Let $a, b \in {\cal P}(\omega)/\mathrm{(fin)}$ with $a<b$. Do we have ${\cal P}(\omega)/(fin)\cong [a,b]$?



Is ${\cal P}(\omega)/\mathrm{(fin)}$ order-isomorphic to its intervals?
The answer is yes.


Without loss of generality, we may choose representatives for $a$ and $b$ such that $a \subseteq b$ when construed as elements of $\mathcal{P}(\omega)$ (by taking arbitrary representatives, and changing these as needed at the finitely many indices where the subset relation may not already hold).


Note then that we have the identity $(S \cap b) \cup a = (S \cup a) \cap b$ for $S \subseteq \omega$; call this operation $f(S)$.


We have that $f$ is a monotonic operator from $\mathcal{P}(\omega)$ to itself, and that $a \subseteq f(S) \subseteq b$, always. Furthermore, the indices at which $f(S)$ and $f(S')$ differ are the intersection of the set of indices at which $S$ and $S'$ differ with the set of indices at which $a$ and $b$ differ. The latter is a co-finite set, and the operation of intersection with a co-finite set both preserves and reflects the property of being co-finite. Thus, we can conclude that $f(S)$ and $f(S')$ are equal at co-finitely many indices if and only if $S$ and $S'$ are equal at co-finitely many indices. From this, we see that $f$ represents an injective homomorphism from $\mathcal{P}(\omega)/(fin)$ to $[a, b]$.


Finally, to see that this is surjective and therefore an isomorphism, consider that $f$ also acts idempotently on those values which are already in $[a, b]$.


