Is every dense complete endless linearly ordered subset of real order-isomorphic to real?

Question

Let $(M,\leq)$ be a non-empty

• dense ($\forall a<b\in M,\exists c\in M,a<c<b$),
• complete (every non-empty subset that is bounded above has a supreme)
• endless (there is no minimal or maximal element)

linearly(totally) ordered subset of $(\mathbb{R},\leq)$. Do we have that $M$ is order-isomorphic to $\mathbb{R}$?

The context is to show that $(\mathbb{R},\leq )$ is the minimal non-empty dense complete endless linearly ordered set up to order-isomorphism, which is a corollary of this problem.

Answer 1

Found it.

The real numbers is the unique non-empty linear order which is complete, dense and separable. It suffices to show that $M$ is separable with the order topology $\tau_{\leq}$.

Since $\mathbb{R}$ is a separable metric space and subspace of separable metric space is separable, it follows that $M$ is separable with the subspace topology $\tau_M$, i.e. there exists a countable sequence $\{x_n:n\in \mathbb{N}\}$ s.t. $\forall U\in \tau_M,\{x_n:n\in \mathbb{N}\}\cap U=\emptyset$.

Also we have the subspace topology is finer than the order topology, i.e. $\tau_{\leq}\subset \tau_M$, it follows that $M$ is separable with the order topology as well. The result follows.