As is well-known, ZFC proves the equipotency of $\mathbb{R}$ and $\mathcal{P}(\mathbb{Q}).$ Is there a nice characterization of those linearly ordered sets $L$ which, like $\mathbb{Q}$, have the property that their completion by Dedekind-cuts $\mathcal{C}(L)$ is equipotent with their powerset $\mathcal{P}(L)$?


**Edit.** At the risk of introducing a non-standard concept into a perfectly standard question, I am especially interested in the case where $L$ is unbounded and "locally homogeneous," in the sense that for all $x,y \in L$ and all $x',y' \in L,$ we have that if $x<y$ and $x'<y'$, then the interval $(x,y)$ is isomorphic to the interval $(x',y')$. (The purpose of this condition is just to cut out all the crazy mish-mash orders involving mixtures of scattered and dense linear orders with complete and incomplete segments all smashed together into some crazy pattern. See Arthur's answer [here](https://math.stackexchange.com/questions/732768/what-are-some-examples-of-homogeneous-linear-orders-other-than-mathbbr-a) for an example of a homogeneous order different from $\mathbb{R}$ and $\mathbb{Q}$).



Which linearly ordered sets have the property that their completion is equipotent with their powerset?
This is a complement to Joel's answer. For uncountable orders, I don't think there will be a nice characterization in terms of embedding a canonical suborder, at least without some assumption about cardinal arithmetic, even if one only considers orders that are locally homogeneous. I'll argue that it's consistent that there are locally homogeneous orders $L\_1, L\_2$ of size $\aleph\_1$ whose completions are of size $2^{\aleph\_1}$ but that have the property that no uncountable order $X$ embeds into both $L\_1$ and $L\_2$. 


First we need some preliminary facts. In general, if $L$ is uncountable, it falls into one of three categories:


1) $L$ embeds a copy of either $\omega\_1$ or $-\omega\_1$ (or both),


2) $L$ does not embed either $\omega\_1$ or $-\omega\_1$ but embeds an uncountable suborder of $\mathbb{R}$, 


3) $L$ does not embed $\omega\_1$, $-\omega\_1$, or any uncountable suborder of $\mathbb{R}$. 


The first is the broadest of the three categories, and I don't have much to say about it. The second and third have been widely studied, however. Lines of type 3 are called *Aronszajn lines*. They exist in ZFC, and it's relatively straightforward to show they necessarily have size $\aleph\_1$. It follows that since they do not embed either $\omega\_1$ or its reverse, that all of their gaps can be approached by countable sequences, and consequently their Dedekind completions always have size $[\omega\_1]^{\omega}=2^{\aleph\_0}$. Thus their completions will have the size of their powerset if and only if $2^{\aleph\_0} = 2^{\aleph\_1}$. 


Lines of type 2 in general can be of size larger than $\aleph\_1$. Yet it's possible to show, by a slightly more complicated argument, that their completions must also have size $2^{\aleph\_0}$. So again, they will have the desired property iff $2^{|L|} = 2^{\aleph\_0}$. 


Now, as to the original claim: assume MA + $2^{\aleph\_0}=\aleph\_2$ + ``all $\aleph\_1$-dense sets of reals are order-isomorphic." It's a result of Baumgartner that this is consistent relative to ZFC. Then $2^{\aleph\_0}=2^{\aleph\_1}=\aleph\_2$. Let $L\_1$ be any $\aleph\_1$-dense subset of $\mathbb{R}$. Then in particular, since every open subinterval of $L\_1$ will also be $\aleph\_1$-dense, $L\_1$ will be locally homogeneous. 


We let $L\_2$ be a particular type of Aronszajn line, called an *$\aleph\_1$-dense nonstationary Countryman line*. The actual properties of such lines are irrelevant for our purposes except that they are Aronszajn and, by a result of Todorcevic, are isomorphic under MA$\_{\aleph\_1}$ to their restriction to every open interval (and so locally homogeneous). 


Then $L\_1$ and $L\_2$ are as claimed. Both have size $\aleph\_1$, and their completions are both of size $2^{\aleph\_1}$. Further, there can be no uncountable order $X$ that embeds into both $L\_1$ and $L\_2$, since that would yield an uncountable suborder of $L\_2$ isomorphic to an uncountable suborder of $\mathbb{R}$, contradicting the definition of being Aronszajn. 


The relevant references are: 


Baumgartner, "All $\aleph\_1$-dense sets of reals can be isomorphic," Fund. Math. 79 (1973) 101-106; 


Todorcevic, Walks on ordinals and their characteristics, Vol. 263 of Progress in Mathematics, Birkhäuser (2007). 


