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Intersection of intervals of bounded length?
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.
Thank you for your effort!
Your midpoints idea made me think of something that gives a significantly stronger result, at least in the single interval case. Make a sequence out of the midpoints of the original sequence [where "midpoint" means a point where >= d/2 of the interval is both above and below that point]. From Bolzano-Weierstrass, this sequence has a convergent subsequence [which in this case is easy to construct--make a monotonic subsequence] that converges to some point x. For any e > 0, cutting off finitely many of the terms of this subsequence gives an infinite sequence of intervals which intersect on [x - (d/2 - e), x + (d/2 - e)]. That is, the set of points which are in the intersection of infinitely many of the original intervals has measure d, at least, so is in particular nonempty and in fact uncountable.
I'll have to think about it to see if this can be extended to the Lebesgue case....
I can also take the sets to be open, and use open intervals instead. Each open set is the at most countable union of disjoint open intervals. The sum of the lengths of these intervals is then >= d, so finitely many of them together have length >= d/2. So, the non-emptiness of their intersection reduces to showing the finiteness property you've already alluded to, in the case of open sets.
I haven't been able to show it, but I haven't tried for long yet.
1 Answer
- Rich JLv 610 years agoFavorite 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...