Below I am writing a sketch of how the proof goes. Since the proof itself got a bit long, we wrote a note: you can find it here.

(1) Consider the following set system of boolean subcubes, let’s call it . Each element of the ground set is identified with a boolean vector . We get each set from fixing some of the coordinates and leaving the other ones free. That is, each set is associated with a vector and consists of those for which whenever .

(2) We construct a set of positive integers by associating an integer to each boolean vector so that each corresponds to a HAP restricted to . This is not hard and is similar to the construction in the post. We pick the first primes and associate to each . The set corresponds to the HAP with difference .

(3) We show that

This proceeds in two steps:

3.1 Use the determinant lower bound of Lovasz, Spencer, and Vesztergombi to show that the matrix of Fourier characters of of weight ( a constant) has hereditary discrepancy . Call this matrix

3.2. Show that . This goes by writing a character of weight as a linear combination of indicator vectors of subcubes. This is possible since a character of weight is a function of components.

We can choose so that 3.1 and 3.2 imply (3).

Dear GIl, yes it seems to be almost the same. Two Beurling’s integers are also allowed to correspond to the same real number, and that corresponds to the statement that we can have pseudo-integers a and b such that ab^{n}$ for all n. The only difference seems to be that pseudointegers have a complete ordering. The ordering of the Beurling’s integers come from a (possible not injective) map to the reals, so it might not be complete. But that is just a little extra structure so in the following I will use “Buerling’s integers” to mean what I used to call “pseudo integers”

I think that if you have a set of Beurling’s integers then for all n, the set of the first n of these Beurling’s integers will be isomorphic to a subset of , and I think you lose this property if you generalise further.

So I still haven't found a set of pseudointeger where I can prove EDP.

Proposition 5: For every there is a set of the natural numbers of density such that restricted to this set the discrepancy of HAPs in {1,2,…,n} is at most polylog (n).
Proof: Simply consider natural numbers with at most prime factors for sifficiently large , and apply Proposition 2

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Some notes on the operation on hypergraphs. The short version of this comment is that Prop 2 can be tightened, and also generalized to show that the operation on hypergraphs increases hereditary discrepancy by at most a polylog factor.

1) Prop 2 holds with the bound (assuming the number of sets of is ). This follows from the following theorem by Banaszczyk:

For any convex set of gaussian measure at least 1/2 and any set of vectors each of euclidean norm at most 1/10, one can find a such that .

Let be the incidence matrix of (as you define it in your proof). Scale down the columns of by for a large enough constant so that each column has norm at most 1/10 and let the scaled columns be .

Let be the incidence matrix of and be such that . As you argue, each row of can be taken to be a 0-1 vector with at most 1′s, so each row of can be taken to have euclidean norm . Define . Clearly is convex and for large enough has gaussian measure at least 1/2. By Banaszczyk’s theorem, we can find such that , and scaling back up we have found such that .

2) If is the result of applying the transformation on and has sets, .

The proof is similar to yours. Take be the result of replacing each set with and ; : the result of replacing with for all and so forth. This gives us for . Since each is the union of *disjoint* restrictions of , for all we have . Then and by a recent result of Matousek, . Finally, as you argue, .

Some notes on the operation on hypergraphs. The short version of this comment is that Prop 2 can be tightened, and also generalized to show that the operation on hypergraphs increases hereditary discrepancy by at most a polylog factor.

1) Prop 2 holds with the bound latex H$ is ). This follows from the following theorem by Banaszczyk:

> For any convex set of gaussian measure at least 1/2 and any set of vectors each of euclidean norm at most 1/10, one can find a such that .

Let be the incidence matrix of (as you define it in your proof). Scale down the columns of by for a large enough constant so that each column has norm at most 1/10 and let the scaled columns be .

Let be the incidence matrix of and be such that . As you argue, each row of can be taken to be a 0-1 vector with at most 1′s, so each row of can be taken to have euclidean norm . Define . Clearly is convex and for large enough has gaussian measure at least 1/2. By Banaszczyk’s theorem, we can find such that , and scaling back up we have found such that .

2) If is the result of applying the transformation on and has sets, .

The proof is similar to yours. Take be the result of replacing each set with and ; : the result of replacing with for all and so forth. This gives us for . Since each is the union of *disjoint* restrictions of , for all we have . Then and by a recent result of Matousek, . Finally, as you argue, .

We can also construct examples where the multiplicative EDP is true, by having arbitrarily long runs of consecutive squares.

2) No. I remember that I didn’t understood Tao’s proof, at least not well enough to check if it worked for the pseudointegers. I had hoped Tao could tell if it works!

Then your is just , and .

It could be interesting to think specifically about where the zero values should be placed in order to minimize the discrepancy in this context.

It seems that I was a bit too sloppy both when I read the definition of E and when I posted the above comment. I though the definition of E(n) was “the smallest possible discrepancy of a sequence of length n”.

What I meant was: One way to investigate EDP is to fix some integer d and try to find the longest possible sequence with discrepancy below d. The sequences we get are then optimized to keep the discrepancy at most d for as long as possible, and it might be that for all extensions of the sequence the discrepancy grows to, say, 3/2d, within the next few terms. This is what I mean by “postponing the problems”.

Instead we might try to keep the discrepancy below d while defining the sequence on a subset of with positive density (and setting all other terms to 0). If the positions of these 0′s are periodic, as in the low-discrepancy examples we know, then it will be possible to fill out more terms without increasing the discrepancy by much.