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For a cardinal $\kappa$ by $[\kappa]^{<\kappa}$ we denote the family of all subsets of cardinality $<\kappa$ in $\kappa$.

Question. Assume that for an infinite cardinal $\kappa$ there exists a family $\mathcal B\subset[\kappa]^{<\kappa}$ of cardinality $|\mathcal B|=\kappa$ such that for any $A\in[\kappa]^{<\kappa}$ the set $\{B\in\mathcal B:A\not\subset B\}$ has cardinality $<\kappa$. Is $\kappa$ a regular cardinal?

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  • $\begingroup$ I don't know about characterization, I think that under MA or something we can get this to fail with $\kappa=\aleph_1$. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 21, 2018 at 16:51
  • $\begingroup$ @AsafKaragila It seems that I have found an affirmative answer to this question. But the proof of a bit lengthy. I needed this property for the proof of normality of balleans on groups and it turns out that I can prove this property using the technique of balleans. But maybe there are direct proofs. $\endgroup$
    – Taras Banakh
    Commented Oct 21, 2018 at 18:06
  • $\begingroup$ I can believe that. I just think that this is not a characterization per se, just a condition that implies regularity. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 21, 2018 at 18:09
  • $\begingroup$ @AsafKaragila No, it is a characterization: if a cardinal $\kappa$ is regular then the family $\mathcal B$ consiting of the initial intervals of $\kappa$ has the required property. $\endgroup$
    – Taras Banakh
    Commented Oct 21, 2018 at 19:12
  • $\begingroup$ Oh. You're absolutely right! $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 21, 2018 at 19:13

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Yes. The assumption implies that $\mathcal{B}$ is cofinal in the poset $[\kappa]^{<\kappa}$, and hence that this poset has cofinality $\le\kappa$. This implies that $\kappa$ is regular, by this answer.

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