Sets with certain property concerning density of sumsets

Question

I am working with subsets of $[n]$ of the form $(A+B)\cap A$, where $A+B$ is a sumset. Namely, I am interested if there are nonempty sets $B$ such that whenever $A$ covers a positive proportion of $[n]$, the set $(A+B)\cap A$ also covers a positive proportion of $[n]$. In fact it would be nice if the proportion is the same, or tends to the same limit. (What I mean by "positive proportion" only really makes sense when we take $n$ to infinity; I suppose if $A$ and $B$ are regarded as (possibly infinite) subsets of ${\bf N}$, then this is their asymptotic density.)

For instance $B = \{1\}$ is not an example, since if $A$ is the set of even integers in $[n]$, then $A$ covers one half of $[n]$ but $(A+B)\cap A = \emptyset$. The specific example I'm considering at the moment is the set $B = \{1, 2, 6, 24, 120,\ldots,\}$ of factorials, so information particular to this example would also be appreciated, if more general information is not known. But this problem seems natural enough that I thought it might have a name, I just didn't know what to search online. As always, thank you all in advance for the help!

Edit. I thought I should illustrate a nontrivial example where $B$ is the set of factorials and the densities of $A$ and $(A+B)\cap A$ are the same. Let $p$ be a prime and let $a$ be a residue modulo $p$. Then the set $A$ of all integers congruent to $a$ modulo $p$ has asymptotic density $1/p$. Now since only finitely many members of $B$ are not divisible by $p$, if I'm not wrong the set $(A+B)\cap A$ should have the same asymptotic density as $A$. Is this true for all $A$, not just this example?

Second edit. Following Terry Tao's hint below, we recall the definition of intersective: $B$ is intersective if for all $A$ of positive upper density, we have $(A-A)\cap B \ne\emptyset$. Here is a proof that this is equivalent to my condition on $B$:

Proof. Let $B\subseteq {\bf N}$ and suppose that there is $A$ of positive upper density such that $(A+B)\cap A$ has density strictly less than that of $A$. Then $A\setminus (A+B)$ has positive upper density. Now let $x,y\in A\setminus(A+B)$ and set $z = x-y$, so $z\in A\setminus(A+B)-A\setminus(A+B)$. Since $x\notin A+B$ and $y\in A$, from $x=y+z$ we conclude that $z\notin B$. Thus we find that $$\bigl(A\setminus(A+B)-A\setminus(A+B)\bigr) \cap B = \emptyset;$$ in other words, $B$ is not intersective.

On the other hand, if $B$ is not intersective, then there exists $A$ of positive upper density such that $(A-A)\cap B = \emptyset$. For such an $A$, we must have $(A+B)\cap A = \emptyset$, since if $a\in (A+B)\cap A$, then $a = a'+b$ for $a'\in A$, $b\in B$, and then $a-a'=b$. ▮

So one part of my question is settled. But I am still interested in knowing whether the set of factorials is intersective or not (and would happily accept this as an answer to the question). Forgive me if this is trivial but I am not able to see it right away. The set $S$ given by the greedy algorithm (suggested by mathworker21 below) seems dense, but this is because for a long time it consists of elements congruent to one of $\{0,3,7,10,14,17,21\}$ modulo $25$, which is fine (and has density $7/25=0.28$) until we get to $5040$, which is congruent to $15 = 0-10$. The phenomenon is more obvious when we start the greedy algorithm with $0$ instead of $1$, in which case for a long time we get numbers congruent to either $0$ or $3$ modulo $7$ (this has density $2/7 = 0.2857\dots$), and this works because the factorials $<5040$ are all congruent to one of $\{1,2,3,6\}$ modulo 7, and $\{0,3\} - \{0,3\} = \{0,3,4\}$ in ${\bf Z}/(7)$. What happens after $5040$ is still somewhat mysterious to me, but the proportion dips below $0.28$ and I cannot convince myself that it won't go to $0$.

Answer 1

The set of factorials $B$ is not intersective, hence there is a set $A$ of positive density for which $A$ and $A+B$ are disjoint.

Indeed, every natural number $n$ has a unique "factorial base" representation $$ n = \sum_{k=1}^\infty a_k k!$$ with $0 \leq a_k \leq k$ and all but finitely many of the $a_k$ non-zero. If we let $E$ be the set of such numbers for which we do not have $a_k=k$ for two consecutive values of $k$, then $E$ has positive density (this is a simple sieve theoretic calculation using the fact that $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ converges). Now we partition $E$ into four classes depending on the parity of the two quantities $$ \sum_{k \geq 1, \hbox{ even}} (k+1) a_{k+1} + a_k$$ and $$ \sum_{k \geq 1, \hbox{ odd}} (k+1) a_{k+1} + a_k.$$ Note that adding $k!$ to $n$ with $k$ even would flip the parity of the first sum (noting that $(k+1) a_{k+1} + a_k$ is strictly less than $k(k+1)$ by definition of $E$), and adding $k!$ to $n$ with $k$ odd would flip the parity of the second sum. Thus none of the four classes in $E$ contains a pair $n, n+k!$. By the pigeonhole principle, one of these classes has positive upper density (indeed it looks straightforward to prove that all four classes have positive density), and the claim follows.