# Sums of sets of lower density 0

We say that a set $A\subseteq \mathbb{N}$ has lower density 0 if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 0.$$

Given $A,B\subseteq \mathbb{N}$ we set $A+B = \{a+b: a\in A, b\in B\}$.

Are there $A, B\subseteq \mathbb{N}$ with lower density 0, but $A+B$ does not have lower density 0?

• If this would not exist, how would anybody believe the Goldbach conjecture. – user9072 May 6 '15 at 13:07

It is relatively easy to prove that the set of perfect squares has asymptotic density equal to $0$. Then either the set $Q_2 := \{x^2+y^2: x,y \in \mathbf N\}$ has positive lower asymptotic density, and we're done, or the lower density of $Q_2$ is zero, and then we just consider that $\mathbf N = Q_2 + Q_2$ (by Lagrange's four-squares theorem).
By the way, it follows, e.g., from E. Landau, Über die Einteilung der positiven ganzen Zahlen in vier Klassen nach der Mindeszahl der zu ihrer additiven Zusammensetzung erforderlichen Quadrate, Arch. Math. Phys. 13 (1908), 305-312 that the asymptotic density of $Q_2$ is actually zero, but this is more than what you need to answer the question in the OP.
• Very elegant - reminds me of the proof that $\sqrt{2}$ is irrational – Dominic van der Zypen May 6 '15 at 11:25
• @Dominic van der Zypen - Do you perhaps mean the proof that there are irrational numbers $a$ and $b$ such that $a^b$ is rational? That proof typically goes by choosing $a = b = \sqrt{2}$ and then splitting into two cases based on whether or not $\sqrt{2}^\sqrt{2}$ is rational. – Nathaniel Johnston May 6 '15 at 14:11
• For what it's worth, here is a proof that the asymptotic density of $Q_2$ is zero, which doesn't depend on Landau's result: mathoverflow.net/a/205866/16537. – Salvo Tringali May 6 '15 at 14:21
Let $A$ be the set of all natural numbers having digit $0$ in every odd-numbered place (counting from the decimal point), and let $B$ be the set of all numbers having a $0$ in every even-numbered place. Then $A$ and $B$ have density $0$, while $A+B=\mathbb N\setminus(A\cup B)$ has density $1$.
Partition $\mathbb N$ into two sets $S$ and $T$, each having lower density $0$ and upper density $1$; for example, by taking $$S=[1!,2!)\cup[3!,4!)\cup[5!,6!)\cup[7!,8!)\cup\cdots,$$ $$T=[2!,3!)\cup[4!,5!)\cup[6!,7!)\cup[8!,9!)\cup\cdots.$$ Now define $$A=\{n\in\mathbb N:n+1\in S\}\cup\{1\},$$ $$B=\{n\in\mathbb N:n+1\in T\}\cup\{1\}.$$ Then each of the sets $A,B$ has lower density $0$ (and upper density $1$), and $A+B=\mathbb N\setminus\{1\}.$