# Give an example or state it's impossible

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...

• What are your thoughts ? – krirkrirk Mar 8 '18 at 16:11
• Hint: They cannot be nested… – Gono Mar 8 '18 at 16:12
• Welcome to stackexchange. You are more likely to get answers rather than downvotes or votes to close if you edit the question to show what you tried and where you are stuck. – Ethan Bolker Mar 8 '18 at 16:14
• To comment on your example, you can't let the intervals depend on $N$. That's backwards. You must first give a fixed collection of intervals which do not change, and then you show that the intersection is non-empty for any $N$. – Arthur Mar 8 '18 at 16:22
• Hint: suppose this were possible. Then $J_n = \bigcap_{k=1}^{n}I_k$ is a decreasing sequence of nonempty closed bounded sets. What can you say about $\bigcap_{n=1}^{\infty}J_n$? – Bungo Mar 8 '18 at 16:40

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.