Does there exist a sequence $\{I_n\}$ of intervals such that $I_{n+1} \subsetneq I_n , \forall n \ge 1$ and $\cap_{n=1}^\infty I_n=(0,1)$ ?

Question

Does there exist a sequence $\{I_n\}$ of intervals such that $I_{n+1} \subsetneq I_n , \forall n \ge 1$ and $\cap_{n=1}^\infty I_n=(0,1)$ ? I know that not all the intervals can be closed , as otherwise their intersection would be closed , but $(0,1)$ is not closed . I have also done that assuming $I_n$'s are bounded , let $b_n:=\sup I_n$ , then $(b_n)$ is a decreasing bounded below sequence . So let $l=\lim b_n=\inf \{b_n\}$ ; now $(0,1) \subseteq I_n$ , so $1\le b_n$ always , then by taking limit , $1 \le l \le b_n , \forall n \ge1$ . Similarly considering $a_n:=\inf I_n$ ,$s:=\lim a_n= \sup \{a_n\}$ , we get $a_n \le s \le 0$ . But then I am stuck . I don't even have any idea what happens if some of the $I_n$'s are unbounded (above or below or both) . Moreover , is there any other way around without sequences ? Please help . Thanks in advance .

Answer 1

As $0$ is not in the intersection, there exists some $n_0$ with $0\notin I_{n_0}$. Likewise there exists $n_1$ with $1\notin I_{n_1}$. By the nesting property, $n\ge\max\{n_0,n_1\}$ implies $0\notin I_n$ and $1\notin I_n$ (whereas of course $\frac12\in I_n$) so that $I_n\subseteq (0,1)$. But then $$ (0,1)=\bigcap_{k=1}^\infty I_k\subseteq I_{n+1}\subsetneq I_n\subseteq (0,1)$$ contradiction!