TITLE: Sets with no asymptotical density over $\mathbb N$
QUESTION [10 upvotes]: Let's consider the Natural density on $\mathbb N$ defined by: 

Take $ A\subset \mathbb N$; define the sequence $x_n= \dfrac{|A\cap[1,n]|}{n}$, and then if $\lim\limits_{n\to\infty} x_n$ exists, call it $d(A)$, the density of $A$ over $\mathbb N$.

My teacher said that the collection of sets that have a natural density is not an algebra, but I don't know why. I can't even find a set that does not have a natural density.
Please help me )=

REPLY [9 votes]: Your teacher is right. For every subset $A$ of $\mathbb N$, call $d^-(A)=\liminf\limits_{n\to\infty}\frac1n|A\cap[1,n]|$ the lower density of $A$ and $d^+(A)=\limsup\limits_{n\to\infty}\frac1n|A\cap[1,n]|$ the upper density of $A$. Note that $d^-(A)$ and $d^+(A)$ always exist and that $A$ has a natural density $d(A)$ if and only if $d^-(A)=d^+(A)$, and this common value is then equal to $d(A)$. 
For every $n$, consider the sets $I_n=[2^n,2^{n+1})$, $O_n=I_n\cap(2\mathbb N+1)$ (where "$O$" stands for odd) and $E_n=I_n\cap(2\mathbb N)$ (where "$E$" stands for even). You are ready for a host of counterexamples, to begin with:

The set $C=\bigcup\limits_nI_{2n}$ has upper density $\frac23$ and lower density $\frac13$. Exercise: Modify the example to find a set with upper density $1$ and lower density $0$.
The sets $A=\bigcup\limits_nE_{2n}\cup\bigcup\limits_nO_{2n+1}$ and $B=2\mathbb N$ both have density $\frac12$ but $A\cap B=\bigcup\limits_nE_{2n}$ has upper density $\frac13$ and lower density $\frac16$.

This proves that the collection of sets that have a natural density is not an algebra.