5
$\begingroup$

$\newcommand{\H}{\mathcal{H}}$ $\newcommand{\A}{\mathcal{A}}$ Recall that a coideal $\H$ over $\omega$ is selective if for every $\{A_n : n < \omega\} \subseteq \H$, where $i < j \implies A_i \supseteq A_j$, there exists some $B \in \H$ such that $B/n := \{m \in B : m > n\} \subseteq A_n$ for all $n \in B$. There are two canonical examples of selective coideals, both of which can be constructed in $\mathsf{ZFC}$:

  • The entire $[\omega]^\omega$ is of course a selective coideal.

  • Let $\A$ be a mad (maximal almost disjoint) family of subsets of $\omega$. Let $\H$ be the set of all infinite subsets of $\omega$ which cannot be covered up to a finite set by finitely many members of $\A$. Then $\H$ is a selective coideal.

Are there any more examples of selective coideals that $\mathsf{ZFC}$ can prove to exist? Note that I'm not looking for selective coideals that exist with further assumptions (e.g. under $\mathsf{CH}$, there are $2^{2^{\aleph_0}}$ many Ramsey ultrafilters, but I'm not asking for such examples).

$\endgroup$
1
  • $\begingroup$ @bof By definition $\mathcal{H}$ is a coideal if $\mathcal{I} := \mathcal{P}(\omega) \setminus \mathcal{H}$ is an ideal. This is not the same as a filter. $\endgroup$ Sep 30, 2022 at 14:08

1 Answer 1

3
$\begingroup$

See Section 12 in S. Todorcevic, Topics in Topology, Lecture notes in Mathematics Vol. 1652, Springer-Verlag Berlin Heidelberg 1997

$\endgroup$
1
  • 2
    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 30, 2022 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.