Is there a totally ordered set we can map any other totally ordered set to?

Question

Are there any totally ordered sets $O$ with $|O|\ge|\mathbb N|$ such that for each totally ordered set $S$ with $|S|\le|O|$ there is function $f_S\colon S\mapsto O$ such that $\forall_{x,y\in S}x\le y\leftrightarrow f_S(x)\le f_S(y)$?

Answer 1

(I'm posting this as an answer because it's too long for a comment, but I'm out of my depth here:)

It seems from $\S6$ of Normal J. Alling's Conway's field of surreal numbers, Trans. AMS 287, no. 1 (1985) that for every ordinal $\alpha$, if $\omega_{\alpha + 1}$ denotes the least ordinal of power $\aleph_{\alpha + 1}$, and $$ S_{\alpha + 1} = \{f \in \{0, 1\}^{\omega_{\alpha + 1}}: \exists \beta < \omega_{\alpha + 1} \text{ s.t. } f(\beta) = 1, f(\gamma) = 0 \text{ if } \beta < \gamma < \omega_{\alpha + 1} \}, $$ then any totally ordered set of power at most $\aleph_{\alpha + 1}$ has an order-preserving map into $S_{\alpha + 1}$. By $\S4$, under GCH, $S_{\alpha + 1}$ may be replaced by a set of power $\aleph_{\alpha + 1}$, which would answer your question.

(I see the question has been edited. The condition was originally $\left\lvert{O}\right\rvert > \left\lvert\mathbb{N}\right\rvert = \aleph_0$.)