Order-automorphisms of countable total orders

Question

Background. These are the last two questions of a problem (I've already proved that $|\operatorname{Aut}(\mathbb Q,<)| = |\operatorname{Aut}(\mathbb R,<)| = 2^{\aleph_0}$ and that if $|\operatorname{Aut}(S,<_S)| < \aleph_0 \implies \operatorname{Aut}(S,<_S) = \{\operatorname{id}_S\}$, where $(S,<_S)$ is totally ordered, if it can be useful). We're working in $\mathsf{ZFC}$.

Questions.

1. Is there a total order $(S,<_S)$ with the following proprieties: $|S| = \aleph_0$, $S$ has a subset (ordered with the restriction of $<_S$) isomorphic to $\mathbb Q$ and $\operatorname{Aut}(S,<_S) = \{\operatorname{id}_S\}$?
2. Given a countable total order $(S,<_S)$ prove that if $|\operatorname{Aut}(S,<_S)|>\aleph_0$, then $|\operatorname{Aut}(S,<_S)| = 2^{\aleph_0}$.

My attempt.

1. I think the answer is negative, let's take a strictly increasing function $f : \mathbb Q \to S$, I already know that $|\operatorname{Aut}(\mathbb Q,<)| = 2^{\aleph_0}$, so given $g \in\operatorname{Aut}(\mathbb Q,<)$, I can define:

$$h_g : S \to S : x \mapsto \begin{cases} f \circ g \circ f^{-1} (x) &\text{if $x \in f[\mathbb Q]$} \\\\ x &\text{otherwise} \end{cases}$$

I'd like this to be an automorphism of $S$ in order to have $\forall g \in \operatorname{Aut}(\mathbb Q,<)$ $h_g \in \operatorname{Aut}(S,<_S)$ and so it'll contradict the fact that the identity is the only automorphism of $(S,<_S)$. The problem is that if I take $x \in f[\mathbb Q]$ and $y \in S \setminus f[\mathbb Q]$, where $x <_S y$, I don't know it's true that $f \circ g \circ f^{-1} (x) <_S y$. Is this the correct way? Or should I use something else (for example the fact that $(S,<_S)$ is itself isomorphic to a subset of $\mathbb Q$, as a consequence of Cantor's isomorphism theorem)?

2. Obviously $\operatorname{Aut}(S,<_S) \subseteq S^S \implies |\operatorname{Aut}(S,<_S)| \leq 2^{\aleph_0}$, so I just need to find $2^{\aleph_0}$ automorphisms. The only idea that I had is trying to "encode" an automorphism of $S$ composing other automorphisms, but I don't know if it actually produces enough distinct functions/and how write this down properly.

Answer 1

1. Yes. Take an injection $f:\mathbb Q\to\mathbb N=\{1,2,3,\dots\}$ and let $S=\{(x,y)\in\mathbb Q\times\mathbb N:y\le f(x)\}$ ordered lexicographically.

2. Give $S$ the discrete topology and give $S^S$, the set of all maps from $S$ to $S$, the Tychonoff product topology. If $|S|=\aleph_0$ then $S^S$ is a Polish space (separable and completely metrizable) homeomorphic to the space $\mathbb R\setminus\mathbb Q$ of irrational numbers. Observe that $\operatorname{Aut}(S,\lt_S)$ is a $G_\delta$ subset of $S^S$. Now use the fact that an uncountable $G_\delta$ subset of a Polish space contains a nonempty perfect set and therefore has cardinality $2^{\aleph_0}$.

To see that $\operatorname{Aut}(S,\lt_S)$ is a $G_\delta$ subset of $S^S$, observe that $$\operatorname{Aut}(S,\lt_S)=\left(\bigcap_{x\in S}U_x\right)\cap\left(\bigcap_{x,y\in S,\,x\lt_S y}V_{x,y}\right)$$ where $U_x=\{f\in S^S:x\in f[S]\}$ and $V_{x,y}=\{f\in S^S:f(x)\lt_S f(y)\}$ are open sets.

Answer 2

This is a supplement to bof's great answer (which you should accept over mine), with a direct construction to show that there are continuum-many automorphisms. It's a nontrivial argument, and it's closely related to the proof of the theorem bof quotes. I have made some effort to frame it in an order-theoretical sort of way. It's not easy, but I think it's possible to come up with an argument like this yourself given some time, if you frame the problem in the right way. I don't know if there is an easier argument - as far as I know this fact is not totally obvious. The only proof I know besides this one is the one bof gives.

The key fact I'll use is this:

If $(S, <)$ is a countable total order and $\mathcal A$ is an uncountable set of automorphisms of $S$, then there is some $x$ and some distinct $y_1, y_2$ in $S$ such that both $\{f \in \mathcal A : f(x) = y_1\}$ and $\{f \in \mathcal A : f(x) = y_2\}$ are uncountable.

