Transform a totally ordered set to a structure that is isomorphic to (R,+,.,≤)

Question

So let $(M,\le_M)$ be a totally ordered set.

Can we define $+$ and $.$ to make $M$ isomorphic to $(\mathbb{R},+,.,\le)$? I mean the well known axioms.

To let this possible:

1. $M$ is not bounded above and not bounded below.
2. $|M|$ and $|\mathbb{R}|$ should be equal. (So $M=\mathbb{Q}$ or $M=\mathcal{P}(\mathbb{R})$ are not ok.)
3. $\le_M$ has to be complete
4. And for $x,y \in M$ with $x<_My$ there should be a $z \in M$ with $x<_M z <_M y$ . ($<_M$ is defined as usual.)

Are these properties sufficient?

1. If so: how can we define $+_M$ and $._M$ on $M$ to get the isomorphism to $(\mathbb{R},+,.,\le)$? (Of course $._M$ and $+_M$ should be compatible with $\le_M$.)
2. If not: which other properties are needed too?

Answer 1

Your properties are not sufficient: Let $M=[0,1]\times\Bbb R$ with lexicographic order. Then 1.-4. are satisfied, but $M$ does not have a countable dense subset because any countable subset will leave out at least some $\{a\}\times\Bbb R$. Hence $M$ is not order-isomorphic to $\Bbb R$ (which of course has the countable dense subset $\Bbb Q$).

In fact this is the only addionally needed property, i.e., every nonempty, separable, complete, dense, endless total order is isomorphic to $(\Bbb R,\le)$ (and once we have an order-isomorphism, we can transfer the field operations fom $\Bbb R$ to$M$ via that isomorphism). Note that we can drop the cardinality condition.

Answer 2

"What other properties are needed too?" Here is an example of an additional property that may (or may not) be adequate.

See Souslin's problem . Instead of requiring "separable" as in Hagen's link, we require "countable chain condition":

Given a non-empty totally ordered set R with the following four properties:
$\bullet$ R does not have a least nor a greatest element;
$\bullet$ the order on R is dense (between any two elements there is another);
$\bullet$ the order on R is complete, in the sense that every non-empty bounded subset has a supremum and an infimum;
$\bullet$ every collection of mutually disjoint non-empty open intervals in R is countable (this is the countable chain condition for the order topology of R).

Is R necessarily order-isomorphic to the real line?

This question cannot be answered with the ZFC axioms of set theory!