Automorphisms countable set

Question

I am stucked with the following exercise from an elementary set theory course. Any help, even just a small advice on how to start, would be appreciated.

Let $(S,<)$ be a total order and suppose $S$ to be a countable set. If the cardinality of the automorphisms of $(S,<)$ is more than countable, prove that then it is exactly $2^{\aleph_0}$. This is of course without assuming CH.

I edited the orginal post by writing the correct hypothesis, namely $\vert \operatorname{Aut}(S) \vert>\aleph_0$.

Answer 1

The edited post is correct, and is an instance of the "definable continuum hypothesis" theme common in descriptive set theory.

Fixing a bijection between the (underlying set of the) linear order $\mathcal{S}=(S;\le)$ in question and the natural numbers, the set of automorphisms of $\mathcal{S}$ can be reconstrued as a set of points in Baire space $\mathbb{N}^\mathbb{N}$. Being injective and order-preserving are each closed properties, and surjectivity is a $G_\delta$ property, so the set of (sequences corresponding to) automorphisms is $G_\delta$. Now use the fact that every uncountable $G_\delta$ set has size continuum.

In case you haven't proved that fact already, here's the general strategy:

Suppose we have open sets $U_i\subseteq\mathbb{N}^\mathbb{N}$ ($i\in\mathbb{N}$) such that $\bigcap_{i\in\mathbb{N}}U_i$ is uncountable. Recursively build a function $$\rho: 2^{<\mathbb{N}}\rightarrow\mathbb{N}^{<\mathbb{N}}$$ such that $(i)$ $\rho$ is prefix-order-preserving and injective, and $(ii)$ for every finite string $\sigma$, every infinite $f\succ\rho(\sigma)$ lies in $U_i$ for every $i<\vert\sigma\vert$. You'll use uncountability to ensure that this can actually be done, and then "applying $\rho$ to the infinite binary sequences" gives an explicit size-continuum subset of $\bigcap_{i\in\mathbb{N}}U_i$.

In fact a more advanced analogue of this ("unfolding" - see Kechris, Classical descriptive set theory, chapter II.21.B) applies to arbitrary analytic sets, but that goes a bit far afield.

Answer 2

This is false (as originally written) because $\Bbb Z$ has only countably many order-automorphisms.

For any $n \in \Bbb Z, \varphi_n: k \mapsto k+n$ is an order-automorphism.

However, because an order-automorphism must preserve successors and predecessors and every element of $\Bbb Z$ is both a successor and a predecessor that can be "reached" from $0$ in finitely many steps, it's easy to prove by induction on $\vert k \vert$ that any automorphism $\varphi$ of $\Bbb Z$ is uniquely determined by $\varphi(0)$. Thus, each order-automorphism of $\Bbb Z$ is of the form $\varphi_n$ for some $n \in \Bbb Z$.