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 $\aleph\_n$ ?



Poset as union of posets of lower cofinality
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.)


