Smallest ordinal $\mu$ not embeddable in ${\cal P}(\omega)$ The motivation for this question is the startling fact that there is an order-preserving injective map (embedding) from $\mathbb{R}$ into ${\cal P}(\omega)$. (Think Dedekind cuts.) I am wondering how "large" a linear order can become and still be embeddable in ${\cal P}(\omega)$. To be able to formalize this, I would like to focus on well-orders.
Question. What is the smallest ordinal $\mu$ such that there is no injective order-preserving map $\varphi:\mu\to{\cal P}(\omega)$?
 A: This is just $\omega_1$. The naive argument ("pick the least new thing") that no uncountable linear order embeds into $\mathcal{P}(\omega)$ actually establishes that no uncountable well-order embeds into $\mathcal{P}(\omega)$: if $f$ is an injection of an ordinal $\theta$ into $\mathcal{P}(\omega)$, the map $$\hat{f}:\theta\rightarrow \omega: \alpha\mapsto\min(f(\alpha+1)\setminus f(\alpha))$$ is an injection of $\theta$ into $\omega$. This doesn't contradict the injectibility of $\mathbb{R}$, since there is no injection from $\omega_1$ into $\mathbb{R}$ either.
(Interestingly, if we order $\mathcal{P}(\omega)$ by mod-finite containment instead of genuine containment, we can indeed inject $\omega_1$.)

EDIT: As bof pointed out below, we can improve the above drastically: there is an order preserving (but not reflecting of course) injection $\mathcal{P}(\omega)\rightarrow\mathbb{R}$, so the linear orders embeddable in $\mathcal{P}(\omega)$ are exactly those embeddable in $\mathbb{R}$. Of course that still leaves plenty of open questions (e.g. are any two $\aleph_1$-dense suborders of $\mathbb{R}$ order-isomorphic?) but I think it lets us stop thinking about $\mathcal{P}(\omega)$ as such.
