A different ordering on ${\cal P}(\omega)$ For $A, B \subseteq \omega$ we set $A \leq_{\text{inj}} B$ if there is an injective and order-preserving map $f:\omega\to \omega$ , such that $f(A)$ is a down-set of $B$. It is easy to see that $\leq_{\text{inj}}$ is an ordering relation on ${\cal P}(\omega)$. (This is different from the lexicographic ordering discussed in another post.)
Let us compare the two posets $({\cal P}(\omega), \subseteq)$ and $({\cal P}(\omega), \leq_{\text{inj}})$: Are there surjective order-preserving maps between them, in either direction?
EDITED: Where down-set is printed above, I had the weaker (and confusing) condition $f(A)\subseteq B$ which doesn't make sense, as Andreas Blass pointed out.
 A: Yes there is a surjective order-preserving map from $\leq_{inj}$ to $\subseteq$, no in the reverse direction.
If you restrict the $\leq_{inj}$ order to the family of all infinite subsets of $\omega$ the result is isomorphic to the set of all functions from $\omega$ to $\omega$ with the coordinatewise order ($f \leq g$ if $f(n) \leq g(n)$ for all $n$). Namely, if $(a_n)$ is an infinite subset of $\omega$ given in increasing order, map it to the sequence $(a_{n+1} - a_n - 1)$. I.e. map an infinite set to the sequence of gaps.
It's easy to see that there is an order-preserving surjection from this space to $\mathcal{P}(\omega)$. (Identify the latter with the set of functions from $\omega$ to ${0,1}$.) Any finite subset corresponds to a function from an initial segment of $\omega$ to $\omega$; fill it out with a sequence of zeros and map it to the image of that sequence. That gives you one direction.
For the nonexistence of a map with the desired properties in the other direction, just note that the standard order had a greatest element (namely, $\omega$) but the inj order does not. So any map has to take $\omega$ somewhere and then nothing could consistently map to any greater element.
