An empty intersection of decreasing sequence of closed sets

What is an example of a family of closed subsets $F_0 \supset F_1 \supset F_2 \supset \dots$ of $\mathbb{R}$ so that $F_n \neq \emptyset$ for each $n$ and $\bigcap_{i=1}^n F_i = \emptyset$?

Thanks!

• Is a possible answer $\bigcap_{n=1}^{\infty} [1-\frac{1}{n}, 1]$?
– user26069
Commented Mar 5, 2012 at 23:04
• @user26069 That wouldn't work, since $1$ is in that intersection. More generally, I know something has to be in there because of the finite intersection property of compact sets. Commented Mar 5, 2012 at 23:05
• isn't this infinite intersection @DylanMoreland? Commented Nov 18, 2020 at 7:22
• @user26069 had you wanted finite intersection from 1 to n or infinite intersection which is something I observe in your answer, in the comments to your question? Commented Nov 18, 2020 at 7:26

You should take $F_n=[n,+\infty)$ then the intersection is empty.

• How do you prove that the infinite intersection is empty
– wizz
Commented Sep 4, 2020 at 13:18
• @KaanYolsever Show that for every $x \in \mathbb{R}$ there exists $F_n$ with $x \notin F_n$. Commented Apr 7, 2022 at 15:17

If $F_{n+1}\subseteq F_n$ then $\bigcap_{i=1}^n F_i = F_n$. This means that the family $\{F_n\mid n\in\mathbb N\}$ has the finite intersection property. In a compact space, this would mean that $\bigcap_{i=1}^\infty F_n\neq\varnothing$.

By that a decreasing sequence of sets whose intersection is empty it would have to be non-compact. In $\mathbb R$ this would mean that the sets are unbounded, so examples of the form $F_n=[a_n,+\infty)$ are essentially the only form of examples you can find (of course $(-\infty,b_n]$ is equally valid).

You should take $F_n=[n,+\infty)$ then the intersection is empty.

This is correct. Every finite intersection has at least one point, but the infinite intersection is empty. As n goes from 1 to infinity, each integer at the start cannot be in the intersection of all. With n=2, the number 1 cannot be in the infinite intersection. With n=3, the number 2 cannot be in the infinite intersection, and so on. Choose any integer m. Then there is an n greater than m such that m is not in the infinite intersection.

I was going to post here, complete metric space $X$ and nested sequence of closed sets $A_m \subset X$ where $\bigcap_{n=1}^\infty = \emptyset$, but it got closed as I was writing my answer...

Anyways, for a more interesting example in which the sets are bounded $($for the link above, which has a slightly different problem statement$)$, let $X$ be the set of all bounded infinite sequences of real numbers $x = (x_1, x_2, \dots)$ and let $d(x,y) = \sup_k |x_k - y_k|$. This is evidently a complete metric space. Take $F_n$ the set of all sequences that converge to $1$, have terms $[-2, 2]$ but have $0$ as each of the first $n$ terms. The sets are obviously nested, and this sequence is manifestly not in any of the sets $F_n$ since the terms do not converge to $1$, so $\bigcap F_n = \emptyset$. Finally for each $x = (x_1, x_2, \dots) \notin F_n$, define$$r_n(x) = \sup(|x_1|, |x_2|, \dots, |x_n|, \limsup_{k \to \infty}|x_k - 1|, |x_1| - 2, |x_2| - 2, |x_3|-2, \dots).$$We have $r_x > 0$ if $x \notin F_n$, and by construction every point in the neighborhood of $x$ of radius $r_n(x)$ fails to meet one of the three criteria for belonging to $F_n$. Thus $F_n$ has an open complement for all $n$, so each $F_n$ is closed. We conclude that the sets $F_n$ have all the desired properties.