Sets with no asymptotical density over $\mathbb N$

Solution 1:

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.

Solution 2:

To find a set that doesn't have a natural density, you need it to have lots of elements, then have lots missing, then have lots, and so on so $x_n$ oscillates and does not approach a limit. An example would be $\{1\}\cup [2^3,2^5] \cup [2^7,2^9] \cup \dots [2^{4k-1},2^{4k+1}]\dots$

Solution 3:

Just to get you started.

To find a set $A$ with no density, you can proceed as follows:

You can have increasing-length blocks of integers which are in $A$, interweaved with increasing-length blocks of integers not in $A$, for example:

$$A = \{ 1,4,5,6,7,16,17,\dots,31\}$$

Then look at $[1,n] \cap A$ where $n$ is a power of 2, and you should find that the fractions $\frac{|A\cap[1,n]|}{n}$ oscillate, and do not converge to a limit.