Can infinite subsets of the reals be defined by their order alone? This is a follow up to a previous question. Consider the structure $(\mathbb{R}, <)$. We adjoin to it a subset $S$ of $\mathbb{R}$. Is there a first-order formula $F$ in the expanded langauge, such that $F$ is true precisely when $S$ is an infinite subset of $\mathbb{R}$? If not, can anyone prove it? This is different from the previous question because I am now restricting the language from $(+,*,0,1,<)$ to just $<$.
 A: A subset $S\subseteq \mathbb{R}$ is infinite if and only if it is unbounded or has a limit point in $\mathbb{R}$. 
Why? It is clear that a finite set is bounded and has no limit points. Conversely, supposes $S$ is infinite. If $S$ is unbounded, we're done. If it is bounded, then it is contained in some closed interval $[a,b]$. Since $[a,b]$ is compact, $S$ has a limit point in this interval.
So $S$ is infinite if and only if $(\mathbb{R};<,S)$ satisfies the following sentence: $$(\forall x \exists y\, (S(y)\land x < y))\lor (\forall x\exists y\, (S(y)\land y < x))\lor (\exists x\forall w\forall z\, ((w < x \land x < z) \rightarrow \exists y\, (S(y)\land \lnot y = x\land w < y \land y < z))).$$

Note that the fact that this works is really something special about the structure $(\mathbb{R};<)$, which doesn't generalize to other models of its complete theory (the theory of dense linear orders without endpoints). 
For example, if $R$ is a proper elementary extension of $\mathbb{R}$, then $\mathbb{Z}\subseteq R$ is bounded (since any such extension must contain elements which are larger than any integer) and infinite but has no limit point. On the other hand, $(\mathbb{Q};<)$ is an elementary substructure of $(\mathbb{R};<)$, and taking $S$ to be an infinite sequence of rational numbers that approximates $\pi$ from below, $S$ is infinite and bounded but has no limit point. 
