# Math Help - Sequence proof

1. ## Sequence proof

Q: Consider a sequence $(I_{n})_{n}$ of closed nested intervals, $I_{n} = [a_{n} , b_{n} ]$. Assume also that
the sequence of the lengths of these intervals, $(l_{n})_{n}$ with $l_{n}=b_{n}-a_{n}$ , converges to zero. Show that the
intersection $\bigcap_{n}\\I_{n}$ consists of just one point.

A: My thinking is to use the axiom of completeness to conclude $(I_{n})$ is bounded. Then, by the nested interval property we know the intersection is non-void and that we have a monotone sequence. And, since $(I_{n})$ is bounded and monotone, by the monotone convergence theorem we know $(I_{n})$ is convergent. We know that $(I_{n})$ 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, $(I_{n})$ 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.

2. I would look at the sequences of endpoints. The left-hand endpoints $a_n$ form an increasing sequence of numbers that is bounded above. The right-hand endpoints $b_n$ form an decreasing sequence of numbers that is bounded below. Show that these two sequences converge to the same limit c. Then deduce that $\textstyle\bigcap I_n = \{c\}$.