1
$\begingroup$

A nice theorem of Dushnik and Miller (from 1940) states that if $(\Omega,\leq)$ is a countable total order, then either

  • there is an element $\omega \in \Omega$ such that $(\Omega \setminus \{\omega\},\leq)$ and $(\Omega,\leq)$ are order isomorphic, or
  • there exists an embedding $(\mathbb{Q},\leq_u) \hookrightarrow (\Omega,\leq)$ (with $\leq_u$ the usual order on the rationals).

My question: is there a similar result for uncountable total orders ?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

I may be missing something, but it seems to me that the exact same statement holds for all infinite total orders.

Indeed, if $\mathbb Q$ does not embed in $\Omega$, then $\Omega$ is scattered, hence it can be constructed from the singleton order by repeatedly using well-ordered and inverse well-ordered sums. Then we can prove by induction on this construction that once the order is infinite, it satisfies the first clause: if $\Omega=\sum_{\alpha<\gamma}\Omega_\alpha$ is infinite, then either some $\Omega_\alpha$ is already infinite, or not. In the first case, $\Omega_\alpha\simeq\Omega_\alpha\smallsetminus\{x\}$ for some $x\in\Omega_\alpha$ by the induction hypothesis, hence $\Omega\simeq\Omega\smallsetminus\{x\}$ as well. In the second case, $\Omega$ itself is an infinite well order, hence it is isomorphic to itself minus its first element. The argument for inverse well ordered sums is completely analogous.

As YCor pointed out, this argument actually shows a bit more: if $\Omega$ is any infinite total order, then

  • $\mathbb Q$ embeds in $\Omega$, or

  • $\Omega$ has a convex subset of type $\omega$ or $\omega^*$.

BTW, I had a look at the Dushnik and Miller paper (Concerning similarity transformations of linearly ordered sets, Bull. AMS 46 (1940), no. 4, pp. 322–326) to see where did countability come from. It turns out they do not state the result formulated in the question at all. Rather, it only implicitly follows from their proof of Theorem 1, which actually states that every countable infinite total order is isomorphic to a proper subset of itself. This statement does not hold for uncountable orders (the paper includes a counterexample).

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ If I'm correct, essentially the same argument proves that every infinite scattered total order has a convex subset isomorphic to either $\omega$ or $-\omega$ (which implies the result). $\endgroup$
    – YCor
    Commented Apr 5, 2019 at 13:19
  • $\begingroup$ Yes, you are right. $\endgroup$
    – Emil Jeřábek
    Commented Apr 5, 2019 at 13:22
  • 1
    $\begingroup$ In fact, the proof of Theorem 1 in Dushnik and Miller can be factored into two parts: (1) The proof of the theorem stated by @THC in the question, but for an arbitrary linear order $\Omega$, not using countability [here one has to modify the proof slightly with the observation that $\mathbb{Q}$ embeds in any dense linear order by a "forth" argument], and (2) the observation that if $\Omega$ is countable and $\mathbb{Q}$ embeds in $\Omega$, then $\Omega$ embeds as a proper suborder of itself. $\endgroup$
    – Alex Kruckman
    Commented Apr 5, 2019 at 14:57

You must log in to answer this question.

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