Poset as union of posets of lower cofinality Let $ \mathbb{P}$ be any directed, well-founded poset of cofinality $ \aleph_{n+1}$, where $n$ is a natural.
Can we write it as an increasing union $ \mathbb{P} = \bigcup_{\alpha < \omega_{n+1} } \mathbb{P}_\alpha $
where each $\mathbb{P}_\alpha $ is a directed poset of cofinality $\aleph_n$ ?
 A: Yes, and well-foundedness is irrelevant (this is working in ZFC). Let $\mathbb{P}$ be the poset, with ordering $\leq$. Let $f:[\mathbb{P}]^2\to\mathbb{P}$ be a function such that for all $x,y\in\mathbb{P}$, we have $x\leq f(\{x,y\})$ and $y\leq f(\{x,y\})$.
Let $c:\aleph_{n+1}\to\mathbb{P}$ be cofinal in $\mathbb{P}$, and injective. For each limit ordinal $\alpha<\aleph_{n+1}$, let $\bar{\mathbb{Q}}_\alpha$ be the restriction of $\mathbb{P}$ to the $f$-closure of $c``\alpha$, and let $\mathbb{Q}_\alpha$
be the $\leq$-downward closure of $\bar{\mathbb{Q}}_\alpha$.
Certainly $\mathbb{P}$ is the monotone increasing union of the $\mathbb{Q}_\alpha$'s, and each $\mathbb{Q}_\alpha$ is  a directed partial order of cofinality $\leq\aleph_n$. Also the sequence of $\mathbb{Q}_\alpha$'s is continuous (at limits of limits). And $\bar{\mathbb{Q}}_\alpha$ is a cofinal subset of $\mathbb{Q}_\alpha$ of cardinality $\leq\aleph_n$. No $\mathbb{Q}_\alpha$ is cofinal, so there is a cofinal set $C\subseteq\aleph_{n+1}$ such that for all $\alpha\in C$, we have (i) $\bar{\mathbb{Q}}_\alpha\not\subseteq\mathbb{Q}_\beta$
for all $\beta<\alpha$, and (ii) $\mathrm{cof}(\alpha)=\aleph_n$.
But note that for each $\alpha\in C$,
$\mathbb{Q}_\alpha$ has cofinality $\aleph_n$. (If $X\subseteq\mathbb{Q}_\alpha$
has cardinality $<\aleph_n$ and is cofinal in $\mathbb{Q}_\alpha$,
then since $\bar{\mathbb{Q}}_\alpha$
is cofinal in $\mathbb{Q}_\alpha$,
note that $X\subseteq$ the $\leq$-downward closure of $\bar{\mathbb{Q}}_\alpha$, but $\bar{\mathbb{Q}}_\alpha=\bigcup_{\beta<\alpha}\bar{\mathbb{Q}}_\beta$,
so there is $\beta<\alpha$
such that $X\subseteq$ the $\leq$-downward closure of $\bar{\mathbb{Q}}_\beta$,
i.e. $X\subseteq\mathbb{Q}_\beta$, but then $\mathbb{Q}_\beta=\mathbb{Q}_\alpha$, contradiction.)
