nested intervals and least upper bound question?

Question

assuming the least upper bound property, how can we prove the nested intervals property?

Answer 1

Let [a₁, b₁], [a₂, b₂], [a₃, b₃]... be a sequence of closed bounded intervals such that for every n, [a_n, b_n] ⊇ [a_(n+1), b_(n+1)]. Note that this implies that a_n ≤ a_(n+1) for every n, so the a_n form a monotone increasing sequence. Further, b_n ≥ b_(n+1) for every n, so the b_n form a monotone decreasing sequence.

Now, let m, n be any two natural numbers. If n≤m, then a_n≤a_m (since the a_n are increasing), and a_m≤b_m, so a_n≤b_m. Conversely, if m
Since {a_n} is a nonempty set bounded above by b₁, it has a least upper bound, s. Let m be any natural number. Since s is an upper bound, we have a_m≤s. Suppose b_m < s -- per the previous paragraph b_m is an upper bound for all the a_m, but since it is smaller than the smallest upper bound, this is a contradiction. Therefore, b_m≥s. It follows that s∈[a_m, b_m]. And since this holds for every m∈ℕ, it follows that s∈[m=1, ∞]⋂[a_m, b_m]. Therefore, the intersection of any sequence of nested closed bounded intervals has at least one element. Q.E.D.