Hausdorff theorem for any linear orders I read this question:
Any good decomposition theorems for total orders?
and the answers. I like very much the Hausdorff theorem for scattered linear order. I repeat it here :
Theorem (Hausdorff). A linear order is scattered iff its order type is in some $S_\alpha$ ($\alpha$ an ordinal), where:


*

*$S_0=\{0,1\}$ 

*For $\alpha>0$, $S_\alpha$ is the smallest class obtained as follows: If $\gamma$ is an ordinal, $I$ is $\gamma$, $\gamma^*$, or $\gamma^*$+$\gamma$, and for each $i \in I$ the linear order $L_i$ is in $\bigcup_{\beta < \alpha} S_\beta$, then $\sum_{i \in I} L_i \in S_\alpha$.
I was wondering whether a similar characterization is possible for any Linear order (say countable to make it simpler), by adding for example $\eta$ (The order-type of the rationals) to $S_0$.
 A: It’s actually sufficient to allow a dense order only at the final stage.
Let $\langle X,\le\rangle$ be a linear order. I’ll quote a construction from an old paper of mine:

Suppose that $R$ is a convex equivalence relation on $X$, i.e., an equivalence relation with convex equivalence classes. For $x\in X$, let $R(x)=\{y\in x:x\mathbin{R}y\}$; since $R$ is convex, $X/R=\{R(x):x\in X\}$ is the image of $X$ under an order-homomorphism whereby if $R(x)\ne R(y)$, then $R(x)\le R(y)$ iff $x\le y$. We define from $R$ a new equivalence relation, $R^*$, on $X$ as follows. If $x,y\in X$ with $x\le y$, put $x\mathbin{R^*}y$ iff there is a finite family $\{x_0,\ldots,x_n\}\subseteq X$ such that $x=x_0\le x_1\le\ldots\le x_n=y$, $[x_i,x_{i+1}]/R$ is a well-order if $i<n$ is even, and $[x_i,x_{i+1}]/R$ is an inverted well-order if $i<n$ is odd; and if $y<x$, put $x\mathbin{R^*}y$ iff $y\mathbin{R^*}x$. Clearly $R\subseteq R^*$, $R^*$ is a convex equivalence relation on $X$.
Now let $R_0$ be the identity relation on $X$, and, for each ordinal $\alpha>0$, let $R_\alpha=\left(\bigcup\{R_\xi^*:\xi<\alpha\}\right)^*$; then each $R_\alpha$ is a convex equivalence relation on $X$, and $R_\alpha\subseteq R_\beta$ whenever $\alpha<\beta$. Define $r(X)=\inf\{\alpha:R_\alpha=R_{\alpha+1}\}$; clearly $r(X)$ exists, since it must in fact be less than $|X|^+$. Note also that if $\alpha=r(X)$, then either $|X/R_\alpha|=1$, (i.e., $R_\alpha=X\times X$), or $X/R_\alpha$ is a dense linear order; and that $X$ is scattered iff $|X/R_\alpha|=1$.

Suppose that $X/R_\alpha$ is a dense linear order, and let $x\in X$. Then $R_\alpha(x)$ is a convex subset of $X$, and $|R_\alpha(x)/R_\alpha|=1$, so $R_\alpha(x)$ is scattered, and $X$ is therefore a densely ordered sum of scattered orders. In particular, if $|X|=\omega$, and $X$ is not scattered, then $X/R_\alpha$ must be one of the four countable dense order types, $\eta$, $\eta+1$, $1+\eta$, or $1+\eta+1$.
One can extract more information from the foregoing construction, but there is a simpler way to reach the same conclusion. Let $\mathscr{D}$ be the set of densely ordered subsets of $X$. If $\mathscr{D}=\varnothing$, $X$ is of course scattered. Otherwise we can use Zorn’s lemma to get a maximal densely ordered subset $D$ of $X$. For each $x\in D$ let $R(x)=\bigcap\{(u,v):u,v\in D\text{ and }u<x<v\}$; each $R(x)$ is a scattered, convex subset of $X$, and $X=\sum_{x\in D}R(x)$ is a densely ordered sum of these scattered orders.
