Nested Closed Intervals

Th UK team for this year’s International Mathematical Olympiad in Santa Marta, Colombia, has just been selected. For details, see the BMOC website.

During the selection camp, which was hosted at Oundle School near Peterborough, we spent a while discussing analytic questions that typically lie outside the scope of the olympiad syllabus. Furthermore, incorrect consideration of, for example, the exact conditions for a stationary point to be a global maximum, are likely to incur very heavy penalties if a candidate has attempted a solution using, for example, Lagrange multipliers. As a result we have a dilemma about how much analysis to teach during the training process: we want the students to be able to use sophisticated methods if necessary; but we don’t want to spoil the experience of learning this theory in a serious step-by-step manner as first year undergraduates.

This post does not present a solution to this dilemma. Rather, I want to discuss one question that arose on the last day of exams. Because the statement of the question is currently classified, I will have to be oblique in discussion of the solution, but this shouldn’t distract from the maths I actually want to talk about.

The setup is as follows. We have a sequence of nested closed intervals in the reals, that is:

[a_1,b_1]\supset [a_2,b_2]\supset [a_3,b_3]\supset\ldots

We want to demonstrate that there is some real number that lies in all of the intervals, that is \cap_{n\geq 1}[a_n,b_n]\neq \varnothing. This feels intuitively obvious, but some form of proof is required.

First, what does closed mean? Well a closed interval is a closed set, but what about, for example, [a,\infty)\subset\mathbb{R}? It turns out that it is most convenient to define an open set, and then take a closed set to be the complement of an open set.

The best way of thinking about an open set is to say that it does not contain its boundary. This is certainly the case for an open interval or an open ball. It is not immediately clear how to extend this to a general definition. But note that if no point in the set X can be on the boundary, then in all the natural examples to consider there must some finite distance between any point x\in X and the boundary. So in particular there is a small open ball centred on x that is entirely contained within X. This is then the definition of an open set in a metric space (in particular for some \mathbb{R}^d).

Note that it is not a sensible definition to say that a closed set has the property that there is a closed ball containing each point. Any open set has this property also! For if there is an open ball of radius R around a point x, then there is a closed ball of radius R/2 around that same point. So we really do have to say that a set is closed if the complement is open. Note that in \mathbb{R}, a closed interval is closed, and a finite union of closed intervals is closed, though not a countable union as:

(0,1]=\cup_{n\geq 1}[\frac{1}{n},1].

Now we know what a closed set is, we can start thinking about the question.

First we remark that it is not true if we allow the interval to be unbounded. Obviously \cap_n [n,\infty)=\varnothing. Note that even though it appears that these sets have an open upper boundary, they are closed because the complement (-\infty,n) is open. This will not be a problem in our question because everything is contained within the first interval [a_1,b_1] and so is bounded.

Secondly we remark that the result is not true if we move to a general host set. For example, it makes sense to consider open and closed sets in the rationals. For example, the open ball radius 1 either side of 1/2 is just all the rationals strictly between -1/2 and 3/2. We could write this as (-\frac12,\frac32)\cap\mathbb{Q}. But then note that

\cap_{n\geq 1}[\pi-\frac{1}{n},\pi+\frac{1}{n}]\cap\mathbb{Q}

cannot contain any rational elements, so must be empty.

There was various talk that this result might in general be a consequence of the Baire Category Theorem. I think this is overkill. Firstly, there is a straightforward proof for the reals which I will give at the end. Secondly, if anything it is a lemma required for the proof of BCT. This result is often called Cantor’s Lemma.

Lemma (Cantor): Let X be a complete metric space, and let F_1\supset F_2\supset\ldots be a nested sequence of non-empty closed subsets of X with \text{diam}(F_n)\rightarrow 0. There there exists x\in X such that

\cap F_n=\{x\}.

Translation: for ‘complete metric space’ think ‘the reals’ or \mathbb{R}^d. The diameter is, unsurprisingly, the largest distance between two points in the set. For reasons that I won’t go into, the argument for the olympiad question gave \text{diam}(F_{n+1})\leq \frac12\text{diam}(F_n) so this certainly holds here.

Proof: With reference to AC if necessary, pick an element x_n\in F_n for all n. Note that by nesting, x_m\in F_n\;\forall n\leq m. As a result, for m>n the distance d(x_n,x_m)\leq \text{diam}(F_n). This tends to 0 as n grows. The definition of complete is that such a sequence then has a limit point x.

Informally, completeness says that if a sequence of points get increasingly close, they must tend towards a point in the set. This is why it doesn’t work for the rationals. You can have a sequence of rationals that get very close together, but approach a point not in the set, eg an irrational. We use the definition of closed sets in terms of sequences: if the sequence is within a closed set, then the limit is too. This could only go wrong if we ‘leaked onto the boundary’ in the limit, but for a closed set, the boundary is in the set. This shows that x\in F_n for each n, and so $x\in\cap_n F_n$. But if there is another point in \cap_n F_n, then the distance between them is strictly positive, contradicting the claim that diameter tends to 0. This ends the proof.

Olympiad-friendly version: I think the following works fine as a fairly topology definition-free proof. Consider the sequence of left-boundaries

a_1\leq a_2\leq a_3\leq \ldots <b_1.

This sequence is non-decreasing and bounded, so it has a well-defined limit. Why? Consider the supremum. We can’t exceed the sup, but we must eventually get arbitrarily close, by definition of supremum and because the sequence is non-decreasing. Call this limit a. Then do the same for the upper boundaries to get limit b.

If a>b, then there must be some a_n>b_n, which is absurd. So we must have some non-empty interval as the intersection. Consideration of the diameter again gives that this must be a single point.

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