if 𝑓: 𝑋 →𝑌 𝑎𝑛𝑑 𝐴, 𝐵 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝑋, 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡I )𝑓(𝐴 ∪ 𝐵) = 𝑓(𝐴) ∪ 𝑓(𝐵)II) 𝑓(𝐴 ∩ 𝐵) ⊆ 𝑓(𝐴) ∩ 𝑓(𝐵)

To prove these properties of functions, we will proceed step by step. Let's break down each part of the proof.

### Problem Recap:
We are given a function \( f: X \to Y \), and two subsets \( A \) and \( B \) of \( X \). We need to prove the following two properties:

**I.** \( f(A \cup B) = f(A) \cup f(B) \)

**II.** \( f(A \cap B) \subseteq f(A) \cap f(B) \)

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### Proof of I: \( f(A \cup B) = f(A) \cup f(B) \)

To prove this equality, we need to show two things:
1. \( f(A \cup B) \subseteq f(A) \cup f(B) \)
2. \( f(A) \cup f(B) \subseteq f(A \cup B) \)

#### 1. \( f(A \cup B) \subseteq f(A) \cup f(B) \)

- Let \( y \in f(A \cup B) \). Then, by the definition of image, there exists some \( x \in A \cup B \) such that \( f(x) = y \).
- Since \( x \in A \cup B \), this means that \( x \in A \) or \( x \in B \).
- If \( x \in A \), then \( f(x) \in f(A) \), and if \( x \in B \), then \( f(x) \in f(B) \).
- Therefore, \( y = f(x) \in f(A) \cup f(B) \).

This shows that \( f(A \cup B) \subseteq f(A) \cup f(B) \).

#### 2. \( f(A) \cup f(B) \subseteq f(A \cup B) \)

- Let \( y \in f(A) \cup f(B) \).
- This means that either \( y \in f(A) \) or \( y \in f(B) \).
- If \( y \in f(A) \), then there exists some \( x \in A \) such that \( f(x) = y \).
- Similarly, if \( y \in f(B) \), there exists some \( x \in B \) such that \( f(x) = y \).
- In either case, \( x \in A \cup B \), so \( f(x) = y \in f(A \cup B) \).

This shows that \( f(A) \cup f(B) \subseteq f(A \cup B) \).

Since both inclusions hold, we have \( f(A \cup B) = f(A) \cup f(B) \).

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### Proof of II: \( f(A \cap B) \subseteq f(A) \cap f(B) \)

To prove this inclusion, we need to show that if \( y \in f(A \cap B) \), then \( y \in f(A) \cap f(B) \).

- Let \( y \in f(A \cap B) \). Then, there exists some \( x \in A \cap B \) such that \( f(x) = y \).
- Since \( x \in A \cap B \), we know that \( x \in A \) and \( x \in B \).
- Therefore, \( f(x) \in f(A) \) and \( f(x) \in f(B) \), which means \( y \in f(A) \cap f(B) \).

Thus, we have \( f(A \cap B) \subseteq f(A) \cap f(B) \).

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### Conclusion:
We have proved the following two properties:
1. \( f(A \cup B) = f(A) \cup f(B) \)
2. \( f(A \cap B) \subseteq f(A) \cap f(B) \)

Do you want more details or have any questions about the proof?

Here are some related questions you might explore:
1. What conditions would make \( f(A \cap B) = f(A) \cap f(B) \)?
2. Can you provide a counterexample where \( f(A \cap B) \neq f(A) \cap f(B) \)?
3. How does the proof change if \( f \) is injective?
4. What happens to the properties if \( f \) is surjective?
5. Can these properties be extended to infinite unions or intersections?

**Tip:** In proving set equalities, always check for two inclusions: one subset of the other and vice versa.

if f: X → Y and A, B are two subsets of X, prove that I) f(A ∪ B) = f(A) ∪ f(B) II) f(A ∩ B) ⊆ f(A) ∩ f(B)

Learn how to prove the properties of functions on set operations: f(A ∪ B) = f(A) ∪ f(B) and f(A ∩ B) ⊆ f(A) ∩ f(B). This problem explores set theory and functions, demonstrating key properties of unions and intersections. Suitable for university-level discrete mathematics students.

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