Q:

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

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

A:

Suppose $a<b$ in $\mathcal P(\omega)/\text{fin}$. Pick representatives $A,B\in\mathcal P(\omega)$ for these equivalence classes mod fin, and, by modifying $A$ by finitely many elements if necessary (which doesn't change the equivalence class), arrange that $A\subseteq B$. The strict inequality $a<b$ implies that $B-A$ is infinite, so you can fix a bijection $f:\omega\to B-A$. I'll use $f$ to define the required isomorphism $g$.
Given any element $x\in\mathcal P(\omega)/\text{fin}$, choose a representative $X\subseteq\omega$. Its image $f[X]$ under $f$ is a subset of $B-A$, so $A\subseteq A\cup f[X]\subseteq B$. So the equivalence class of $A\cup f[X]$ is an element $g(x)$ of the interval between $a$ and $b$ in $\mathcal P(\omega)/\text{fin}$. Note that it doesn't depend on the choice of the representative $X$, since a finite change in $X$ will change $f[X]$ only finitely.
It remains to check that $g$ is an isomorphism, which is not difficult, but I'll have to leave that to the reader (or to an editor or commenter) because I have to return from MO to real life in a couple of minutes.

A:

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 $[a, b]$.

