Is the set of totally ordered sets totally ordered?

Question

This seems super intuitive, but I can't seem to prove it. Is the set of equivalence classes of totally ordered sets totally ordered?

More precisely, given two totally ordered sets, $F$ and $G$, does there always exist an order preserving injection from one into the other?

I would think there is some adaptation of Zermelo's theorem that can fix the problem, but again, I can't seem to find it.

Answer 1

The natural numbers, $\omega$ and their inverse order (i.e. the negative integers) are incomparable.

But this is not even a partial order. The rational numbers embed into $[0,1]\cap\Bbb Q$ and vice versa, so it's not even antisymmetric.

Answer 2

The answer is no, for example there is no order preserving injection $\omega_1\to\Bbb R$ (every ordinal embedding into the reals is countable) as well as no order preserving injection $\Bbb R\to\omega_1$ (since $\Bbb R$ has infinite descending chains, while $\omega_1$ doesn't)