Incompressible countable Total-ordering implies well-ordering i

Question

A totally-ordered set (S,<) is incompressible if $(S,<) \cong (T,<)$ and $S \supseteq T$ implies $S = T$.

Is it true that if $S$ is incompressible countable and totally-ordered then $S$ is well-ordered?

Answer 1

Say that a linear order $\langle X,\le\rangle$ is cushioned iff every order-preserving injection $f:X\to X$ has the property that $x\le f(x)$ for each $x\in X$. Let $\langle X,\le\rangle$ be an incompressible linear order. If $X$ is not cushioned, there are an order-preserving injection $f:X\to X$ and an $a\in X$ such that $f(a)<a$. Define

$$g:X\to X:x\mapsto\begin{cases} f(x),&\text{if }x\le a\\ x,&\text{if }x>a\;; \end{cases}$$

then $g$ is an order-isomorphism from $X$ to $g[X]$. But $a\notin g[X]$, so $g[X]\subsetneqq X$, contradicting the incompressibility of $X$. Thus, $X$ must be cushioned. It follows from Theorem $3.0$ of Brian M. Scott, A characterization of well-orders, Fundamenta Mathematicæ, $\mathbf{111 (1)}$, $71$-$76$, MR0607921 (82i:06001), that $X$ is a well-order.