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Let $S(\omega)$ denote the collection of "sparse" infinite subsets of $\omega$, that is, $X\subseteq \omega$ is a member of $S(\omega)$ if and only if both $X$ and $\omega\setminus X$ are infinite.

Is there ${\cal S}\subseteq S(\omega)$ such that for all $a\neq b \in \omega$ we have $|\{s\in {\cal S}: \{a,b\} \subseteq s\}| = 1$?

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    $\begingroup$ I assume you mean $\{a,b\}\subseteq s$, since both $\{a,b\}$ and $s$ are subsets of $\omega$. $\endgroup$
    – Andreas Blass
    Commented Jul 8, 2022 at 21:07
  • $\begingroup$ Right @AndreasBlass - sorry for the notational error -> will correct this. $\endgroup$
    – Dominic van der Zypen
    Commented Jul 9, 2022 at 10:32

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Sure, let $K$ be any countably infinite field and let $P$ be the projective plane (or a higher-dimensional projective space) over $K$. Let $S'$ be the set of lines in $P$ (where a line is regarded as a set of points), and transport the family $S'$ of subsets of $P$ to a family $S$ of subsets of $\omega$ via your favorite enumeration of $P$.

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    $\begingroup$ Affine plane would work just as well, right? $\endgroup$
    – bof
    Commented Jul 8, 2022 at 21:29
  • $\begingroup$ @bof Yes. (I just happen to like projective geometry.) $\endgroup$
    – Andreas Blass
    Commented Jul 9, 2022 at 1:44

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