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To prove the two statements involving the supremum (sup) and infimum (inf) of the union of two non-empty subsets \( A \) and \( B \) of real numbers, let's go through the details step-by-step.

### 1. Proving that:
\[
\sup(A \cup B) = \max(\sup A, \sup B)
\]

- **Supremum Definition:** The supremum \( \sup A \) of a set \( A \) is the least upper bound of \( A \), meaning it is the smallest real number that is greater than or equal to every element in \( A \).

- **Step 1:** Since \( \sup A \) is the least upper bound of \( A \), and \( \sup B \) is the least upper bound of \( B \), both \( \sup A \) and \( \sup B \) are upper bounds for \( A \cup B \). Therefore, the supremum of \( A \cup B \) must be less than or equal to \( \max(\sup A, \sup B) \), as \( \max(\sup A, \sup B) \) is also an upper bound of \( A \cup B \).

- **Step 2:** Now, let's check the reverse inequality. If \( x \in A \cup B \), then either \( x \in A \) or \( x \in B \). In both cases, \( x \leq \sup A \) or \( x \leq \sup B \). Therefore, \( x \leq \max(\sup A, \sup B) \). This shows that \( \max(\sup A, \sup B) \) is the least upper bound for \( A \cup B \), implying:
  \[
  \sup(A \cup B) = \max(\sup A, \sup B).
  \]

### 2. Proving that:
\[
\inf(A \cup B) = \min(\inf A, \inf B)
\]

- **Infimum Definition:** The infimum \( \inf A \) of a set \( A \) is the greatest lower bound of \( A \), meaning it is the largest real number that is less than or equal to every element in \( A \).

- **Step 1:** Since \( \inf A \) is the greatest lower bound of \( A \), and \( \inf B \) is the greatest lower bound of \( B \), both \( \inf A \) and \( \inf B \) are lower bounds for \( A \cup B \). Therefore, the infimum of \( A \cup B \) must be greater than or equal to \( \min(\inf A, \inf B) \), as \( \min(\inf A, \inf B) \) is also a lower bound of \( A \cup B \).

- **Step 2:** For the reverse inequality, if \( x \in A \cup B \), then either \( x \in A \) or \( x \in B \). In both cases, \( x \geq \inf A \) or \( x \geq \inf B \). Therefore, \( x \geq \min(\inf A, \inf B) \). This shows that \( \min(\inf A, \inf B) \) is the greatest lower bound for \( A \cup B \), implying:
  \[
  \inf(A \cup B) = \min(\inf A, \inf B).
  \]

### Conclusion:
We have demonstrated that:
1. \( \sup(A \cup B) = \max(\sup A, \sup B) \),
2. \( \inf(A \cup B) = \min(\inf A, \inf B) \).

Would you like further details or clarifications? Feel free to ask!

Here are five related questions to expand on the topic:
1. What is the relationship between supremum and maximum for finite sets?
2. How can we use the properties of infimum and supremum to prove similar results for intersections of sets?
3. In what way would the proof change if \( A \) or \( B \) were empty?
4. How do these results relate to the concept of closure in topology?
5. Can the supremum and infimum be defined for sets that are not subsets of real numbers?

**Tip:** When dealing with sets in analysis, always verify whether the set is bounded to ensure the existence of supremum and infimum.

Let A and B be non-empty subsets of real numbers. Show that sup(A ∪ B) = max{sup A, sup B} and inf(A ∪ B) = min{inf A, inf B}.

This proof shows that for any two non-empty subsets of real numbers A and B, the supremum of their union is equal to the maximum of their suprema, and the infimum of their union is equal to the minimum of their infima. This problem is part of real analysis and set theory.

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