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Let $X$ denote a complete metric space. Suppose $\mathcal{C}$ is a collection of closed and bounded subsets of $X$ with the finite intersection property.

Question. Is $\bigcap\mathcal{C}$ necessarily non-empty?

A couple of remarks.

Firstly:

  • To see that closedness is necessary, consider the sequence of open intervals $(0,1/n)$ as subsets of the real line.

  • To see that boundedness is necessary, consider the sequence of closed intervals $[n,\infty)$ as subsets of the real line.

Secondly: I'm also interested in the converse; that is, whether the above condition implies completeness. If so, this gives us a notion of completeness for any set $X$ equipped with both a topology and a bornology.

goblin GONE
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    I'd start by looking at the closed unit sphere of an infinite dimensional Banach space, and using some counterexample to compactness to get a counterexample. – Asaf Karagila Nov 26 '17 at 12:00
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    As Asaf said, this can't be true, because the (bounded) unit sphere of a real Banach space $X$ is not compact iff $X$ is infinite-dimensional -- but if this property held for all $\mathcal{C}$, it would have to be compact (it's an equivalent def. of compactness). So you don't need to exhibit anything, actually. – Mr. Chip Nov 26 '17 at 12:09
  • I just posted a question related to the last line of your question: https://math.stackexchange.com/q/3046536/71829 – Keshav Srinivasan Dec 20 '18 at 02:48
  • You may also be interested in my question here: https://math.stackexchange.com/q/3047070/71829 – Keshav Srinivasan Dec 20 '18 at 15:37

1 Answers1

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So, it turns out there's a really easy counter-example. Put a metric space structure on $\mathbb{N}$ by declaring that the distance between any two distinct points is exactly $1$. This is complete, because the only Cauchy-sequences are the constant sequences, which are obviously convergent. Also, every subset of $\mathbb{N}$ has diameter at most $1$, and is therefore bounded; further, the induced topology is the discrete topology, so every subset is closed. So in this case, the question reads:

Does every collection $\mathcal{C}$ of subsets of $\mathbb{N}$ with the finite intersection property have non-empty intersection?

Of course, the answer is "no": consider, for example, $$\mathcal{C} = \{[n,\infty)_{\mathbb{N}} : n \in \mathbb{N}\}.$$

goblin GONE
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  • The same example works in $\mathbb{R}$ with the complete (!) metric $d(x,y) = \min(|x-y|, 1)$, the truncated standard metric. – Henno Brandsma Nov 26 '17 at 14:06
  • This counterexample has the form of a nested sequence of sets $C_n \supset C_{n+1}$. Somewhat counter-intuitively, if one requires $\operatorname{diam} C_n\to 0$ as an additional hypothesis (i.e., the sets are "small"), then the intersection must be nonempty. –  Nov 27 '17 at 17:14
  • @6'whitemale, yes. Furthermore I think if we take the original conjecture can replace "the elements of $\mathcal{C}$ are bounded" with "the elements of $\mathcal{C}$ have arbitrarily small diameter", it becomes equivalent to completeness. – goblin GONE Nov 28 '17 at 02:41