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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!

user126743
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2 Answers2

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

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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.

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.

5xum
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  • You're right, I think I need to add a condition on the sets that they be closed... – user126743 May 20 '14 at 07:47
  • @user126743 I added an example using closed sets which does not work in a non-compact space. In a compact space using closed sets, however, the intersection will not be empty. – 5xum May 20 '14 at 07:54
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    Thanks, so I guess that I must either require closed sets + compactness, or converging diameter + completeness? – user126743 May 20 '14 at 07:58