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I have come across an interesting puzzle and can't seem to come up with a proof or a counterexample. Consider;

Let $a_n$ and $b_n$ be strictly increasing sequences of natural numbers. Define $A_N = \{ a_n: n > N \}$, and $B_N = \{ b_n : n > N\}$. We say the two sequences $a_n$ and $b_n$ are tail-set divergent if

$\liminf\limits_{N \rightarrow \infty} \{|a-b|: a \in A_N, b\in B_N \} = \infty$.

Determine the cardinality of the set of pair-wise tail-set divergent strictly increasing sequences of natural numbers. (i.e. countable or uncountable).

This seems difficult to get a hand on. I believe I can show there are countably many at least, however I can't seem to determine how to show if there are uncountably many or not. It certainly seems plausible since the number of possible subsequences in general is uncountable but I can't seem to find a method of approach for the tail-set divergent sequences.

Martin Sleziak
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Jason
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  • So we are to find the cardinality of the set of ordered pairs $((a_n), (b_n))$ that satisfy the condition? – zhw. Jun 07 '17 at 20:52
  • Correct. It doesn't technically need to be ordered pairs, but since the cardinality is some flavor of infinite, it doesn't matter. – Jason Jun 07 '17 at 20:53
  • Could you please identify the source from which you copied that problem? Is that an exact, word-for-word quotation of the original? I ask because as written it doesn't exactly make sense. "Determine the cardinality of the set of pairwise tail-set divergent strictly increasing sequences." There are many such sets, e.g, any subset of a set of pairwise tail-set divergent s.i. sequences will be a set of pairwise tail-set divergent s.i. sequences. My guess: they want us to find the maximum cardinality of such a set, of the supremum of their cardinalities. – bof Jun 07 '17 at 22:58
  • Apologies. This wasn't a problem directly, but rather occurred to me while considering something fairly unrelated (I went down a bit of a rabbit hole to get here), so the lack of clarity is my fault. What I intended is what you suggested; What is the maximal possible cardinality of such a set. Hopefully that clarifies, but if not let me know. – Jason Jun 09 '17 at 11:34

1 Answers1

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The maximum cardinality of a family of pairwise tail-set divergent strictly increasing sequences of natural numbers is not just uncountable: it is $2^\omega=\mathfrak c=|\Bbb R|$, the cardinality of the continuum.

Two subsets of $\Bbb N$ are said to be almost disjoint if their intersection is finite. A family of subsets of $\Bbb N$ is an almost disjoint family if its members are pairwise almost disjoint.

Proposition: Let $A=\{a_k:k\in\Bbb N\}$ and $B=\{b_k:k\in\Bbb N\}$ be almost disjoint subsets of $\Bbb N$ such that the sequences $\langle a_k:k\in\Bbb N\rangle$ and $\langle b_k:k\in\Bbb N\rangle$ are strictly increasing; then the sequences $\sigma_A=\left\langle 2^{a_k}:k\in\Bbb N\right\rangle$ and $\sigma_B=\left\langle 2^{b_k}:k\in\Bbb N\right\rangle$ are tail-set divergent.

Proof: Since $A$ and $B$ are almost disjoint, there is an $m\in\Bbb N$ such that $A_m\cap B_m=\varnothing$. For each $n\ge m$ let $c_n=\min\{a_{n+1},b_{n+1}\}$. If $\ell>n$, then $a_\ell\ne b_\ell$; without loss of generality suppose that $a_\ell>b_\ell$. Then $a_\ell>b_\ell\ge c_n$, so

$$\left|2^{a_\ell}-2^{b_\ell}\right|=2^{a_\ell}-2^{b_\ell}=2^{b_\ell}\left(2^{a_\ell-b_\ell}-1\right)\ge 2^{b_\ell}\ge 2^{c_n}\ge 2^{n+1}\,,$$

since $a_k,b_k\ge k$ for each $k\in\Bbb N$. Thus, $$\inf\left\{\left|2^a-2^b\right|:a\in A_n\text{ and }b\in B_n\right\}\ge 2^{n+1}$$ and $$\liminf\limits_{n\to\infty}\left\{\left|2^a-2^b\right|:a\in A_n\text{ and }b\in B_n\right\}=\infty\,,$$ as desired. $\dashv$

For each $A=\{a_k:k\in\Bbb N\}$ such that $\langle a_k:k\in\Bbb N\rangle$ is strictly increasing let $\sigma_A=\left\langle 2^{a_k}:k\in\Bbb N\right\rangle$. It follows immediately from the proposition that if $\mathscr{A}$ is an almost disjoint family of infinite subsets of $\Bbb N$, then $\{\sigma_A:A\in\mathscr{A}\}$ is a family of pairwise tail-set divergent sequences. This answer and the answers to this question give a variety of proofs of the fact that there are almost disjoint families of $\Bbb N$ of cardinality $\mathfrak c$, so there are also families of $\mathfrak c$ pairwise tail-set divergent sequences. This is the maximum possible size of such a family, since there are only that many strictly increasing sequences of natural numbers.

Brian M. Scott
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