There are concise and elegant characterisations of the real line as a topological space and as an ordered space in the literature. I am interested in the harder case of characterising subsets of the reals in this manner. There are satisfactory answers to the topological version (e.g., de Groot, Mary Ellen Rudin) which are, as to be expected, more complicated and inticrate in proof than for the whole space. I recall reading a solution for the corresponding result in the category of ordered spaces but the standard search methods have failed to locate it. Can anybody on this site assist me with a reference?



Characterising subsets of the reals as ordered spaces
The suggestion in the comments that a linear order embeds into
$\mathbb{R}$ just in case it has a countable dense set is not
quite true. For example, let $2\times\mathbb{R}$ be the *doubled real line* ($\mathbb{R}$ copies of $2$), the order arising from the reals by replacing each real number with two copies, a
lower one and an upper one, adjacent to each other. This order has a countable dense set (dense in the order topology) — that is, it is separable in the order topology — since every **nonempty** open
interval $(a,b)$ continues to contain (two duplicates of) a rational number. But $2\times\mathbb{R}$
cannot embed into $\mathbb{R}$ because any dense order extending
$2\times\mathbb{R}$ will not be separable.


Meanwhile, there is a nearby characterization that does succeed:


**Theorem.** The following are equivalent for a linear order
$\langle L,<\rangle$.


1. $\langle L,<\rangle$ embeds into the real line
$\langle\mathbb{R},<\rangle$.
2. $\langle L,<\rangle$ is separable and has at most countably
many closed intervals $[a,b]$ with $a<b$ and $(a,b)=\emptyset$.


**Proof.** ($1\to 2$) This is clear, since once you have embedded
$L$ into $\mathbb{R}$, then you can pick an element from $L$ from
each rational interval having any elements from the image of $L$, and this will be dense in $L$, and also $L$ can
have at most countably many discrete intervals $[a,b]$, because
each will have a distinct rational in $\mathbb{R}$.


($2\to 1$) Here, you use the suggestion from the comments. Let
$Q\subset L$ be a countable dense set that also includes all the
endpoints $a$ and $b$ from any discrete closed interval $[a,b]$ in
$L$. This is a countable linear order, which we may map into the
rationals $\mathbb{Q}$ using the usual back-and-forth (but just
forth) construction of Cantor. Now, we may extend this to all of
$L$, since any element of $L$ is the least upper bound in $L$ of
the elements of $Q$ below it. **QED**


Meanwhile, the separable linear orders are characterized as those that embed into $2\times\mathbb{R}$. We just map the countable dense set into the rationals, using always the lower rational number, say. And then we can extend this to the whole order since every element of $L$ is either the LUB or the GLB of a subset of that countable dense set, and so we can map into $2\times\mathbb{R}$ accordingly. 


**Theorem.** The following are equivalent for any linear order$\langle L,<\rangle$.


1. $\langle L,<\rangle$ embeds into the doubled real line
$2\times\mathbb{R}$.
2. $\langle L,<\rangle$ has a countable dense set.


