Sumsets Contained in Sets of Upper Banach Density 1

Abstract

Every set A of positive integers with upper Banach density 1 contains an infinite sequence of pairwise disjoint subsets $(B_i)_{i=1}^{\infty }$ such that $B_i$ has upper Banach density 1 for all $i \in \mathbf {N}$ and $\sum _{i\in I} B_i \subseteq A$ for every nonempty finite set I of positive integers.

Supported in part by a grant from the PSC-CUNY Research Award Program.

Appendix: Subadditivity and Limits

A real-valued arithmetic function f is subadditive if

$$\begin{aligned} f(n_1+n_2) \le f(n_1) + f(n_2) \end{aligned}$$

(4)

for all $n_1,n_2 \in \mathbf {N}$.

The following result is sometimes called Fekete’s lemma.

Lemma 2

If f is a subadditive arithmetic function, then $\lim _{n\rightarrow \infty } f(n)/n$ exists, and

$$ \lim _{n\rightarrow \infty } \frac{f(n)}{n} = \inf _{n\in \mathbf {N}} \frac{f(n)}{n}. $$

Proof

It follows by induction from inequality (4) that

$$ f(n_1+\cdots + n_q) \le f(n_1) + \cdots + f(n_q) $$

for all $n_1,\ldots , n_q \in \mathbf {N}$. Let $f(0) = 0$. Fix a positive integer d. For all $q,r \in \mathbf {N}_0$, we have

$$ f(qd+r) \le qf(d)+ f(r). $$

By the division algorithm, every nonnegative integer n can be represented uniquely in the form $n = qd+r$, where $q \in \mathbf {N}_0$ and $r \in [0,d-1]$. Therefore,

$$ \frac{f(n)}{n} = \frac{f(qd+r)}{n} \le \frac{qf(d)}{qd} + \frac{f(r)}{n} = \frac{f(d)}{d} + \frac{f(r)}{n}. $$

Because the set $\{f(r):r\in [0,d-1]\}$ is bounded, it follows that

$$ \limsup _{n\rightarrow \infty } \frac{f(n)}{n} \le \limsup _{n\rightarrow \infty } \left( \frac{f(d)}{d} + \frac{f(r)}{n} \right) = \frac{f(d)}{d} $$

for all $d \in \mathbf {N}$, and so

$$ \limsup _{n\rightarrow \infty } \frac{f(n)}{n} \le \inf _{d\in \mathbf {N}} \frac{f(d)}{d} \le \liminf _{d \rightarrow \infty } \frac{f(d)}{d} = \liminf _{n\rightarrow \infty } \frac{f(n)}{n}. $$

This completes the proof.