Give an example or state it's impossible

Question

A sequence of closed bounded (not necessarily nested) intervals $I_1,I_2,I_3,\dots$ with the property that $\bigcap_{n=1}^NI_n\neq\emptyset$ for all $N\in\mathbb{N}$, but $\bigcap_{n=1}^{\infty}I_n=\emptyset$.

I tried to come up with some examples but failed. One example I came up with is a "shifting" sequence of closed bounded intervals, $I_n=[a_n,b_n]$ with $(a_n)=(1,2,3,4,5,\dots,N,N+1,\dots)$ and $(b_n)=(N,N+1,N+2,\dots)$. Then $\bigcap_{n=1}^NI_n =\{N\}$ and $\bigcap_{n=1}^{\infty}I_n=\emptyset$. But clearly it's wrong...

Answer 1

Summarizing the comments, there exists no such sequence. Suppose that each $I_n$ is a closed, bounded interval and that for each $n$, we have $\bigcap_{k=1}^{n}I_k \neq \emptyset$. Define $J_n = \bigcap_{k=1}^{n}I_k$. Then:

• $J_n$ is closed, because it is an intersection of closed sets.
• $J_n$ is bounded, because it is contained in $I_1$.
• $J_n$ is nonempty by assumption.
• $J_n$ is an interval, because a nonempty intersection of finitely many intervals is an interval (easy proof by induction).
• $(J_n)_{n=1}^{\infty}$ is a decreasing nested sequence, because for each $n$, we have $J_{n+1} = I_{n+1} \cap J_n \subseteq J_n$.

By the nested interval theorem, we conclude that $\bigcap_{n=1}^{\infty}J_n$ is nonempty. But you can easily verify that $\bigcap_{n=1}^{\infty}J_n = \bigcap_{n=1}^{\infty}I_n$ (show that they are subsets of each other), so $\bigcap_{n=1}^{\infty}I_n$ is nonempty.