# 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!

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I think you mean to write $\bigcap_{i = 1}^\infty F_i$ here. –  Dylan Moreland Mar 5 '12 at 22:58
Oh is it impossible if it's only 1 to n? –  user26069 Mar 5 '12 at 23:01
If $F_1$ and $F_2$ are nonempty, and $F_1$ contains $F_2$, then how could $F_1\cap F_2$ possibly be empty? –  Gerry Myerson Mar 5 '12 at 23:03
Is a possible answer $\bigcap_{n=1}^{\infty} [1-\frac{1}{n}, 1]$? –  user26069 Mar 5 '12 at 23:04
What does this have to do with either unions or compact sets? Looks like intersections and closed sets. –  Thomas Andrews Mar 5 '12 at 23:05
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You should take $F_n=[n,+\infty)$ then the intersection is empty.
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).