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Let $\kappa$ be a cardinal, and let $P$ be a poset. Let $\mathcal{P}_\kappa(P)$ denote the poset of $\kappa$-small subposets of $P$ and let $\mathcal{P}_\kappa^\downarrow(P)\subseteq\mathcal{P}_\kappa(P)$ be the subposet consisting of those subposets that are downward-closed. Then according to a reliable source, when $\kappa$ is regular, we can show that $\mathcal{P}^\downarrow_\kappa(P)$ is $\kappa$-filtered because given some $\kappa$-small family of $\kappa$-small subposets, $$A_i:I\to \mathcal{P}_\kappa(P)\quad |I|<\kappa$$

the downward closure of the union over this family, $\operatorname{Cl}^\downarrow(\bigcup_{i\in I}A_i)$, is $\kappa$-small (which gives a majorant for the family $A_i$).

However, since I have no experience at all working with regular cardinals, I'm not really sure how to make heads or tails of this. Why does the regularity of $\kappa$ imply that the downward-closure of that union is $\kappa$-small?

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A cardinal $\kappa$ is regular if (and only if) the union of fewer than $\kappa$ many sets of size less than $\kappa$ still always has size less than $\kappa$. That seems to be exactly what you have here. Also, the union of downward closed sets remains downward closed, so you don't need to take the downward closure of the union, as it is already downward closed.

Note, however, that the downward closure of a $\kappa$-small family might no longer be $\kappa$-small, if $P$ has large initial segments. For example, $P$ may have no $\kappa$-small downward closed subposets at all (this is true in the reverse ordinal $\kappa^*$, which is $\kappa$ turned upside down).

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  • $\begingroup$ Why doesn't the Wikipedia page have that definition?! It has some silly thing about "a cardinal is called regular if it is equal to is own cofinality", which is completely opaque and unhelpful. Thank you very much! $\endgroup$
    – Harry Gindi
    Commented Aug 1, 2010 at 13:47
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    $\begingroup$ The cofinality definition is equivalent, since a smaller cofinality means there is a small increasing sequence of ordinals below $\kappa$, which would be a counterexample to the characterization I give. $\endgroup$
    – Joel David Hamkins
    Commented Aug 1, 2010 at 13:53
  • $\begingroup$ Re: Your edit: Alright, well it seems that everything works out fine in this case anyway, since the poset $P$ is specifically a $\kappa$-good tree (well-founded order, and the downward closure of any element $\alpha$ is $\kappa$-small, so $P$ is the $\kappa$-filtered union of its $\kappa$-small subsets. $\endgroup$
    – Harry Gindi
    Commented Aug 1, 2010 at 13:57
  • $\begingroup$ $\kappa$-small downward-closed subsets, rather. $\endgroup$
    – Harry Gindi
    Commented Aug 1, 2010 at 13:59
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    $\begingroup$ The smallest non-regular infinite cardinal is $\aleph_\omega$, which is a countable union of the sets $\aleph_n$, each of which are smaller than $\aleph_\omega$. But every infinite successor cardinal is regular. $\endgroup$
    – Joel David Hamkins
    Commented Aug 1, 2010 at 16:12

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