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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.

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  • $\begingroup$ By $A\in\mathbb N$ I think you mean $A\subseteq\mathbb N$. $\endgroup$ – bof Feb 23 at 17:58
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As a counterexample, how about $A_n=\Bbb N \setminus[n]$?

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

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