# Nested family of closed intervals and convergence

Let $I_n = [a_n,b_n]$ be a nested family of closed intervals, that is, $a_n,b_n \in \Bbb R$, $a_n < b_n$, with $a_n ≤ a_n+1$ and $b_n ≥ b_n+1$ for every $n$. Assume that $\lim_{n\to\infty}(b_n − a_n) = 0$. Suppose $\{x_n\}_{n≥1}$ is a sequence of real numbers, where $x_n \in I_n$ for every $n$. Show that $\{x_n\}_{n≥1}$ is convergent.

As the sequence $(a_n)$ and $(b_n)$ are both convergent and converges to the same limit hence the nested interval has a convergent sequence now $(x_n)$ belongs to $(a_n)$ or $(b_n)$ or intersection of both two. Now if it belongs to $a_n$ or $b_n$ then convergent but what if it belongs to its intersection As the intersection will be non empty then how to proceed?

• What do you mean by the intersection of the sequences $(a_n)$ and $(b_n)$? And if the intervals are all nondegenerate, there’s no reason to think that any $x_n$ belongs to either of them. For example, you could have $x_n=(a_n+b_n)/2$.
– MPW
Commented Feb 24, 2018 at 10:14
• What do you mean with "...$(x_n)$ belongs to $(a_n)$..."? Commented Feb 24, 2018 at 10:17
• If (xn) is a subsequence of (an) or (bn) Commented Feb 24, 2018 at 10:20
• Does one need cauchy's theorem on intervals to prove it Commented Feb 24, 2018 at 10:21
• This is just the so-called “squeeze theorem”, since $a_n\leq x_n\leq b_n$
– MPW
Commented Feb 24, 2018 at 10:25

Suppose that $(x_n)_n$ does not converge.

Then it is not a Cauchy-sequence so an $\epsilon>0$ exists such that for all $n\in\mathbb N$ we can find $k,m\geq n$ with $|x_k-x_m|>\epsilon$.

However, we have $a_n\leq x_k\leq b_n$ and $a_n\leq x_m\leq b_n$ so that $|x_k-x_m|\leq b_n-a_n$.

• You can choose the $n$ as large as you want, and for $n$ large enough we have $b_n-a_n\leq\epsilon$. This combined with $|x_k-x_m|\leq b_n-a_n$ contradicts $|x_k-x_m|>\epsilon$. Commented Feb 24, 2018 at 11:02