Sets of natural numbers with finite intersections and divergent sums of reciprocals

Question

Does there exist an uncountable collection $\Lambda$ of infinite subsets of the set of natural numbers such that (i) any two distinct subsets in the collection have a finite intersection and (ii) the sum of the reciprocals is divergent for each $A \in \Lambda$?

Answer 1

This can be done by a fairly simple diagonalization argument. Call a subset of $\mathbb{N}$ large if its sum of reciprocals is infinite. It suffices to prove the following (for we can then construct an uncountable family by induction). Let $\Lambda$ be a countable collection of large sets such that for any finite subset $F\subseteq\Lambda$, $\mathbb{N}\setminus\bigcup F$ is large. Then there exists a large set $C$ that has finite intersection with every element of $\Lambda$ such that for every finite subset $F\subseteq\Lambda\cup\{C\}$, $\mathbb{N}\setminus\bigcup F$ is large.

To prove this, let $\{A_n\}$ be an enumeration of $\Lambda$, and let $B_n=\bigcup_{i=0}^n A_i$; we want to construct a large set $C$ such that $C\cap B_n$ is finite and $\mathbb{N}\setminus (C\cup B_n)$ is large for all $n$. We define the characteristic function $1_C:\mathbb{N}\to\{0,1\}$ of $B$ by the following process: First, make $1_C$ be $1$ at finitely many points such that the reciprocal sum of $C$ is at least $1$. Then, add finitely many $0$s such that the reciprocal sum of $\mathbb{N}\setminus (C\cup B_0)$ is at least $1$. Now declare $1_C$ to be $0$ on all points of $B_0$ where we have not yet defined it. Now, add finitely many $1$s such that the reciprocal sum of $C$ is at least $2$. Now add finitely many $0$s such that the reciprocal sum of $\mathbb{N}\setminus (C\cup B_1)$ is at least $2$. Now declare $1_C$ to be $0$ at all points of $B_1$ where we have not yet defined it. Now add finitely many $1$s such that the reciprocal sum of $C$ is at least $3$. And so on.

Answer 2

There is in fact a perfect family of such sets, given as the branches of a certain perfect binary tree. (By a perfect binary tree I mean a rooted tree in which each element has 1 or 2 successors, such that there is no isolated branch, i.e., above each node of the tree there is at least one "splitting" node, i.e., one with two successors.)

(Via characteristic functions, this family can be viewed as a subset of $\{0,1\}^{\mathbb N}$, the set of all infinite $\{0,1\}$-sequences. Both the space $\{0,1\}^{\mathbb N}$ and the family I constructed are naturally homeomorphic to the Cantor set, and in particular have the same cardinality as the real numbers.)

The root of the tree is the number 1. It has two successors, 2 and 3. Above 2 there is a sequence 4,6,8,...,14 of nonsplitting nodes; note that the sum of their inverses is more than 1. Also above 3 there is a sequence 5,7,9,...,15 of nonsplitting nodes, again with sum of inverses greater than 1.

Above 14 we add a splitting node 16 with successors 18 and 20, and above 15 we add a splitting node 17 with successors 19 and 21. Above each of nodes 18,19,20,21 we add a finite arithmetic sequence of nonsplitting nodes with difference 4 (say: 22,26,... etc above 18; 23,27,... above 19, etc.) in such a way that each respective sum of reciprocals is greater than 1. These arithmetic sequences will be disjoint.

Above each of these 4 arithmetic sequences we add a splitting node; above those we attach 8 disjoint arithmetic sequences with difference 8 which are again so long that their sum of reciprocals is each greater than 1. Etc.

Clearly any two branches of this tree will be almost disjoint sets, and along any branch the sum of reciprocals diverges.

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An alternative (perhaps more transparent) formulation: Let $2^{<\omega}$ be the set of all finite $\{0,1\}$-sequences; this is a countable set. Find a countable family of pairwise disjoint finite sets $(A_s:s\in 2^{<\omega})$ such that $\sum_{n\in A_s}\frac1n \ge 1$ for each $s\in 2^{<\omega}$. Now every infinite $\{0,1\}$-sequence $f$ will define a set $A_f:=\bigcup _{s\vartriangleleft f} A_s$. (Here, $s\vartriangleleft f$ means that the finite sequence $s$ is an initial segment of $f$.)

These sets $A_f$ (for each $f:\mathbb N \to \{0,1\}$) will now be almost disjoint, and the harmonic series diverges along each of them.

A similar construction can be done for any divergent series.

Answer 3

First we partition $\mathbb N$ into a disjoint union of intervals each consisting of finitely many consecutive integers such that the sum of reciprocals in each interval exceeds 1. The collection of these intervals is a countably infinite set, which we denote by $\tilde{\Sigma}$. Next we consider the set $\Sigma'$ of square-free integers. (An integer $ n$ is square-free if $p^2$does not divide $n$ for any prime $p$. ) For every infinite subset $A = \{p_1<p_2<p_3<\cdots \} $ of primes define $A'\subseteq \Sigma'$ by $ A'=\{p_1,p_1p_2,p_1p_2p_3, \cdots ,p_1p_2p_3\cdots p_i,\cdots \}$. If $A \neq B$, then $ A'\cap B'$ is finite. Take any bijection from $\Sigma'$ to $\tilde{\Sigma}$. A subset $A'\subseteq \Sigma'$ gives rise to a subset $\tilde A \subseteq \mathbb N$ by replacing points of $\Sigma'$ by the interval in $\mathbb N $ via the bijection $\Sigma' \rightarrow \tilde {\Sigma}$. The collection of subsets $\tilde A \subseteq\mathbb N$ thus obtained satisfies the requirements.