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For $A\subseteq \mathbb{N}$ we define the lower density by if $$\operatorname{ld}(A)=\liminf_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ If $A\subseteq \mathbb{N}$ satisfies $\operatorname{ld}(A) = 0$, does this imply $\operatorname{ld}(\mathbb{N}\smallsetminus A) = 1$?

Michael Hardy
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Dominic van der Zypen
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  • What if $A$ consists of all numbers between $n!$ and $2(n!)$ for some $n$? I.e., $A = {1, 2, 3, 4, 6, \ldots, 12, 24, \ldots, 48, 120, \ldots, 240, 720, \ldots, 1440, \ldots}$. – Brian Tung Jun 04 '15 at 06:40
  • You have asked several questions about asymptotic density. Maybe some of the properties and references mentioned here can be useful for you: http://math.stackexchange.com/questions/126745/references-on-density-of-subsets-of-mathbbn – Martin Sleziak Apr 14 '16 at 05:23

2 Answers2

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You have :

$$\operatorname{ld}(A)=\liminf_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$$

Hence $$\operatorname{ld}(\mathbb{N}\setminus A)=\liminf_{n\to\infty}1-\frac{|A\cap\{1,\ldots,n\}|}{n}=1-\limsup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$$

Hence we see that $ld(A)=0$ and $ld(\mathbb{N}\setminus A)=1$ if and only if :

$$\limsup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}=\liminf_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}=0$$

This implies in particular that $(\frac{|A\cap\{1,\ldots,n\}|}{n})$ has a limit which is zero, I think it is then easy to find counter-examples of $A$ such the infimum limit is zero but whose density does not converge toward zero.

Clément Guérin
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Let $A_k$ consist of all integers between $2^{2^{2k}}$ and $2^{2^{2k+1}}$, and let $A = \bigcup_{k=1}^\infty A_k$. (Equivalently, $n\in A$ if and only if $\lfloor \log_2 (\log_2 n) \rfloor$ is even.) Then both $A$ and $\Bbb N \setminus A$ have lower density $0$ and upper density $1$.

Greg Martin
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