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Let $(S,\leq)$ be a countable linear order with the following property (H):

for all positive integers $n$, and all increasing sequences $s_1 < s_2 < \ldots < s_n$ and $t_1 < t_2 < \ldots < t_n$, there is an order automorphism $\alpha$ of $S$ such that $\alpha(s_i) = t_i$ for all $i = 1, 2, \ldots, n$.

Then by a result of Cantor, $(S,\leq)$ is order isomorphic to $(\mathbb{Q},\leq)$.

Now suppose that $S$ is not countable, but also having property (H). It seems to me that the order automorphism group of $(S,\leq)$ should share many properties with the order automorphism group of $(\mathbb{Q},\leq)$.

What are some of the important properties that they do not share ? What are the (main) differences ?

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2 Answers 2

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One difference is the number of orbits of increasing length-$\omega$ sequences. This isn't really a difference in the automorphism groups per se, but rather their actions, but I think it's interesting enough to note.


In $Q=(\mathbb{Q},<)$ there are exactly three kinds of increasing sequence (up to automorphism, in the sense that any two sequences of the same type can be swapped by an automorphism of $Q$). Namely, an increasing sequence $(a_i)_{i\in\omega}$ of rationals has exactly one of the following three properties, and if $(a_i)_{i\in\omega},(b_i)_{i\in\omega}$ have the same property then there is an automorphism $\alpha$ of $Q$ such that $\alpha(a_i)=b_i$ for all $i\in\omega$:

  • Unbounded: the set $\{a_i:i\in\omega\}$ has no upper bound.

  • Bounded but no least upper bound: the sequence is bounded but whenever $b\in Q$ is such that $b>a_i$ for all $i$ there is some $c\in Q$ such that $c>a_i$ for all $i$ and $c<b$.

  • Has a least upper bound: there is some $b\in Q$ such that $b>a_i$ for all $i$ and for all $c<b$ there is some $i$ such that $c<a_i$.

For example, taking $a_i=i$ yields a sequence of the first type, taking $a_i=\pi-{1\over i+1}$ yields a sequence of the second type (the "$+1$" addresses the case $i=0$), and taking $a_i=1-{1\over i+1}$ yields a sequence of the third type.

By contrast, this can fail for uncountable orders with property (H). For example, $R=(\mathbb{R},<)$ has property $(H)$ but only two types of increasing $\omega$-sequences (every bounded sequence has a least upper bound).

  • As an aside, this is fundamentally impossible in the countable context: Vaught's never-two theorem says that no complete theory in a countable language can have exactly two countable models up to isomorphism. Gauging the number of models of a theory of a given cardinality is a major theme in model theory - for example, if $T$ is a complete theory in a countable language then the number of countably infinite models of $T$ is either finite, $\aleph_0$, $\aleph_1$, or $2^{\aleph_0}$. Every finite value other than $2$ is possible, as are $\aleph_0$ and $2^{\aleph_0}$; it is unknown whether (in case $\aleph_1\not=2^{\aleph_0}$) the case $\aleph_1$ is possible. It's also worth noting that there's a rephrasing of Vaught's conjecture that doesn't trivialize under the continuum hypothesis (see here).
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  • $\begingroup$ @ Noah Schweber : I don't think I understand your statement about $(\mathbb{Q},\leq)$; let $(a_i)$ be an unbounded increasing sequence in $\mathbb{Q}$ which is dense as a subset of $\mathbb{Q}$, and let $(b_i)$ be the sequence of integers. How can those be order isomorphic by an order automorphism of $(\mathbb{Q},\leq)$ ? $\endgroup$
    – Boccherini
    Oct 16, 2019 at 10:58
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    $\begingroup$ @Boccherini An increasing sequence can't be dense, and the set of integers can't be enumerated by an increasing sequence either... $\endgroup$ Oct 16, 2019 at 13:44
  • $\begingroup$ @AlexKruckman: oops, I am using the wrong definition for (increasing) sequence here, I guess -- is such a sequence by definition indexed over the positive integers ? $\endgroup$
    – Boccherini
    Oct 16, 2019 at 13:51
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    $\begingroup$ @Boccherini Noah was careful to specify that his sequences are indexed by $\omega$: the set of natural numbers. $\endgroup$ Oct 16, 2019 at 13:58
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    $\begingroup$ @Boccherini Here $\omega$ is the first infinite ordinal. But your confusion is understandable: Logicians often forget that other people like to use the notation $\mathbb{N}$. :) $\endgroup$ Oct 16, 2019 at 16:14
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We usually think of automorphism groups not just as groups, but as topological groups, under the topology of pointwise convergence. Then the automorphism group of any countable structure is a Polish group: it is separable and completely metrizable. On the other hand, if $M$ is an uncountable structure in which some point has an uncountable orbit (for example, if $M$ is an uncountable homogeneous linear order in your sense), then $\text{Aut}(M)$ has an uncountable family of disjoint open sets, so it is not separable.

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  • $\begingroup$ @ Alex Kruckman : suppose $(a,b)$ is an open interval in $\mathbb{Q}$; then in the order automorphism group, we can consider automorphisms fixing all points of $\mathbb{Q}$ outside $(a,b)$, and which move an arbitrary point $c \in (a,b)$ to an arbitrary point $c' > c$ which is also in $(a,b)$. I have the feeling that such constructions also should work for linear orders $(S,\leq)$ with $S$ not countable, and where (H) is satisfied. Is there a general way to "transfer" such properties to the uncountable case ? (As you remark, not all "non-obvious" group-theoretical properties transfer.) $\endgroup$
    – Boccherini
    Oct 16, 2019 at 14:33
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    $\begingroup$ @Boccherini I don't have a general transfer principle in mind, but it's easy enough to prove the specific claim you ask about in your comment. Suppose $(S,\leq)$ is a homogeneous linear order (in your sense), with $a<c<b$ and $a<c'<b$ in $S$. Then by (H), there is an automorphism $\sigma$ of $S$ which fixes $a$ and $b$ and moves $c$ to $c'$. Define $\sigma'\colon S\to S$ by $\sigma'(d) = \sigma(d)$ if $d\in (a,b)$ and $\sigma'(d) = d$ otherwise. Then $\sigma'$ is an automorphism of $S$ which fixes all points outside $(a,b)$ and moves $c$ to $c'$. $\endgroup$ Oct 16, 2019 at 16:19

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