Let ${\mathbb N}$ denote the set of positive integers, let $A,B\subseteq \mathbb{N}$. For $n\in\mathbb{N}$ we set $n+A:=\{n+a: a\in A\}$. We say that $A$ *is infinitely often in* $B$ if the set $$\big\{n\in\mathbb{N}:(n+A)\subseteq B\big\}$$ is infinite.
Moreover, for any $S\subseteq {\mathbb N}$ we set $\mu(S) = \lim \inf\_{n\to\infty}\frac{|S\cap\{1,\ldots,n\}|}{n}.$


**Question.** If $B\subseteq \mathbb{N}$ and $\mu(B) > 0$, does the following statement always hold?



> 
> (S): For any $n\in \mathbb{N}$ there is $A\subseteq \mathbb{N}$ with $|A|=n$ and $A$ is infinitely often in $B$.
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> 
> 


If not, is there a positive $r\in\mathbb{R}$ with $r<1$ such that whenever $B\subseteq \mathbb{N}$ has the property that $\mu(B)\geq r$ then (S) holds? (I will accept answers of the first question, but if it is negative, I would be interested in remarks concerning the latter question.)



On the "infinitely often in" relation between subsets of $\mathbb{N}$
Unless I'm missing something, the answer is **yes** (even if you replace "inf" with "sup").




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I'm going to switch from *sets* to *binary sequences* for simplicity. We define $\mu(B)$ for an infinite binary sequence $B$ as $\mu(\{n: B(n)=1\})$. Fix $n\in\mathbb{N}$ and $B$ an infinite binary string with $\mu(B)=\epsilon>0$. In order to have $\mu(B)=\epsilon$ we must have for each $m$ that strings of length $m$ and with at least $\epsilon m$-many $1$s must occur infinitely often in the characteristic function of $B$. But there are only finitely many such strings, so one occurs infinitely often. Now take $m$ large enough that $\epsilon m>n$, and think about the corresponding string $\sigma$ - this is a finite binary string which occurs infinitely often in $B$ and has at least $n$-many $1$s.


Put another way, given a set $B$ with $\mu(B)>0$ and natural number $n$, there is a finite binary sequence $\sigma$ occurring containing at least $n$-many $1$s and occurring infinitely often in the characteristic function of $B$. WLOG $\sigma$ contains *exactly* $n$-many $1$s (otherwise, truncate appropriately). Now the set $$\{x: \sigma(x)=1\}$$ is the desired $A$.


