Problem 1 (Nested interval theorem). Let an, bn ∈ R be numbers such that an ≤ bn foreach n ∈ N. Suppose that [an+1, bn+1] ⊆ [an, bn] for all n ∈ N. Prove that the set⋂n∈N[an, bn] = {x ∈ R : x ∈ [an, bn] for all n ∈ N}is nonempty. (Hint: Try sup{an : n ∈ N}.)

To prove that the set ⋂n∈N[an, bn] is nonempty, we can use the Nested Interval Theorem.

Step 1: Define the set A = {an : n ∈ N}.

Step 2: Notice that A is bounded above by any bn. This is because an ≤ bn for all n ∈ N, and [an+1, bn+1] ⊆ [an, bn] for all n ∈ N.

Step 3: By the completeness axiom, every nonempty set of real numbers that is bounded above has a least upper bound. Therefore, A has a least upper bound, which we'll call x.

Step 4: We claim that x ∈ [an, bn] for all n ∈ N.

Step 5: To prove this claim, we'll use contradiction. Suppose x ∉ [an, bn] for some n ∈ N.

Step 6: If x < an, then an would be an upper bound of A that is less than x, which contradicts the definition of x as the least upper bound of A.

Step 7: If x > bn, then because [an+1, bn+1] ⊆ [an, bn], x would not be an upper bound of A, which also contradicts the definition of x as the least upper bound of A.

Step 8: Therefore, x must be in [an, bn] for all n ∈ N.

Step 9: This means that x is in the intersection of all the intervals [an, bn], so the set ⋂n∈N[an, bn] is nonempty.

Therefore, we have proved that the set ⋂n∈N[an, bn] is nonempty.