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1
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Let $A\subseteq\mathbb N$, as usual we set $d^+(A)=\lim\sup_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}$ and $d^{-}=\lim\inf_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}$. It's very stardard that in general $d^+(A)\neq d^-(A)$. My question is little different: is there any $A$ for which $d^-(A)=0$ and $d^+(A)\neq0$? Equal to $1$? Thanks in advance, Valerio |
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4
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For an explicit example, you could use $A= \{ n: \lfloor \log_2 \log_2{n} \rfloor \text{ is even} \} $. |
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3
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Yes of course. To construct $A$ throw in as many elements as needed to make $|A\cap [1,n]|/n$ within $\epsilon$ of 1, then leave out as many elements as needed to make $|A\cap [1,n]|/n$ within $\epsilon$ of 0. Repeat as you send $\epsilon \to 0$. |
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2
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Here’s an explicit construction: let $A$ be the set of natural numbers $n$ such that $\lfloor\log\log n\rfloor$ is odd. |
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1
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Another explicit example: $A=\lbrace 1!,\dots,2!\rbrace\cup\lbrace 3!,\dots, 4!\rbrace\cup\lbrace 5!,\dots,6!\rbrace\cup\dots$. |
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