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Let $\mathcal{C}$ be the collection of subsets $C$ of $\mathbb{N}$ such that

$$\lim_{m \rightarrow \infty}\frac{ \# \{k \in C\mid 1 \le k \le m \} }{m}$$ exists.

Find $A,B \in \mathcal{C}$ such that $A \cap B \notin C$.

I think I found candidates for $A,B$ but cant calculate the limit for their intersection or find an expression for their common elements; $A=\{3n\}$ and $B=\{2m+1\}$.

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  • $\begingroup$ Well, $A$ consists of all multiples of 3, and $B$ is the set of odd numbers, so their intersection is the collection of odd multiples of 3. That said, I don't think that is going to give you a counterexample. $\endgroup$
    – Xander Henderson
    Commented Aug 26, 2017 at 14:03
  • $\begingroup$ @XanderHenderson was hoping on some "non-stablizing" of the limit. $\endgroup$
    – user123124
    Commented Aug 26, 2017 at 14:10
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    $\begingroup$ The classical example is to choose $A=2\mathbb N$ and that $B$ contains exactly one integer from each $\{2n,2n+1\}$ but that this integer is $2n$ if $4^k\leqslant n<2\cdot4^k$ for some $k$, and $2n+1$ if $2\cdot4^k\leqslant n<4^{k+1}$ for some $k$. Then the densities of $A$ and $B$ both exist and are both $\frac12$ but the partial densities of $A\cap B$ oscillate endlessly (between $\frac13$ and $\frac16$, or something similar). $\endgroup$
    – Did
    Commented Aug 26, 2017 at 14:10
  • $\begingroup$ @Did nice, this what I had in mind. Thanks $\endgroup$
    – user123124
    Commented Aug 26, 2017 at 14:18
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    $\begingroup$ You are welcome. A fact that I find amusing about this example is that it seems to stay in one's head rather easily, and even a long time after one sees it first... Dunno why. $\endgroup$
    – Did
    Commented Aug 26, 2017 at 14:22

1 Answer 1

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The classical example is to choose $A=2N$ and that $B$ contains exactly one integer from each $\{{2n,2n+1}\}$ but that this integer is $2n$ if $4^{k}⩽n<2⋅4^{k}$ for some k, and $2n+1$ if $2⋅4^{k} ⩽n<4^{k+1}$ for some k. Then the densities of A and B both exist and are both a half but the partial densities of A∩B oscillate endlessly.

By Did

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