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Let $S \subset \mathbb N$ be a subset. The natural density is defined as

$$D(S) = \lim_{n \to \infty} \frac{|E \cap \{1, \cdots, n\}|}{n}$$

whenever this limit exists.

So question is the following: Let $S \subset \mathbb N$ be a set of natural density zero. Let $E = \{n_k : k\in \mathbb N\}$ be a subsequence of $\mathbb N$. Assume that $E$ has positive upper density, that is

$$ D^*(E) = \limsup_{n\to \infty} \frac{|E \cap \{1, \cdots, n\}|}{n}>0$$

Does the set $E\cap S$ has natural density zero in $E$? That is, do we have

$$\lim_{k \to \infty} \frac{|\{k :n_k \in S\}\cap \{1, \cdots, k\}|}{k} = 0 \ \ ?$$

This question is related to the following:

a simple question about the density convergence of sequences

(Also not sure if the "Number theory" tag is suitable, feel free to edit)

1 Answers1

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Let $$E=(1!,2!]\cup(4!,5!]\cup(7!,8!]\cup(10!,11!]\cup(13!,14!]\cup\cdots,$$ $$S=(4!,4!+3!]\cup(7!,7!+6!]\cup(10!,10!+9!]\cup(13!,13!+12!]\cup\cdots.$$ Then $E$ has upper density one, $S$ has density zero, but $E\cap S=S$ has upper density one in $E.$

bof
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