A strict total order on $\mathbb{R}$ is a relation $R \subseteq \mathbb{R} \times \mathbb{R}$ such that:
- $\forall x \in \mathbb{R}, (x,x) \not \in R$
- $\forall x,y,z \in \mathbb{R}, (x,y) \in R \text{ and } (y,z) \in R \implies (x,z) \in R$
- $\forall x,y \in \mathbb{R}, x \neq y \implies$ precisely one of $(x,y) \in R$ and $(y,x) \in R$ holds
What is the cardinality of the set of strict total orders on $\mathbb{R}$?
Let $E$ be the considered set. Clearly $|E| \leq 2^{2^{\aleph_0}}$ since $E\subseteq \mathcal{P}(\mathbb{R} \times \mathbb{R})$ by definition. My intuition says that this is actually an equality of cardinals, but I'm having trouble proving that $2^{2^{\aleph_0}} \leq |E|$. I tried to find an injection from $\mathcal{P}(\mathbb{R})$ to $E$ by defining a strict total order $<_S \in E$ as follows: $$\forall x,y \in \mathbb{R}, x <_S y \iff (x < y \text{ and } x,y \in S) \text{ or } (x < y \text{ and } x,y \not \in S) \text{ or } (x \in S \text{ and } y \not \in S)$$ but this doesn't seem to be injective. I've also tried to consider functions from $\mathcal{P}(\mathbb{R} \setminus \{ 0 \})$ to $E$ instead, but again couldn't find injections.
Your idea does work after all, with a tiny change: Consider only $S$ with $\Bbb Z\subseteq S$ (using the fact that $|\Bbb R|=|\Bbb R\setminus\Bbb Z|$). Then $S\mapsto{<_S}$ is injective because we can reconstruct $S$ as $$\{\,x\in\Bbb R\mid \exists n\in\Bbb Z\colon x<_Sn\,\} $$