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?


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.

  • $\begingroup$ Do the reals definitely not have this property? $\endgroup$ – Theo Bendit Aug 2 '15 at 5:24
  • 1
    $\begingroup$ @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 '15 at 5:27
  • 1
    $\begingroup$ How about the long line, in which every countable set is bounded? $\endgroup$ – Chris Culter Aug 2 '15 at 5:32
  • 2
    $\begingroup$ @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 '15 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 '15 at 23:28

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).

| cite | improve this answer | |
  • $\begingroup$ I don't know enough of that subject to follow it,but thanks for the answer. $\endgroup$ – DanielWainfleet Aug 3 '15 at 1:35
  • $\begingroup$ 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 '15 at 23:30
  • $\begingroup$ @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 '15 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.