TITLE: Density of a set of natural numbers whose differences are not bounded.
QUESTION [11 upvotes]: I guess this is quite standard and probably easy for experts or young lovers of number theory.
For $A\subseteq\mathbb N$, denote by $d^+(A)$ its upper density, which is 
$$
d^+(A)=\lim\sup_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}
$$
Now let $A=(a_n)$ be an increasing sequence of natural numbers and let $d_n$ be the $n$-th difference $d_n=a_{n}-a_{n-1}$.

Question: Suppose that $d_n$ is not bounded, is it true that $d^+(A)=0$? 

More generally: if $d_n$ is not bounded, is it true that any (additive) invariant mean on $\mathbb N$ takes value $0$ over $A$?
Thanks in advance,
Valerio

REPLY [11 votes]: Remove from $\mathbb N$ the subset $\cup_{n=1}^\infty \lbrace n!,n!+1,\dots,n!+n\rbrace$.
The result has density $1$ and arbitrarily large gaps.