Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've just started to read about additive combinatorics and I'd like to know how I can use Bohr sets to make a statement about arithmetic progressions in a given subset $A$ of an Abelian group $Z$ (the ambient group).

For example: Green-Tao states that if $A$ is the set of all primes then we can always find an arithmetic progression of length $k$ in $A$ for all $k \in \mathbb N$.

Can I do the following: If $B(S, \rho)$ is a Bohr set with certain properties then it contains an arithmetic progression of length say, $k=10$?

If yes: how exactly does it work?

If no: why not?

A question directly related to this is: why were Bohr sets introduced? I've read that they replace the orthogonal complement $S^\bot$ of a subset $S$ of $Z$ when $S^\bot$ is not a subgroup of $Z$ because for example $Z = Z_p$ for $p$ prime.

And one more question is: if $F$ is a linear space then every linear subspace is a Bohr set (Tao/Vu, p. 166). So for some reason we want to study linear subspaces of $Z$ or subgroups if possible. How do we use this structure or in particular the closedness with respect to addition of subsets? I'm slightly confused about whether we're interested in subsets or subspaces of $Z$.


Let $S \subset \widehat{Z}$ be a set of characters of $Z$ and let $\rho > 0$ be a real number. Then a Bohr set is defined as $$ \operatorname{Bohr}(S, \rho) := \{ z \in Z \mid \sup_{\chi \in S} \left | \chi(z) - 1 \right | < \rho \}$$

Thanks for your help.

share|cite|improve this question
Perhaps you'd like to tell us what a Bohr set is. – Gerry Myerson Apr 1 '12 at 0:38
up vote 1 down vote accepted

Why Bohr sets?

For a given ambient group $Z$ we want to know about arithmetic progressions in a subset $A$ of $Z$. If the set $A$ is a group it means that $A$ has a high degree of additive structure. Consider for example the integers with the subgroup $A = 2 \mathbb Z$. Then clearly, $A$ contains arithmetic progressions of arbitrary length.

Unfortunately, in general, the sets we study aren't (sub-)groups.

This is why we want Bohr sets: They are the next closest thing with a comparable amount of structure. To see how this works let's give the definition of a Bohr set and then consider an example.

For an ambient group $Z$ and a set of characters $S \subset \widehat{Z}$ and a radius $\rho > 0$ we define the Bohr set as follows $$ Bohr ( S, \rho) := \{ z \in Z \mid \sup_{\chi \in S} |\chi (z) - 1| \leq \rho \}$$ Note: Subgroups of $\mathbb Z_n$ are Bohr sets.


For $\mathbb Z_{12}$, $S = \{6\}$ and $\rho = 0.4$ we get $Bohr(S, \rho) = \{0, 2, 4, 6, 8, 10 \}$.

For $S = \{2\}$ we get $Bohr(S, \rho) = \{0, 6\}$.

As you can see, the Bohr set of $S$ is the orthogonal complement $S^\bot$ of $S$ and is also a subgroup of $Z$.

Now we see how we can use Bohr sets to say something about arithmetic progressions in a subset $A$ of $Z$.

share|cite|improve this answer
You may want to include these thoughts in OP as ideas you have for the solution of your problem. Or you may not – Ilya Apr 1 '12 at 11:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.