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It is possible to construct a sequence of integers with lower density 0 and upper density 1? where lower and upper density means asymptotic lower and upper density (cf. References on density of subsets of $\mathbb{N}$)

EDIT: So, if this is true, then one can split $\mathbb{N}$ into to sequeneces of null lower density. I find it paradoxical.

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Yes, it is possible. What follows isn't a completely rigorous proof, but you should be able to make one from that.

The basic idea is to construct a set $A$ such that along one subsequence, $\lim |A \cap I_{a_n}|/a_n = 1$ and along another subsequence, $\lim |A \cap I_{b_n}|/b_n = 0$. We'll do this by adding or skipping consecutive "chunks" of integers.

  • Start by adding $1$ to $A$. We're at density $1$.
  • Then skip $2$. For $n = 2$, we have $|A \cap I_2|/2 = 1/2$.
  • Now add $3$ and $4$, so that $|A \cap I_4|/4 = 3/4$.
  • Now skip $5$ through $12$, so that $|A \cap I_{12}|/12 = 3/12 = 1/4$.

This should give you an idea of how the following works:

  • "Add" enough integers so that you're back to density $\geq (2^n - 1)/2^n$
  • Then "skip" enough integers so that you're below density $\leq 1/2^n$.

Repeat the two steps in order. Along the two subsequences constructed, the limit of the density is either 1 or 0.

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Yes. Let $A$ be the set (still to be constructed), and for $n\in\Bbb Z^+$ let $A_n=\{k\in A:k\le n\}$. Suppose that you’ve constructed $A_n$. For any $\epsilon>0$ choose $m\in\Bbb Z^+$ large enough so that $\frac{n}m<\epsilon$, and let $A_m=A_n$. (In other words, omit from $A$ every integer $k$ such that $n<k\le m$.) Then $\frac{|A_m|}m=\frac{|A_n|}m\le\frac{n}m<\epsilon$. Now choose $r\in\Bbb Z^+$ large enough so that $\frac{m}r<\epsilon$, and let $$A_r=A_m\cup\{k\in\Bbb Z^+:m<k\le r\}\;;$$ then $\frac{|A_r|}r\ge\frac{r-m}r=1-\frac{m}r>1-\epsilon$. By alternating in this fashion, using a sequence of $\epsilon$’s converging to $0$, you can construct a set $A$ such that $\underline{d}(A)=0$ and $\overline{d}(A)=1$.

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Yes. For example, divide $\mathbb{N}$ into intervals $[n_k, n_{k+1})$ where $n_k$ is sufficiently fast growing and take the union of every other interval.

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  • $\begingroup$ I think I'm missing something. "Every other interval" is just $[1,n_1)$ $\endgroup$ – Tito Eliatron Jul 4 '13 at 9:42

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