Let $(P,\leq)$ be a countably infinite poset with the property that whenever $a<b\in P$ then $P\cong [a,b]$.


**Question.** If $P$ does contain elements $a,b$ with $a<b$, does this imply that $P \cong [0,1]\cap\mathbb{Q}$, or that $P$ is isomorphic to the nonzero countable atomless Boolean algebra?


**Note.** Thanks to user @YCor for suggestions to make this a better question.



Countably infinite posets isomorphic to its intervals
How about this. Let $P$ have "levels" indexed by $\mathbb{Q} \cap [0,1]$. Levels $0$ and $1$ each contain a single element, all other levels contain a countably infinite set of elements. An element at level $a$ lies below one at level $b$ if and only if $a <b$. It's clearly not a Boolean algebra, but also clearly has the desired property.


