Intersection of intervals of bounded length?

Question

Take the real unit interval [0, 1]. For some fixed d > 0, you have an infinite sequence of intervals of the form [a, b] in [0, 1], where (b-a) >= d. The question is, does some (infinite) subsequence of these intervals have non-empty intersection?

I've shown that for each natural number n, the intersection of some n intervals is non-empty. More reasoning is needed to extend this arbitrarily high intersection to an infinite intersection, though, and I haven't been able to supply it. More than anything, I just need a fresh perspective.


Note: my actual situation is more complicated. The intervals [a, b] are actually (optionally closed, though I'd prefer if closedness didn't matter) Lebesgue-measurable sets with measure >= d. I imagine the situations are reasonably analogous, though, and using intervals cuts out the measure theory to make it much more accessible.

Rich J · Accepted Answer

If:
"you have an infinite sequence of intervals of the form [a, b] in [0, 1], where (b-a) >= d"
was a general assertion about all the sequences of this form, then just consider the infinite subsequence of intervals
{ [(1-x(n))/2, (1+x(n))/2], n>=1 }
where x(n) = d + (1-d)/n, with n>=1, for which x(n)->d as n->inf.
Their intersection is [(1-d)/2, (1+d)/2].
(and closedness doesn't really matter)

But I suspect you rather meant you already have a specific infinite sequence of intervals of this form, for a fixed 0 < d < 1. So I'll try to answer this below.

1) Consider the case d >= 1/2. Clearly, any interval [a, b] with b-a >= d >= 1/2 must contain the mid-point of the interval (here 1/2). So the mid-point is in the intersection of the infinite subsequence contained in the interval (here [0, 1]).

2) If 0 < 1/4 <= d < 1/2, then we can find an infinite subsequence that contains 1/2, or we cannot.
- If we can, then 1/2 is in the intersection of that infinite subsequence.
- If we cannot, then we must have an infinite subsequence *strictly* contained in either [0, 1/2] or [1/2, 1]. Since d >= 1/4, we have a similar case as in (1), and we have an infinite subsequence that contains 1/4 or 3/4, and one of those points is in their intersection.

I think you see where I'm going with this: since we have a fixed d, there is an integer n such that:
1/2^(n+1) <= d < 1/2^n

3) Divide [0, 1] into the 2^n intervals I(k) = [(k-1)/2^n, k/2^n], k=1, ... , 2^n.
Then we can find an infinite subsequence of our intervals, each of which contain one of the points {k/2^n, k=1, ... 2^n -1}, or we cannot.
- If we can, and since there is only 2^n - 1 (finitely) such points, then one of them is contained in an infinite (sub-)sub-sequence of our intervals, and is in their intersection.
- If we cannot, i.e. if only a finite number of our intervals intersect points like k/2^n, then an infinite number of them is strictly in the union of the interiors of the I(k)'s, and since there's only a finite (2^n) number of these intervals, then an infinite (sub-)sub-sequence of our intervals are in the interior of some particular I(k0). Since that interval measures 1/2^n, and since d>=1/2^(n+1), we have a similar case as in (1), and this infinite subsequence contains (k0-1)/2^n + 1/2^(n+1), which is in their intersection.

Well, I hope that was clear enough, and that I understood your question.

As for the more general Lebesgue-measure version of the problem, I don't know if the above can be adapted to it; my study of Lebesgue is too far away, I'd just be stuttering (hehe... ok, too easy a joke).

//Additional Note:
The above reasoning can be applied in the case where the sets are *finite* unions of disjoint intervals (with sum of lengths >=d), rather than intervals, with some important details to modify.
I've tried generalizing to *countable* union of disjoint intervals, but I'm stuck on some annoying details...

Intersection of intervals of bounded length?

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