Problem 6. Let A, B ⊆ R be nonempty and bounded sets.(a) Prove that if A ⊆ B, then inf B ≤ inf A ≤ sup A ≤ sup B.(b) Prove that sup(A ∪ B) = max{sup A, sup B}. (Hint: A is a subset of A ∪ B, so we canapply part (a) to these two sets.

(a) We need to prove that if A ⊆ B, then inf B ≤ inf A ≤ sup A ≤ sup B.

Proof:

1. Since A ⊆ B, every element of A is also an element of B. Therefore, the infimum of B (inf B) is less than or equal to the infimum of A (inf A), because inf B is the greatest lower bound of B and hence also a lower bound for A.

2. Similarly, the supremum of A (sup A) is less than or equal to the supremum of B (sup B), because sup B is the least upper bound of B and hence also an upper bound for A.

3. Since every element of A is less than or equal to sup A and greater than or equal to inf A, it follows that inf A ≤ sup A.

Therefore, we have inf B ≤ inf A ≤ sup A ≤ sup B.

(b) We need to prove that sup(A ∪ B) = max{sup A, sup B}.

Proof:

1. Since A is a subset of A ∪ B, from part (a), we know that sup A ≤ sup(A ∪ B).

2. Similarly, since B is a subset of A ∪ B, we know that sup B ≤ sup(A ∪ B).

3. Therefore, max{sup A, sup B} ≤ sup(A ∪ B).

4. On the other hand, since every element of A ∪ B is either in A or in B, the supremum of A ∪ B is less than or equal to the maximum of sup A and sup B.

5. Therefore, sup(A ∪ B) ≤ max{sup A, sup B}.

From 3 and 5, we conclude that sup(A ∪ B) = max{sup A, sup B}.