If the intersection of nested intervals is a one-point set, can we conclude the diameters tend to zero?

Question

The theorem of nested intervals is well known:

Let $(I_n)$ be a sequence of closed (non-degenerated) intervals in $\mathbb{R}$. If $I_{n+1}\subseteq I_n$ and $\lim_nD(I_n)=0$ (where $D$ is diameter), then $\bigcap_nI_n$ is a one-point set.

My question now is the following: the $I_n$'s are again nested, but now we (only) know that $\bigcap_nI_n$ is a one-point set. Is it possible to conclude that $\lim_nD(I_n)=0$?

I think the answer is yes. But I couldn't prove it yet.

Answer 1

Yes, it is true. Assume otherwise.

Let $p\in\mathbb R$ be such that $\{p\}=\bigcap_{n\in\mathbb N}I_n$. Since the $I_n$'s are nested, the sequence $\bigl(D(I_n)\bigr)_{n\in\mathbb N}$ is decreasing and therefore it converges to some $r$, which is greater than $0$; since we are assuming that it is not $0$. Since each $I_n$ is an interval, for each $n\in\mathbb N$ one of the numbers $p\pm\frac r2$ belongs to $I_n$. But, since $(I_n)_{n\in\mathbb N}$ is a sequnce of nested intervalas, we deduce from this that at lest one of them belongs to every $I_n$. So, at least one of them belongs to $\bigcap_{n\in\mathbb N}I_n$ too, contradicting our assumption.

Answer 2

Let $a$ be the point. Take $r>0$. Eventually both $a+r\notin I_n$ and $a-r\notin I_n$ and this means that $I_n$ has length $<2r$.

Answer 3

If $I_n=[a_n, b_n] $ then it can be proved that $\bigcap_n I_n=[a, b] $ where $a=\lim a_n, b=\lim b_n$. You can now conclude that your statement is true.

Answer 4

A stronger statement is that for $I_n=[a_n,b_n]$ then $\bigcap_n I_n= [\sup a_n, \inf b_n]$. And since $\lim_n D(I_n)=\lim_n (b_n-a_n)=0$ while note that $0 \le \inf b_n - \sup a_n \le b_n-a_n$, we find $\sup a_n=\inf b_n$, which also equals to the only point in $\bigcap_n I_n$.