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.
Best Answer
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.