An algebraically generated set of linear orders

Notation: Let $L_1,L_2,...$ be linearly ordered sets.

• $L_{i}^{-1}$ denotes the reverse linear order of $L_{i}$,
• $L_1+L_2$ denotes the sum of linear orders, i.e. the disjoint union $L_1\cup L_2$ with its natural linear order,
• $\sum_{n\in\omega}L_n$ denotes the sum of linear orders indexed by the natural numbers, i.e. the disjoint union $\bigcup_{n\in\omega}L_n$ with its natural linear order.

Construction: Let $S$ be the smallest class of linear orders with the following properties:

• $S$ contains the one-point order $L=\{\ast\}$,
• If $L\in S$, then $L^{-1}\in S$,
• If $L_1,L_2\in S$, then $L_1+L_2\in S$
• If $\{L_n|n\in \omega\}\subseteq S$, then $\sum_{n\in\omega}L_n\in S$.

Restricting to isomorphism classes, $S$ becomes something like a free semigroup of linear orders with involution and infinitary product generated by the one-point order. Using transfinite induction, it is easy to show that every element of $S$ is a countable linear order with countably many Dedekind cuts. However, it is not clear to me that this characterizes the elements of $S$ entirely.

Question: Is there a nice characterization of the linear orders in $S$, perhaps in terms of their dedekind cuts?

• I suspect that $S$ does consist of exactly the countable scattered orders, via induction on Hausdorff rank (see math.berkeley.edu/~antonio/slides/montrealh.pdf), but I don't immediately see the details. – Noah Schweber Aug 11 '16 at 16:47
• – Joel David Hamkins Aug 11 '16 at 16:52
• Hausdorff proved that the scattered orders are what you get from the one point orders by closing under well-ordered and reverse-well-ordered sums. Although Jeremy has only sums indexed by $\omega$, you can prove by induction that his class is closed under countable-well-ordered indexed sums (and reversals). – Joel David Hamkins Aug 11 '16 at 17:02
• Countable scattered is the same as countable with countably many Dedekind cuts, so if @NoahSchweber´s suspicion is right, OP´s original characterization works. – Ramiro de la Vega Aug 11 '16 at 17:17
• Rosenstein's book "Linear Orderings" proves Hausdorff's theorem, and also gives his 1908 paper Grundzüge einer Theorie der geordneten Mengen (Math. Ann. 65) as a reference. – Goldstern Aug 11 '16 at 20:01

What I describe as $S$ in my question, Rosenstein calls the class of very discrete linear orderings.
Theorem 5.24: (Hausdoff) A countable linear ordering $L$ is very discrete if and only if it is scattered.