If $(P,\leq)$ is a partially ordered set and $a,b\in P$ we set $[a,b]:=\{x\in P: a\leq x\leq b\}$. We say that $P$ is *fractal* if whenever $a,b\in P$ and $[a,b]$ contains more than one element, then $[a,b]\cong P$.


So for instance, $[0,1]$ and $[0,1]\cap \mathbb{Q}$ are fractal with their usual linear orderings.


For $A,B\in{\cal P}(\omega)$ we say $A\subseteq^\* B$ if $A\setminus B$ is finite (that is, $A$ is "almost contained" in $B$). We write $A\simeq\_{\text{fin}} B$ if $A\subseteq^\* B$ and $B\subseteq^\* A$ (that is, the sets $A, B$ are "almost the same set" except for finitely many elements). It is easy to see that $\simeq\_{\text{fin}}$ is an equivalence relation on ${\cal P}(\omega)$.


We denote the collection of equivalence classes on ${\cal P}(\omega)$ with respect to $\simeq\_{\text{fin}}$ by ${\cal P}(\omega)/(\text{fin})$. Using $\subseteq^\*$ on representatives of equivalence classes, it is easy to see that we can make ${\cal P}(\omega)/(\text{fin})$ into a partially ordered set.


Is ${\cal P}(\omega)/(\text{fin})$ a fractal poset? If yes, is it true that if ${\cal B}$ is a fractal Boolean algebra on more than $2$ points, then ${\cal P}(\omega)/(\text{fin})$ can be order-embedded into ${\cal B}$?



Is ${\cal P}(\omega)/\text{(fin)}$ a fractal poset?
Yes, $P(\omega)/\text{fin}$ is fractal. If $A\subseteq^\* B$ but not equivalent, then the interval $[A,B]$ in $P(\omega)/\text{fin}$ consists of the sets that almost contain $A$ and are almost contained in $B$, and this is isomorphic to $P(B-A)/\text{fin}$, which is isomorphic to $P(\omega)/\text{fin}$, since $B-A$ will be infinite.


