Consider the partial order $(P(\omega),\subseteq)$. Let $L$ be a dense linear suborder. Does $L$ have a countable dense subset?


(Note that it contains a copy of $\mathbb R$, via Dedekind cuts of $\mathbb Q$.)



Linear suborders of $(P(\omega),\subseteq)$
For any $a\neq b\in\omega$ pick a set $A\_{ab}\in L$ such that $a\in A\_{ab},b\not\in A\_{ab}$, if such exists, and let $D$ be the set of all $A\_{ab}$ we have picked. We claim $D$ is dense in $L$.


Let $X,Y\in L$ with $X\subsetneq Y$. Pick any $Z$ such that $X\subsetneq Z\subsetneq Y$ and take $a\in Z\setminus X,b\in Y\setminus Z$. Then $Z$ is an element containing $a$ but not $b$, so $A\_{ab}\in D$ exists. We have $A\_{ab}\not\subseteq X$, so $X\subseteq A\_{ab}$, and $Y\not\subseteq A\_{ab}$, so $A\_{ab}\subseteq Y$. Hence $D$ contains an element $A\_{ab}$ strictly between $X$ and $Y$, so $D$ is a dense countable subset of $L$.


Note that countable choice was used in the proof, but it is unavoidable, since (\mathbb R and hence also) $P(\omega)$ can contain a finite Dedekind-infinite subset, and this will necessarily contain no dense countable subset.


