( B \) have no elements in common, i.e., \( A \cap B = \emptyset \).
  - **(⇐ direction):** If \( A \cap B = \emptyset \), then no elements from \( A \) are in \( B \), so subtracting \( B \) from \( A \) has no effect, hence \( A - B = A \).

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### Do you want more details or have any questions?

#### **5 Related Questions:**
1. Can you prove the distributive property of sets, \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)?
2. What is the relationship between \( A \cap A' \) and \( \emptyset \)?
3. How does De Morgan's Law relate to the complements of union and intersection of sets?
4. Can we generalize the property \( A - B = \emptyset \) to any arbitrary number of subsets?
5. How does the concept of symmetric difference relate to the proof of \( A - B = B - A \)?

#### **Tip:**
When dealing with set operations, Venn diagrams are an excellent tool for visualizing and understanding relationships between sets.

Explore set theory properties with a focus on set difference and complement. Learn proofs for key properties including Α-Α = Φ, A = U - A', and more. Suitable for advanced students and enthusiasts of mathematical set theory.

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