Thread: Sequence proof

Q: Consider a sequence of closed nested intervals, . Assume also that
the sequence of the lengths of these intervals, with , converges to zero. Show that the
intersection consists of just one point.

A: My thinking is to use the axiom of completeness to conclude is bounded. Then, by the nested interval property we know the intersection is non-void and that we have a monotone sequence. And, since is bounded and monotone, by the monotone convergence theorem we know is convergent. We know that has a convergent subsequence and by our hypothesis (and Bolzona Weierstrass Theorem). Also, theorem 2.5.2 states, "subsequences of a convergent sequence converge to the same limit as the original sequence". So, also must converge to zero.

From here I am not sure what to do. We just got into Cauchy sequences and how to do basic proofs with some Cauchy theorems. I am not sure if I need to use any of those theorems to prove the above.

I would appriciate some guidence. I am having a really hard time with this one.

I would look at the sequences of endpoints. The left-hand endpoints form an increasing sequence of numbers that is bounded above. The right-hand endpoints form an decreasing sequence of numbers that is bounded below. Show that these two sequences converge to the same limit c. Then deduce that .