# Uncountable homogeneous linear orders

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 ?

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.

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

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).
• @ 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 $(\mathbb{Q},\leq)$ ? Oct 16, 2019 at 10:58
• @Boccherini An increasing sequence can't be dense, and the set of integers can't be enumerated by an increasing sequence either... Oct 16, 2019 at 13:44
• @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 ? Oct 16, 2019 at 13:51
• @Boccherini Noah was careful to specify that his sequences are indexed by $\omega$: the set of natural numbers. Oct 16, 2019 at 13:58
• @Boccherini Here $\omega$ is the first infinite ordinal. But your confusion is understandable: Logicians often forget that other people like to use the notation $\mathbb{N}$. :) Oct 16, 2019 at 16:14

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.

• @ 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.) Oct 16, 2019 at 14:33
• @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 $c'$. Oct 16, 2019 at 16:19