4
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ 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. $\endgroup$
    – Noah Schweber
    Commented Aug 11, 2016 at 16:47
  • $\begingroup$ See also en.wikipedia.org/wiki/Scattered_order $\endgroup$
    – Joel David Hamkins
    Commented Aug 11, 2016 at 16:52
  • 3
    $\begingroup$ 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). $\endgroup$
    – Joel David Hamkins
    Commented Aug 11, 2016 at 17:02
  • 2
    $\begingroup$ Countable scattered is the same as countable with countably many Dedekind cuts, so if @NoahSchweber´s suspicion is right, OP´s original characterization works. $\endgroup$
    – Ramiro de la Vega
    Commented Aug 11, 2016 at 17:17
  • 2
    $\begingroup$ 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. $\endgroup$
    – Goldstern
    Commented Aug 11, 2016 at 20:01

1 Answer 1

Reset to default
1
$\begingroup$

Thanks to Goldstern for the reference. Very helpful! I finally found time to look through a copy of "Linear Orderings" by Joseph Rosenstein, which is a very nice book. It turns out that Rosenstein addresses precisely what I'm asking in Chapter 5 using iterations of the "condensations" developed in Chapter 4. I was excited to see that he works through the countable case in great detail.

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.

The more general version of Hausdorff's Theorem appears as Theorem 5.26. The proof is only sketched out since it is similar to proof in the countable case.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What is the more general version of Hausdorff's Theorem which appears as Theorem 5.26? $\endgroup$
    – bof
    Commented Oct 28, 2016 at 0:18
  • 1
    $\begingroup$ I believe there is also a discussion of Hausdorff's theorem, and some kind of generalization to higher cardinals, in R. Laver: On Fraïssé's order type conjecture, Ann. of Math. (2), 93 (1971), 89–111. $\endgroup$
    – bof
    Commented Oct 28, 2016 at 0:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .