Characterising subsets of the reals as ordered spaces

Question

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?

Answer 1

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.

Answer 2

Just for the record, the result seems to be due to Isodore Fleischer ("Numerical representation of utility", Jour. Soc. Ind. Appl. Math, 9 (1961) 48-50). As the title indicates, it is useful in determining when a suitable ordering on a state space is induced by a numerical function (price, utility function, temperature, entropy) and unifies many such results in fields such as economics, thermodynamics, philosophy ... This (surprisingly late) reference is the earliest one that I can trace.