A few years back I was playing with the strong law of large numbers (essentially I wanted to understand pointwise convergence for averages over sets other than ) following Etemadi’s proof. This didn’t lead anywhere in the end (essentially because almost surely pointwise convergence is not a topological convergence, see example 40 on page 111 of this book), but somehow along the way I was faced with the following question:

Is there an infinite set such that for all , the intersection is finite? And what if, instead of we only ask for ?

After asking this question around I got an answer from (then a high school student) Miguel Santos and this prompted me to generalize the result. In this post I will summarize the observations I made on this subject. Later I learned that most of what follows is *well known* among some circles. In particular the question above is exercise I-1.24 in this book; exercises I-1.25, I-1.26 and II-1.25 of that book also follow from the results below; as well as exercise 9.12 of this other book.

** — 1. Continuous functions — **

Let be a function. It is a classical fact that if and only if **for every** sequence with we have .

In this post I’m are interested in relaxing the `for every’ condition, reducing the set of sequences one needs to test. To make any progress I will only deal with continuous functions. As an example, we have:

Theorem 1Let be a continuous function and let . For any non-empty open interval , the following conditions are equivalent:where under the limits we consider and .

This will follow from Theorem 2 below.

Remark 1In conditions (1)-(1), if we replace the quantifier with the weaker condition `for almost every ‘ (in the sense of the Lebesgue measure) the theorem no longer holds. In fact, Lesigne showed that for every continuous integrable function and almost every one has . On the other hand, it is well known that there are continuous integrable functions which do not converge at infinity.

Recall that a set is called nowhere dense if its closure contains no open set, and a set is called meagre (or of the first category) if it is the countable union of nowhere dense sets.

Theorem 2Let and let be a function such that

- For each , the map is increasing and unbounded.
- For each , the function is continuous.
- For any pair , if is large enough depending on and , we have .
Then, for any continuous function and , either or the set

is meagre.

Remark 2If we combine Theorem 2 for the function with Lesigne’s result mentioned in Remark 1, we conclude that given any continuous integrable function that doesn’t converge to at infinity, the setis both meagre and has full Lebesgue measure.

*Proof:*

Let be a function under the conditions of the theorem and let be a continuous function. Assume that, for some , the set

is not meager. We will prove that . For each let . The second condition on implies that each is a continuous function.

Fix and for , let

Each is in some , in other words . Since we are assuming that is not meager, some is somewhere dense, i.e. the closure must contain an open interval.

The crucial observation now is that each is a closed set, because each is continuous and is a closed set. We conclude that some is somewhere dense, and hence it contains an open interval, say .

By the third condition on , there is some such that, for , we have . By the first condition on , the function is increasing and unbounded. This implies that if we have for some . By continuity we get that for some .

We conclude that if and then we have , for some . Thus, , and because was arbitrary, this implies that as desired.

In order to prove Theorem 1, it will be convenient to relax the conditions for in Theorem 2 to make it easier to check in concrete examples. The next corollary gives us such conditions.

Corollary 3Let be a real number and let be a continuously differentiable function such that for and both partial derivatives are positive, and such that for we have the two conditions:(here’s what uniformly in compact sets means). Then for any continuous function and , either or the set

is meagre.

*Proof:* Let be a function that satisfies the conditions of this corollary. We need to show that satisfies the conditions of Theorem 2. The first condition follows from the fact that for all we have and the partial derivative is positive. Since is continuous, the second condition holds as well.

To check the third condition let and fix some positive integer . We need to show that . Let . Note that and . The Mean Value Theorem now implies that for some . We have

From the hypothesis on we can choose such that for and all we have

Since both partial derivatives of are positive we have that . Therefore, if , then .

We can now easily deduce Theorem 1 from this corollary.

*Proof:* (of Theorem 1) Condition (1) trivially implies the others. Now consider the functions , , and . Each of those functions satisfy the conditions of Corollary 3.

By Baire’s category theorem, the open interval is not meagre. Therefore, it follows from Corollary 3 that any of the conditions (1) to (1) imply that .

** — 2. Combinatorial consequences — **

The techniques used to prove Theorem 2 can be adapted to study convergence of sequences. This in turn can be used to show that certain sparse sets of have infinite intersection. We use the classical notation to denote the largest integer no greater than the real number .

Theorem 4Let be an infinite set and let be a non-empty open interval. Then

Taking or , part 3 of this theorem answers the question raised in the introduction of this post.

Remark 3The strength of Theorem 4 lies in the fact that can be any infinite set. Thus we can bootstrap the theorem and get, for instance, that there exist and in any given small non-empty open interval such that the intersectioncontains infinitely many prime numbers or infinitely many square numbers (say). In fact, it follows from Corollary 6 below that we can choose and such that the set (1) contains infinitely many primes

andinfinitely many squares.

In order to prove Theorem 4 we need to establish an analogue to Theorem 2 for functions which are not necessarily continuous but which are constant on each interval . To do that we need to assume that increases with for each fixed . This is not a very strong condition, in particular it is also present in Corollary 3.

Theorem 5Let and let be a function satisfying the conditions of Theorem 2 with the additional property that, for each , the function is increasing. Then for any sequence taking values on and any , either or the setis meagre.

To make the analogy with the Theorem 2 take . This function is usually not continuous so we can not apply the Theorem 2 directly, but it is constant on each interval , so we can modify the argument a little to cover this case.

*Proof:* Let be a function that satisfies the conditions and let be a real valued sequence. Assume that, for some , the set

is not meagre. We will prove that .

Fix . For let

and for , let

By construction, each is in some , in other words is a countable covering of . Since is assumed to be not meagre, some must be somewhere dense and thus must contain an open interval, say .

Unlike in the proof of Theorem 2, the set is not necessarily closed, but we claim that if an open interval is contained in , then it is actually contained in . Since

we have for all . Define . Clearly is the union of intervals of the form with integer end points. We can write and, moreover, the function is continuous and increasing, so is also the union of some intervals of the form . This implies that, since , we actually have that . Since this is true for each we showed that , proving the claim.

By the hypothesis, there is some such that for we have . Thus for we have for some . This implies that for some and so if we have . Since was arbitrary, we conclude that as desired.

We are now ready to give a quick proof of Theorem 4. In fact, we can prove a stronger statement.

Corollary 6Let be a countable family of infinite subsets of . Let be a non-empty open interval. Then

- There exists such that for each , the intersection is infinite.
- For each real there exists such that for each , the intersection is infinite.
- There exists such that for each , the intersection is infinite.
- There exists such that for each , the intersection is infinite.

*Proof:* (of Theorem 4) We prove only the part (4) since the same proof works for the other cases with a different (but obvious) choice of .

Let , let be infinite subsets of and fix an open interval . Observe that satisfies the conditions of Theorem 5. Define the indicator sequence of by letting if and if . For each define

For each and each we have

Since is infinite, the limit is not . It follows from Theorem 5 that is a meager set. Thus the union is also meager.

By the Baire Category theorem, is not meagre, thus . Taking any we have that, for every , the intersection is infinite.