This fact will enable us to make infinitely many binary choices when constructing an automorphism. Some care is required to make sure that we end up with an automorphism, and that different sequences of choices end up giving different automorphisms. But that's the core idea. I will prove the fact and finish the construction inside a spoiler tag, in case you want to have a go yourself first. Please don't be alarmed by the length - I have included some commentary and some general flowery language, adding to the length.

Proof of the Key Fact. Suppose not. This means that for each $x \in S$, there is a unique $y_x \in S$ such that $\{f \in \mathcal A : f(x) = y_x\}$ is uncountable. (There must be at least one such $y$, otherwise $\mathcal A = \bigcup_y \{f \in \mathcal A : f(x) = y\}$ would be countable). I'll show that this implies $\mathcal A \setminus \{x \mapsto y_x\}$ is countable. Indeed, $\mathcal A \setminus \{x \mapsto y_x\} = \bigcup_x \bigcup_{y \ne y_x} \{g \in \mathcal A : g(x) = y\}$ is a countable union of countable sets, so the proof is completed.

Now we will use the key fact to build some automorphisms. The idea is that we'll inductively construct lots of finite partial functions (meaning partial functions whose domain is finite), using the key fact to keep a "dangling hypothesis" from which we can continue to profit. At each stage, we'll get two new finite partial functions for each finite partial function from the previous stage. You can think of this as a sort of nondeterministic process, where you flip a coin to determine what choice you make at each stage.

First, I'll introduce a bit of notation. If $p: S \to S$ is a finite partial function, then write $\mathcal A_p$ for the set of all automorphisms of $S$ that extend $p$. Now for each finite binary sequence $\mathbf b \in \{0, 1\}^n$, we will construct a finite partial function $p_{\mathbf b}$ such that $\mathcal A_{p_{\mathbf b}}$ is uncountable. I'll use $++$ to denote concatenation of sequences.

A naïve approach might be to simply say "having defined $p_{\mathbf b}$, apply the key fact to $\mathcal A_{p_{\mathbf b}}$ to obtain $x, y_1, y_2$. Then define $p_{\mathbf b ++ (0)}(x) = y_1$ and $p_{\mathbf b ++ (1)}(x) = y_2$". However, this won't quite work, as while it will guarantee they are each injective and order-preserving, it doesn't guarantee that in the limit they will be surjective and total. We can fix this with a bit more book-keeping, as in a typical "back-and-forth" argument when dealing with countable total orders. To help with this, assume henceforth that the underlying set is $S = \Bbb N$.

For book-keeping we will use these facts:

1. If $q$ is a finite partial function such that $\mathcal A_q$ is uncountable, then if $y$ is not in the image of $q$, we can extend $q$ to a finite partial function $q'$ whose image does contain $y$, such that $\mathcal A_{q'}$ is still uncountable. (This is because $\mathcal A_q$ is a countable union $\bigcup_x \{f \in \mathcal A_q : f(x) = y\}$).
2. If $q$ is a finite partial function such that $\mathcal A_q$ is uncountable, then if $x$ is not in the domain of $q$, we can extend $q$ to a finite partial function $q'$ whose domain does contain $x$, such that $\mathcal A_{q'}$ is still uncountable. (This is because $\mathcal A_q$ is a countable union $\bigcup_y \{f \in \mathcal A_q : f(x) = y\}$).

So, off we go. First define $p_{()}$ to be the empty function. Now suppose $\mathbf b$ is of length $n$ and we have constructed $p_{\mathbf b}$. First, apply both of the book-keeping facts in turn to obtain an extension $p_{\mathbf b}'$ such that $n$ is in both the domain and image. Now apply the key fact to $\mathcal A_{p_{\mathbf b}'}$ to obtain $x, y_1, y_2$, and define extensions $p_{\mathbf b ++ (0)}, p_{\mathbf b ++ (1)}$ of $p_{\mathbf b}'$ by $p_{\mathbf b ++ (0)}(x) = y_1$ and $p_{\mathbf b ++ (1)}(x) = y_2$.

When all this is said and done, we obtain a coherent system of finite partial functions $p_{\mathbf b}$. If $b \in \{0, 1\}^{\Bbb N}$ is an infinite binary sequence, then we can define $f_b(n) = p_{(b_0, \dotsc, b_n)}(n)$, since we guaranteed that $n$ would be in the domain after the $n$th stage. The function $f_b$ is surjective, since we also guaranteed that $n$ would be in the image after the $n$th stage, and it's injective and order-preserving, because each $p_{\mathbf b}$ extends to an automorphism. Hence $f_b$ itself is an automorphism.

Lastly, if $b, b'$ are distinct sequences, let $i$ be the least index such that $b_i \ne b_i'$. Then for the $x$ chosen at the $i$th stage (when $\mathbf b = (b_0, \dotsc, b_{i - 1})$), we must have $p_{(b_0, \dotsc, b_i)}(x) \ne p_{(b_0', \dotsc, b_i')}(x)$. So the $f_b$ are all distinct.