TITLE: Sets of natural numbers with finite intersections and divergent sums of reciprocals
QUESTION [5 upvotes]: 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 $A \in \Lambda$?

REPLY [2 votes]: 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.