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](https://en.wikipedia.org/wiki/Dedekind_cut).) 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 $\varphi:\mu\to{\cal P}(\omega)$?



Smallest ordinal $\mu$ not embeddable in ${\cal P}(\omega)$
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 $\omega\_1$.)




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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.


