Recognizing linear orders embeddable in $\Bbb R^2$ ordered lexicographically Let $(C, \leq_C)$ be any chain: binary relation $\leq_C$ on $C$ is reflexive, antisymmetric, transitive, and total.
Give $\mathbb{R}^2$ the lexicographic order: for all real numbers $a,b,c,d$, $(a,b) \leq_L (c,d)$ if and only if $a < c$ or ($a = c$ and $b \leq d$).
Question: what are necessary and sufficient conditions for there to be an order-isomorphism from $(C, \leq_C)$ to a subset of $(\mathbb{R}^2, \leq_L)$, i.e., a function $f: C \to \mathbb{R}^2$ with $x \leq_C y$ if and only if $f(x) \leq_L f(y)$?
Motivation: I know what characterizes order-isomorphisms to subsets of $\mathbb{R}$ with its usual order. Birkhoff's Lattice Theory (3rd ed, p. 200), for instance, shows that such an isomorphism exists if and only if $C$ has a countable "order-dense" subset $D$: a countable subset $D$ of $C$ such that for all $a, b \in C \setminus D$ with $a <_C b$ there is a $d \in D$ with $a <_C d <_C b$.
But I find lexicographic orders much more difficult to grasp. My initial guess was that there might be an appealing extension of the order-density-condition to answer my question (for $\mathbb{R}^2$ and perhaps with an inductive argument for $\mathbb{R}^n$), but I have not been able to find one.
 A: Corrected version. Let $\pi_1:\Bbb R_{\text{lex}}^2\to\Bbb R:\langle x,y\rangle\mapsto x$ be the projection to the first coordinate. Let $A\subseteq\Bbb R_{\text{lex}}^2$ be well-ordered or inversely well-ordered. Clearly $A\cap(\{x\}\times\Bbb R)$ is countable for each $x\in\Bbb R$, and $\pi_1[A]$ is also countable, so $A$ is countable. Thus, the first requirement is that $C$ have no uncountable subset well-ordered or inversely well-ordered by $\le_C$, and I will henceforth assume that $C$ has satisfies this requirement.
Define an equivalence relation $\sim$ on $C$ as follows: for $x,y\in C$ with $x\le_C y$ let $x\sim y$ and $y\sim x$ iff the interval $[x,y]$ in $C$ has a countable order-dense subset. Let $\mathscr{E}$ be the set of $\sim$-equivalence classes; clearly each $E\in\mathscr{E}$ is order-convex. Let $E\in\mathscr{E}$. Then either $E$ has a $\le_C$-last element, or there is an ordinary sequence $\langle y_n:n\in\Bbb N\rangle$ cofinal in $E$: if not, $E$ would contain an uncountable well-ordered set. Similarly, either $E$ has a $\le_C$-first element, or it has an ordinary co-initial sequence $\langle x_n:n\in\Bbb N\rangle$. If $E$ has a last element $y$, let $y_n=y$ for all $n\in\Bbb N$, and if $E$ has a first element $x$, let $x_n=x$ for all $n=\Bbb N$. Without loss of generality $x_0\le y_0$. Then $E=\bigcup_{n\in\Bbb N}[x_n,y_n]$, so $E$ has a countable order-dense subset and is therefore order-isomorphic to a subset of $\Bbb R$. Moreover, since the $\sim$-classes are convex, the order $\le_C$ induces a natural linear order $\preceq$ on $\mathscr{E}$.

Theorem. $\langle C,\le_C\rangle$ order-embeds in the lexicographically ordered plane iff $\langle\mathscr{E},\preceq\rangle$ has a countable order-dense subset.
Proof. For $x\in C$ let $E(x)$ be the $\sim$-equivalence class of $x$. Suppose first that $\langle\mathscr{E},\preceq\rangle$ is separable. Then there is an order-embedding $\varphi:\mathscr{E}\to\Bbb R$, and for each $E\in\mathscr{E}$ there is an order-embedding $\varphi_E:E\to\Bbb R$; clearly the map $$\psi:C\to\Bbb R^2_{\text{lex}}:x\mapsto\left\langle\varphi\big(E(x)\big),\varphi_{E(x)}(x)\right\rangle$$ is an order-embedding of $C$ into the lexicographically ordered plane.
Conversely, suppose that there is an order-embedding $\psi:C\to\Bbb R^2_{\text{lex}}$. For $\alpha\in\Bbb R$ let $$C_\alpha=\big\{x\in C:\psi(x)\in\{\alpha\}\times\Bbb R\big\}\;;$$ and let $A=\{\alpha\in\Bbb R:C_\alpha\ne\varnothing\}$. Clearly $x\sim y$ whenever there is an $\alpha\in A$ such that $x,y\in C_\alpha$, so $\mathscr{C}=\{C_\alpha:\alpha\in A\}$ is a refinement of $\mathscr{E}$ by order-convex subsets of $C$. And $\mathscr{C}$ in the order induced by $\le_C$ clearly has a countable order-dense subset, so $\mathscr{E}$ has as well. $\dashv$

