Thread: Sequence proof

1. Sequence proof

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

A: My thinking is to use the axiom of completeness to conclude $\displaystyle (I_{n})$ is bounded. Then, by the nested interval property we know the intersection is non-void and that we have a monotone sequence. And, since $\displaystyle (I_{n})$ is bounded and monotone, by the monotone convergence theorem we know $\displaystyle (I_{n})$ is convergent. We know that $\displaystyle (I_{n})$ 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, $\displaystyle (I_{n})$ 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 $\displaystyle a_n$ form an increasing sequence of numbers that is bounded above. The right-hand endpoints $\displaystyle b_n$ form an decreasing sequence of numbers that is bounded below. Show that these two sequences converge to the same limit c. Then deduce that $\displaystyle \textstyle\bigcap I_n = \{c\}$.