In ergodic Ramsey theory, one often wants to prove that certain dynamically defined sets in a probability space intersect (or “recur”) in non-trivial ways. Typically, this is achieved by studying the long term behavior of the sets as the dynamics flow. However, in certain situations, one can establish the desired intersection (or recurrence) using purely combinatorial arguments, and without using the fact that the sets are dynamically defined. In such cases, one ends up obtaining a “static” (as opposed to dynamical) statement. An instance of this situation is the following intersectivity result of Bergelson, first used in this paper, and which I have mentioned before in this blog.
A different kind of intersection property is the following static modification of Poincaré’s recurrence theorem.
Observe that if the events are independent, then the lower bound is essentially achieved, and so this lemma is that sense optimal. The purpose of this post is to present a proof of the following common strengthening of Lemmas 1 and 2.
Observe that the bound is optimal, again by considering the case of independent sets.
One could ask whether something more can be said about the set in Theorem 3 other than being infinite . Something one can not hope for is that the set has positive density, as showed in my previous post, using Forest’s theorem that not all sets of recurrence are sets of nice recurrence. On the other hand, one can indeed obtain certain combinatorial structures inside . For instance, assuming the index set of the sets is given the structure of a homogeneous tree, one can choose to be a strong subtree; this is Theorem 3 in this paper of Dodos, Kanellopoulos and Tyros. Results of this kind are related to density versions of the Hales-Jewett and the Carlson-Simpson theorems due respectively to Furstenberg and Katznelson, and to Dodos, Kanellopoulos and Tyros.
— 1. A truncated version —
We start with a proof of Lemma 2 which we will need to prove Theorem 3. In fact, we prove the following strengthening of Lemma 2 which I have mentioned before in this blog (but without a proof) and can be seen as a “truncated” version of Theorem 3.
Proof: We partition the collection of all subsets of with size into two pieces, according to whether is bigger or smaller than . We then use the infinite Ramsey’s theorem to find an infinite set such that every is in the same cell of the partition. We now need only to show that it is impossible to have for every with .
In order to do this, let be the set of the first elements of , where is very large and will be determined later. Also let . Then by Jensen’s inequality
It is clear that contains tuples of distinct elements, each appearing times, and hence contains tuples of elements with some repetition. Thus we get
and so, using the fact that , for some with we have
Since as , if follows that if was large enough, then
as claimed, and this finishes the proof.
Observe that this proof would give the full Theorem 3 if the following extension of Ramsey’s theorem were true.
Statement Let denote the collection of all finite non-empty subsets of . For every finite coloring of there exists an infinite set such that for every , the collection
Unfortunately, this statement is false, as seen by the following example. In this sense it is perhaps surprising that Theorem 3 is true.
Example 1 Let be the coloring given by
Then given any infinite set , let and find a finite subset with and (which then satisfies ) and another finite subset with but (which thus satisfies ).
— 2. Proof of Theorem 3 —
Since the proof of Theorem 3 involves some annoying parameters, we first outline the steps in the proof without full details.
By refining the sequence of sets we can assume that they all have similar measure for all . Applying Lemma 2 we find a set with for every . After refining , if needed, we can assume that, in fact, for every distinct we have . Let be arbitrary.
Now comes the tricky part: we condition the measure on . In other words, we consider the probability measure on defined by . Observe that for every . Now use Lemma 2 to find an infinite set such that for every distinct .
Let be arbitrary. Now we condition the measure on , letting and noting that, since , for every we have . Therefore by applying Lemma 2 we can find an infinite such that for every distinct we have . Before we can choose , we need to also consider the situation conditional on (which we recall, has measure because both and are in ). Thus, letting we obtain that, for every , . Therefore, applying Lemma 2 again, we can further refine to an infinite subset such that for every distinct also . We can now chose arbitrarily.
We continuing building the elements of the eventually infinite set , at each step making sure we have an infinite set such that for every non-empty subset we have
and that every we have
In each stage we can keep moving by conditioning in each new subset of and applying Lemma 2.
To make everything work out, we need to introduce a refinement step at each stage, to make sure all the sets in have similar measure, for all the conditional measures . To this end we make use of the following version of the pigeonhole principle.
Proof: Let be large enough that the intervals , with cover the interval . Since for every , the pigeonhole principle implies that there exists for which the result holds with .
We are now ready to prove Theorem 3.
Proof of Theorem 3: For each , denote by the intersection , with the convention that , and let denote the measure on defined by whenever . Let be such that for every .
Suppose now that and that we have found , satisfying (1) and (2). Enumerate the subsets of as . Let and apply successively Lemma 2 for each in to obtain an infinite set such that for every distinct we have
We run another refinement cycle, setting and successively using Lemma 6 for each to find , where , and an infinite set such that for every ,
and (1) follows by setting , which will be greater that since both and are.