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Let $\Omega=\mathbb{N}$. For each $E\subset\Omega$ let $N_n(E)$ be the cardinality of the set $E\cap [1,2,\ldots,n]$. Define $C=\left\{E: \lim_{n\rightarrow \infty} \frac{N_n(E)}{n} \text{ exists}\right\}$. Show that $C$ isn't a field.

I already know that it isn't closed to finite union of non-disjoint set but I can't see why. I saw a post that somebody said "the set of natural numbers whose first digit is 1 doesn't have this limit".

I'm in trouble to see why this limit wouldn't exists in finite unions.

Thanks in advance.

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    If the set of naturals with first digit 1 is not a finite union of elements of $C$, then how do you plan to use it for the original problem?2012-04-19
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    Oh, I'm not planning...I just said because I can't even see why that is true.2012-04-19
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    Rodolfo: When $n=999\dots 9$, less than $\frac{2}{10}$ of numbers in $\{1,2,...,n\}$ start with 1 (since all numbers in [$200\dots 0$,$999\dots 9]$ do not start with 1). When $n=1999\dots 9$, over a half of them start with 1. Therefore the sequence $N_n(E)/n$ has no limit.2012-04-19

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We show that the intersection of two sets that have "natural density" does not necessarily have natural density. Then union can be dealt with by taking complements. It is basically an example involving decimal expansions, though for smoothness we use base $4$.

Let $A$ be the set of even integers. The natural density exists and is equal to $1/2$.

Let $B=B_0 \cup B_1$, where $B_0$ is the set of all even numbers in intervals of the shape $[2^{2m}, 2^{2m+1}]$ and $B_1$ is the set of all odd numbers in intervals of the shape $[2^{2m+1},2^{2m+2}]$. Again, $B$ has natural density $1/2$.

Now look at $A\cap B=E$. These are all the even numbers in intervals of the shape $[2^{2m},2^{2m+1}]$. There is very considerable fluctuation in $\frac{N_n(E)}{n}$. For large $m$, if $n$ has shape $2^{2m+1}$, then $\frac{N_n(E)}{n}$ is very close to $1/3$, while if $n$ has the shape $2^{2m+2}$, then $\frac{N_n(E)}{n}$ is very close to $1/6$.

Somewhat less prettily, we can play the same game with decimal representations. The set $A$ could be all numbers that end in a $0$. For $B$, between $10^{2m}$ and $10^{2m+1}$, use the numbers that end in a $0$, and between $10^{2m+1}$ and $10^{2m+2}$ use the numbers that end in a $1$. Each of $A$ and $B$ has natural density $1/10$. But $A\cap B$ has huge relatively thick regions, followed by huge empty gaps, and the natural density does not exist.

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    Thank you very much for your answer. But is it possible to do this without this decimal expansion? I'm not familiar with it.2012-04-19
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    Well, the first solution does not use decimal expansion. The decimal expansion one is slightly more intuitive, since we are more familiar with decimal expansions. For the first solution, all we need to do is to make sure that $B$ consists of the even numbers for a while, followed by the odd numbers for much longer, followed by even numbers for still much longer, and so on. That makes sure that the asymptotic density of $B$ is $1/2$, but when we restrict to the even numbers, there are large fluctuations of density.2012-04-19
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    Very well explained. Now I can see what is happening with this limit. Thank you very much!2012-04-20