When does a descending sequence of nonempty sets have a non empty intersection?

Question

Let $\langle F_n\rangle_{n\in\Bbb{N}}$ be a descending sequence of nonempty sets in a Metric Space - $F_1\supset F_2\supset\cdots$. What are the conditions on the underlying space so that $\bigcap_{n=1}^{\infty}F_n\ne \emptyset$?

On the one hand I know compactness of the space is sufficient, but I think it's not necessary. It seems that if I'm dealing with a metric space, and the space isn't complete then this doesn't hold either (simple counter examples in $\mathbb{Q}$). But is completeness sufficient? How is this formally shown?

Thanks!

Answer 1

The completness is sufficient if the diameters of $F_n$ tends to $0$. Find a Cauchy sequence which has a limit in the intersection.

Answer 2

Compactness of the space is not sufficient. Take, for example, $$F_1=\{1,\frac12,\frac13,\dots\}\\ F_2=\{\frac12,\frac13,\dots\}\\ \vdots\\ F_n=\{\frac1n,\frac{1}{n+1},\dots\}\\ \vdots$$

for which the intersection is an empty set.

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But what about if the sets are closed? Well then completeness will not be enough to provide you with an empty intersection, as the example of $F_n=[n,\infty)\subset\mathbb R$ clearly shows.