In this short post I will present some curious facts I came across recently.

Given a real number construct the set , where, as usual, denotes the floor function. Note that if is an integer, then is an (infinite) arithmetic progression starting at , so we can think of the sets as generalized arithmetic progressions. In this post I will consider two types of questions about this sets: “can be disjoint from ?” and “Can the union of finitely many contain all sufficiently large integer?”.

To try to get some intuition, let’s assume first that is a rational number, say with coprime. Then it should be clear that is a periodic set with period , because and since is integer, taking floors we get . Thus for any finitely many with , if we make be the least common multiple of we have that each multiple of is contained in each , answering the first question in the negative. Moreover it’s not hard to see that , so is not in any of the and, more generally, no number of the form is in the union , answering the second question in the negative.

This gives some intuition about what we could expect for a general (not necessarily rational). Note that if and are very close then the sets and agree in the beginning, so one can hope to approximate an arbitrary with rational numbers. After this reasoning the following result should come as a surprise:

Theorem 1Let be irrational numbers such that . Then and are disjoint and moreover .

*Proof:* Assume , so that . Since , the set has only “holes” of size , i.e., if is a natural number not in then both and are in .

For intuition it may be useful to think of as being close to , although all the equations remain true for any . We have for small natural numbers that , until the first hole, which happens at the smallest natural number such that . Solving this for we get and adding we get , so is the first hole in and also , so is the first element in .

Using the same procedure we find that the -th hole in happens at , where is the least natural number such that . Solving this for we get and adding we get , so that .

We conclude that the -th hole in is the -th element in , so in other words and are disjoint and the union .

This can be used to provide simple counter-examples to some other combinatorial questions. Recall the notion of IP-set: Given an increasing sequence of natural numbers, we define the set of finite sums . A subset is called an IP-set if it contains the set for some increasing sequence .

Maybe more important than IP-sets are the sets that intersect non-trivially every IP-set, such sets are called IP:

Definition 2{IP} Let . We say that is an IP set if for any IP-set the intersection is nonempty.

For instance it is not hard to see that the set of all multiples of is IP sets for any . One of the most important features of the IP sets is that the intersection of two such sets is still an IP, this follows almost immediately from Hindman’s theorem. From this we see that for any IP set and any , the intersection is not only non-empty but actually an IP itself. Moreover, if we divide each element in that intersection by we still get an IP set, in other words, the set is an IP as long as is an IP (I will not prove this here, but this is a non-trivial result).

This result implies that for any IP set and any rational number , the set is an IP, because contains . Thus it is natural to ask if for any real number on has that is an IP itself. Using the theorem 1 one concludes promptly that this is not the case, because if so, then both and would be IP, but they are disjoint when (note that ).

Seems like an interesting topic. I’ll be waiting for new posts!

Joel, by the way, you have a typo in the beginning of the proof of Thm 1: the x+2 should be an x+1.

Thanks for noticing the typo, Pedro, it is now corrected.