# Upper density, and nested sets

For a given $$A \in \mathbb{N}$$, we define the upper desnity $$\bar d(A)$$ by $$\begin{equation*} \bar d(A) = \limsup_{N \to \infty} \frac{A \cap [N]}{N} \end{equation*}$$ where $$[N]$$ denotes the set $$\{1,2,\dots, N\}$$. My question is, if we have a sequence of sets $$\begin{equation*} A_1 \supseteq A_2 \supseteq \dots \supseteq A_n \supseteq \dots \end{equation*}$$ such that $$\begin{equation*} \bigcap_{n \geq 1} A_n = \varnothing \end{equation*}$$ then do we we have $$\begin{equation*} \liminf_{n \to \infty} \bar d(A_n) = 0 \end{equation*}$$ I'm not sure if this is obvious or not. I was thinking that a way to prove this statement is to first define $$\frac{1}{2} A_1 \subseteq \mathbb{N}$$ to be such that it includes every other element of $$A_1$$. Then we would have $$\bar d(\frac{1}{2} A_1) = \frac{1}{2} \bar d(A_1)$$, and there is an $$m \in \mathbb{N}$$ such that $$A_m \subseteq \frac{1}{2} A_1$$, so $$\bar d(A_m) \leq \frac{1}{2} \bar d(A_1) \leq \frac{1}{2}$$. Carrying on like this we get a subsequence which decays as $$\frac{1}{2^k}$$ which implies the claim. This argument does feel a bit fishy to me though.

• By $A\in\mathbb N$ I think you mean $A\subseteq\mathbb N$. – bof Feb 23 at 17:58

As a counterexample, how about $$A_n=\Bbb N \setminus[n]$$?
No. $$A_n=\mathbb{N}-[n]$$ is a counterexample. There is no reason why $$A_m$$ must be a subset of your $$\frac12A_1$$.