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Nested interval theorem. Suppose we have two sequences $$a,b : \mathbb{N} \rightarrow \mathbb{R}$$ such that $a \leq b$ and $|b-a| \rightarrow 0$.

Then the intersection $$\bigcap_{i \in \mathbb{N}}[a,b]$$

has exactly one element.

Wikipedia regards this as a version of completeness.

Suppose we want to generalize to arbitrary metric spaces. The following definition seems appropriate.

Nested set property. Let $X$ denote a metric space. Then $X$ is said to have the nested set property iff for all sequences $A : \mathbb{N} \rightarrow \mathcal{P}(X)$, we have that if $A$ is order-reversing, and if each $A_i$ is closed and non-empty, and if the sequence $\operatorname{diam}(A)$ converges to $0$, then $$\bigcap_{i \in \mathbb{N}} A_i$$ has exactly one element.

Question. Is the nested set property equivalent to Cauchy completeness?

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goblin GONE
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  • Where it says $\mathcal P(\mathbb R),$ was it supposed to be $\mathcal P(X)\text{?} \qquad$ – Michael Hardy Aug 08 '17 at 04:21
  • @MichaelHardy, thank you, and yes. – goblin GONE Aug 08 '17 at 04:21
  • Suppose all except finitely many of the closed sets in the sequence are empty. Then the intersection is empty even when $X$ is complete. – Michael Hardy Aug 08 '17 at 04:24
  • ok, You've now edited to take care of that. I remark that $\operatorname{diam}(\varnothing) = 0$ because the supremum of the empty set of non-negative real numbers is $0. \qquad$ – Michael Hardy Aug 08 '17 at 04:26
  • @MichaelHardy, yes, sure, indeed they could all just be empty. I've edited to eliminate this condition. Originally there was a connectedness assumption too, which implicitly included non-emptiness, but I ended up deciding it probably wasn't too relevant and removing it. Thanks for your help so far. – goblin GONE Aug 08 '17 at 04:27

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Suppose $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence. Let $A_n$ be the sequence $a_n, a_{n+1},a_{n+2},\ldots$ If this Cauchy sequence does not converge, then the set $A_n$ is closed. The fact that it is a Cauchy sequence implies the diameter approaches $0.$ But $\bigcap_{n=1}^\infty A_n = \varnothing.$ Thus if a metric space is not complete, then it does not have the nested set property.

Now suppose the space is complete. Let $a_n\in A_n.$ Then the fact that the diameter approaches $0$ means $a_1,a_2,a_3,\ldots$ is a Cauchy sequence. It converges. Since the intersection of arbitrarily many closed sets is closed, the limit is a member of $A_n.$ Thus the intersection contains at least one point. But it cannot contain more than one: Suppose $a\ne b$ are both members of the intersection. Let $\varepsilon = d(a,b)/2.$ For some $n,$ you have $\operatorname{diam} A_n < \varepsilon.$ Hence $A_n$ must exclude either $a$ or $b.$

Thus equivalence holds.

Michael Hardy
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