Can I embed every complete linear order $(A,<^A)$ with $\vert A\vert=2^{\aleph_0}$ into $(\mathbb{R},<)$. If not, what could be conditions on $(A,<^A)$ to make this work?
I know that any separable complete dense linear order without endpoints is isomorphic to $(\mathbb{R},<)$. So similar as you prove that any countable linear order embeds into $(\mathbb{Q},<)$ by densifying it and using the isomorphism of the expansion and $(\mathbb{Q},<)$, I thought about densifying $(A,<^A)$, but I am not sure of how to proceed, if possible.
A well-ordered set embeds into $(\Bbb R,<)$ iff it is countable. Assuming ZFC, there is a well-ordered set of cardinality $2^{\aleph_0}$. This cannot be embedded into $(\Bbb R,<)$.