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Is there any literature on (and a standard name for) the proposition that for any arbitrary-cardinality collection of closed intervals in the reals that is nested (in the sense that, given any two of the intervals, one of them is a subset of the other), the intersection of all the intervals is non-empty?

Note that the standard nested interval theorem assumes that the collection of intervals is countable, and that it comes with an ordering such that the $j$th interval is a subset of the $i$th interval if $i < j$. In the non-standard version I'm asking about, the collection can be uncountable (e.g., the collection of all intervals $[-t,t]$ for $t>0$).

Of course this proposition isn't hard to prove, but I'm interested in other facts about the proposition, e.g., whether the proposition implies Dedekind completeness in the presence of the ordered field axioms.

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You want at least one of the intervals compact, right? Gerhard "Ask Me About System Design" Paseman, 2011.08.09 –  Gerhard Paseman Aug 9 '11 at 16:40
The nested interval theorem for reals follows from the compactness of closed bounded intervals, so perhaps generalizations of compactness might be more fruitful to pursue in the context of other ordered fields; for this purpose the following paper might be useful; it also has a brief description of Sikorski fields in the introduction, which came up in my solution to your other question: Cowles, John; LaGrange, Robert Generalized Archimedean fields. Notre Dame J. Formal Logic 24 (1983), no. 1, 133–140. –  Ali Enayat Aug 9 '11 at 17:56
@Gerhard Paseman: I don't see why I should stipulate that at least one of the intervals is compact. Can you explain? –  James Propp Aug 9 '11 at 18:15
Some consider an interval to be potentially nonbounded, e.g. (a, infinity). There are collections of these intervals which are closed subsets of the reals and are nested and have empty intersection. Requiring one interval to be compact, thus bounded, removes these collections from consideration. Gerhard "Ask Me About System Design" Paseman, 2011.08.09 –  Gerhard Paseman Aug 9 '11 at 19:41
Alternatively, make sure that the reader knows that all closed intervals you consider are bounded/compact. I still think you want at least one to be compact. Gerhard "Ask Me About System Design" Paseman, 2011.08.09 –  Gerhard Paseman Aug 9 '11 at 19:46
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2 Answers

In the case of the reals, the uncountable version reduces easily to the countable version, since there will be a countable subfamily of the given uncountable family that is cofinal in the inclusion order of the intervals. Basically, one can find a countable subfamily with the same intersection.

But the nested interval property (allowing arbitrary families) is interesting in a general linear order, where one cannot always find such countable subfamilies. In this context, the nested interval property is not equivalent to the completeness of the order. This can be seen by considering the order $\omega+\omega_1^*$, that is, the order consisting of an increasing $\omega$-sequence with a decreasing $\omega_1$ sequence on top of it. This linear order is not complete, since the initial $\omega$-sequence has no least upper bound, but it has the nested interval property for arbitrary nested sequences of intervals as a result of the mis-match in the cofinality of the gap. That is, any nested sequence of intervals $[a_\alpha,b_\alpha]$ will have nonempty intersection if the interval is eventually on one side of the gap, by compactness, and if the intervals straddle the gap, then if the sequence has countable cofinality, then the $b_\alpha$'s will be bounded in the $\omega_1^*$ part of the order, leading to a nonempty intersection; and if the sequence has uncountable cofinality, then the $a_\alpha$'s must eventually stabilize, and so again the intersection will be nonempty.

Meanwhile, the countable-NIT for ordered fields does not imply completeness, since if $F$ is, say, a nonprincipal ultrapower of the real field, then $F$ will be saturated for countable types, and the type of "being inside the intervals of a given countable nested sequence" is finitely consistent, hence realized inside the model. So any such $F$ satisfies the countable-NIT.

I am unsure about the general situation of NIT in incomplete ordered fields.

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See related math.stackexchange.com/questions/46489/… –  Joel David Hamkins Aug 9 '11 at 17:23
I had posted an answer using Sikorski's theorem (from Ali's comment) to produce an incomplete ordered field with NIT, but the argument was flawed in the case $b_\alpha-a_\alpha$ is bounded from $0$, and so I have retracted it. It isn't necessarily true that all such nested sequences have nonempty intersection; consider a nonstandard model whose standard cut has cofinality $\omega$ from each side. –  Joel David Hamkins Aug 9 '11 at 19:24
I think that Shelah's Theorem 1.1 (if I understand it correctly) comes close to providing a negative answer to the question at the end of my original post (namely, whether my "different nested intervals property" implies Dedekind completeness in the presence of the ordered field axioms). However, Shelah deals with an ordered collection of intervals (indexed by ordinals) while mine deals with an unordered collection of intervals (e.g. the set of intervals $[0,1-1/n]$, which has no largest member to serve as the first). Does anyone see how to bridge this gap? –  James Propp Aug 31 '11 at 16:36
Perhaps I should not have called the property in question a version of the nested intervals property, but rather a version of the Helly property. (Recall what Helly's theorem tells us in 1 dimension.) –  James Propp Aug 31 '11 at 16:39
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I will suggest you take a look at the paper "Quite Complete real closed fields" by Shelah. The abstract of this paper says the following.

We prove that any ordered field can be extended to one for which every decreasing sequence of bounded closed intervals, of any length, has a nonempty intersection; equivalently, there are no Dedekind cuts with equal cofinality from both sides.

The paper was published in 2004 (Israel Journal of Mathematics), but the Arxiv version seems to be improved in comparison with the published one.

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This is great! And it seems that Shelah's results also answer my related question at mathoverflow.net/questions/72612/…. –  Joel David Hamkins Aug 11 '11 at 2:05
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