The poset of k-small downward-closed subposets of a poset P is k-filtered when k is a regular cardinal? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:15:41Z http://mathoverflow.net/feeds/question/34107 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34107/the-poset-of-k-small-downward-closed-subposets-of-a-poset-p-is-k-filtered-when-k The poset of k-small downward-closed subposets of a poset P is k-filtered when k is a regular cardinal? Harry Gindi 2010-08-01T13:13:29Z 2010-08-01T13:46:36Z <p>Let $\kappa$ be a cardinal, and let $P$ be a poset. Let <code>$\mathcal{P}_\kappa(P)$</code> denote the poset of $\kappa$-small subposets of $P$ and let <code>$\mathcal{P}_\kappa^\downarrow(P)\subseteq\mathcal{P}_\kappa(P)$</code> 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, $$A_i:I\to \mathcal{P}_\kappa(P)\quad |I|&lt;\kappa$$</p> <p>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 $A_i$). </p> <p>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?</p> http://mathoverflow.net/questions/34107/the-poset-of-k-small-downward-closed-subposets-of-a-poset-p-is-k-filtered-when-k/34108#34108 Answer by Joel David Hamkins for The poset of k-small downward-closed subposets of a poset P is k-filtered when k is a regular cardinal? Joel David Hamkins 2010-08-01T13:40:52Z 2010-08-01T13:46:36Z <p>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.</p> <p>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).</p>