13
$\begingroup$

A linear order $<$ on a set $S$ is countably transitive iff, whenever $A$ and $B$ are order-isomorphic countable subsets of $S,$ there is an order-automorphism of $S$ which maps $A$ onto $B.$ Does such an order exist on some infinite set?

Background

One book defined $n$-transitive for finite $n>0$ to mean that whenever $A$ and $B$ are $n$-element subsets, there is an order-automorphism that maps $A$ onto $B.$ It is easy to show that $2$-transitive implies $n$-transitive for all finite $n>2.$ I thought of extending this. I easily found an infinite $S$ ,with no largest member, where, whenever $A$ and $B$ are subsets of $S$, each order-isomorphic to $\omega$, there is an order-aut of $S$ that maps $A$ onto $B$ . Then I tried to extend it to this question. A solution must have the following properties : It is order-dense in itself, as no endpoints, has no $(\omega,\omega^*)$ gaps, and every countable subset is bounded and closed in the order topology.

$\endgroup$
6
  • 3
    $\begingroup$ @TheoBendit The reals definitely do not have this property. Consider $A$ as the rational numbers, and $B$ as the rational numbers in $[0,1]$. They are order-isomorphic, but any order-preserving function mapping $A$ onto $B$ must have image within $[0,1]$. $\endgroup$
    – user231101
    Aug 2, 2015 at 5:27
  • 1
    $\begingroup$ How about the long line, in which every countable set is bounded? $\endgroup$
    – Chris Culter
    Aug 2, 2015 at 5:32
  • 1
    $\begingroup$ Note that my former comment should use $(0,1)$ instead of $[0,1]$: to be order-isomorphic to the rationals, you can't have endpoints. $\endgroup$
    – user231101
    Aug 2, 2015 at 5:33
  • 4
    $\begingroup$ @ChrisCulter The long line is not an example. Consider subsets of the form $L\cup R$ where $L$ is a strictly increasing sequence of points, and $R$ a strictly decreasing sequence, and every point in $L$ is less than any point in $R$. All such subsets are order isomorphic, but in some the two parts $L$ and $R$ have two distinct limit points, and in some the limit points coincide. I think possibly there is no simpler example than the ones in hot_queen's answer. $\endgroup$
    – Colin McLarty
    Aug 2, 2015 at 14:41
  • 1
    $\begingroup$ Assuming CH, two examples are: modest mouse, ultrapowers of rationals, $2^{<\omega_1}$. (ed ajf) $\endgroup$
    – hot_queen
    Aug 4, 2015 at 23:28

1 Answer 1

Reset to default
10
$\begingroup$

Yes. Every infinite structure has a strongly $\omega_1$-homogeneous elementary extension. So you can start with the rationals and find an $\omega_1$-homogeneous elementary extension $(L, <)$ which is countably transitive being $\omega_1$-homogeneous. You can find a construction here. If you also assume CH, then there is a saturated DLO without end points of size $\omega_1$ (which is clearly strongly $\omega_1$-homogeneous).

$\endgroup$
3
  • $\begingroup$ I don't know enough of that subject to follow it,but thanks for the answer. $\endgroup$
    – DanielWainfleet
    Aug 3, 2015 at 1:35
  • 1
    $\begingroup$ If you read the contents of the link you'd see that this is really simple. I just didn't want to type much here. $\endgroup$
    – hot_queen
    Aug 4, 2015 at 23:30
  • 1
    $\begingroup$ @user254665 $\omega_1$ says this concerns countable substructures. "Homogeneity" says: substructures that look like each other, have identical relations to the whole. E.g. for linear order, if one strictly increasing sequence has a limit point, then all must. Notice we can embed any linear order $P$ in a discrete linear order, $P\times\mathbb{Z}$ with lexicographic order. No strictly increasing infinite sequence has a limit point in $P\times \mathbb{Z}$, so $P\times \mathbb{Z}$ achieves one bit of $\omega_1$-homogeneity. Full $\omega_1$-homogeneity is a much stronger demand. $\endgroup$
    – Colin McLarty
    Aug 5, 2015 at 6:30

You must log in to answer this question.

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