2
$\begingroup$

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 $|\{s\in {\cal S}: \{a,b\} \subseteq s\}| = 1$?

$\endgroup$
2
  • 2
    $\begingroup$ I assume you mean $\{a,b\}\subseteq s$, since both $\{a,b\}$ and $s$ are subsets of $\omega$. $\endgroup$ Commented Jul 8, 2022 at 21:07
  • $\begingroup$ Right @AndreasBlass - sorry for the notational error -> will correct this. $\endgroup$ Commented Jul 9, 2022 at 10:32

1 Answer 1

4
$\begingroup$

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 $P$.

$\endgroup$
2
  • 1
    $\begingroup$ Affine plane would work just as well, right? $\endgroup$ Commented Jul 8, 2022 at 21:29
  • $\begingroup$ @bof Yes. (I just happen to like projective geometry.) $\endgroup$ Commented Jul 9, 2022 at 1:44

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.