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Let $A\subseteq\mathbb N$, as usual we define the Beurling density $d(A)=\lim_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}$, when it exists. It seems to me it is well-known that the family of subsets which have density is not closed under intersection, but I have been not able to find an explicit counter-example so far. Could anybody give me one?

Thanks in advance, Valerio.

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Let $A$ be the set of even numbers, and construct $B$ as follows. Let $B\cap[2^n, 2^{n+1})$ consist of the even numbers in $[2^n, 2^{n+1})$ if $n$ is even, and the odd numbers in $[2^n, 2^{n+1})$ for $n$ odd. Then both $A$ and $B$ have density $1/2$, but $A\cap B$ consists of even numbers whose binary representation has odd length; it is easy to check that this set has no density.

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To phrase it slightly differently : consider C a set with no density. Then set A=2\mathbb{N} and B=2C+(2(\mathbb{N}∖C)+1). Then A and B both have density 1/2. But A∩B=2C. Therefore you just have to pick for C your favorite subset with no density. –  camomille Apr 7 '11 at 8:18
    
Sure, though I think concreteness can't hurt. –  Daniel Litt Apr 7 '11 at 17:24
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Let $A$ be the set of odd integers $\geq 0$, and let $B$ be the set of those integers $n$ which are odd if $2^m \leq n < 2^{m+1}$ for an odd $m$ and even if $2^m \leq n < 2^{m+1}$ for an odd $m$. Both sets $A$ and $B$ have naive density $\frac12$ (you call it Beurling density).

The intersection $A\cap B$ has no density: its upper density (limsup...) is $\frac 12$ while its lower density is zero.

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nah, now you have it twice. –  Xandi Tuni Apr 7 '11 at 7:35
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We think alike, you and I (and apparently have very similar typing speed). –  Daniel Litt Apr 7 '11 at 17:25
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Take $A$ the set of even numbers, $B$ the set of odd numbers. Than $d(A) = d(B) = \frac{1}{2}$, while $d(A \cap B) = 0$ since $A \cap B = \emptyset$.

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$0$ is a density. –  Daniel Litt Apr 7 '11 at 7:25
    
I'm sorry, maybe my question was not clear, but this is not an example: I would need a pair of sets $A,B$ with density such that $d(A\cap B)$ does not exist. –  Valerio Capraro Apr 7 '11 at 7:33
    
Oh, I'm sorry, I understood the question wrong... –  felix Apr 7 '11 at 8:21
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