Let $(S,<)$ be a totaly ordered set under the strict order relation $<$. Suppose that, for any $a,b\in S$, if $a<b$, then there exists $c\in S$ such that $a<c<b$. We also assume that $S$ is countable (countably infinite, for clarification). Is it true that $S$ is order-isomorphic to one of the following totally ordered sets: $\mathbb{Q}$, $\mathbb{Q}\cup\{-\infty\}$, $\mathbb{Q}\cup\{+\infty\}$, and $\mathbb{Q}\cup\{-\infty,+\infty\}$ (equipped with their natural total orders, of course)? If not, what is a counterexample?
An order isomorphism from a partially ordered set $(A,<)$ to another partially ordered set $(B,<)$ is a bijection $f:A\to B$ such that, for $x,y\in A$ with $x< y$, we have $f(x)<f(y)$. Two partially ordered sets are order-isomorphic if there exists an order isomorphism from one to the other. This question is inspired by Order preserving bijection from $\mathbb{Q}$ to $\mathbb{Q}\backslash\lbrace{0}\rbrace$ (I just saw that somebody had posted an answer to my question in this link).