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Let $A$ be a set of positive integers $\{a_n: n\ge 1\}$ (in increasing order) with positive upper density, that is, $$ \limsup_{n\to \infty}\frac{1}{n}|A \cap [1,n]|>0. $$ Let also $B=\{b_n: n\ge 1\}$ a set of positive upper density.

Is it true that $\{a_{b_n}: n\ge 1\}$ has positive upper density?

1 Answers1

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Counterexample based on my answer to this question.

The set $$A=(1!,2!]\cup(4!,\ 5!]\cup(7!,\ 8!]\cup(10!,\ 11!]\cup(13!,\ 14!]\cup\cdots$$ has upper density $1$, since $$\frac{|A\cap[1,\ (3n+2)!]|}{(3n+2)!}\ge\frac{(3n+2)!-(3n+1)!}{(3n+2)!}=\frac{3n+1}{3n+2}\to1.$$

The subset $S\subset A$ defined by $$S=(4!,\ 4!+3!]\cup(7!,\ 7!+6!]\cup(10!,\ 10!+9!]\cup(13!,\ 13!+12!]\cup\cdots$$ has density $0,$ since $$\frac{|S\cap[1,\ (3n+1)!+(3n)!]|}{(3n+1)!+(3n)!}=\frac{3!+6!+\cdots+(3n)!}{(3n+1)!+(3n)!}\lt\frac2{3n+1}\to0.$$

However, $S$ has upper density $1$ in $A,$ since $$\frac{|S\cap[1,\ (3n+1)!+(3n)!|}{|A\cap[1,\ (3n+1)!+(3n)!|}\ge\frac{(3n)!}{(3n-1)!+(3n)!}=\frac{3n}{3n+1}\to1.$$

I hope this is good enough. If you insist on putting it in your awful $a_n,b_n,a_{b_n}$ notation, I'm afraid it's going to get ugly.

bof
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  • This is perfect. The curious thing is that it works under the (a bit stronger) assumption the lower density of $B$ is positive –  Jun 01 '17 at 10:51
  • I wouldn't say positive lower density is a bit stronger that positive upper density, I'd say it's a lot stronger. – bof Jun 01 '17 at 10:56
  • I know, I know :P In that case, taking a suitable subset of $B$, it can be assumed without loss of generality that $B$ admits positive density –  Jun 01 '17 at 10:57
  • @Nduccio Wait a minute. Are you saying that, if $B$ has positive lower density, then there is a subset of $B$ whose density exists and is positive? Are you sure about that? – bof Jun 01 '17 at 11:04
  • @Nduccio In some old notes of mine I find the statement: for any given $\alpha\in[0,1)$ there is a subset $A$ of $\mathbb N,$ with lower density $\alpha$ and upper density $1,$ such that there is no subset of $A$ whose density exists and is positive. – bof Jun 01 '17 at 12:29
  • You are right. There are two results about the Darboux property of upper and lower densities (at least, the ones which I know): every upper and lower density has the Darboux property (see here http://www.sciencedirect.com/science/article/pii/S0022314X17300033); and another one (which is related to your result) is the PhD thesis of Georges Grekos, about the joint Darboux property of the lower and upper asymptotic density, published in JNT here http://www.sciencedirect.com/science/article/pii/0022314X78900343 . –  Jun 01 '17 at 13:07
  • In particular, the closed-convex subsets of $[0,1]^2$ (with the obvious constraints due to monotonicity) cover all the possibilities. Hence, your example has to exist :D –  Jun 01 '17 at 13:07