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

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

• dense ($$\forall a),
• 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.

• I think that the answer is no: see the long line.
– Joe
Commented Jun 8, 2023 at 18:17
• @Joe, the long line is not a subset of $\mathbb{R}$, which is what makes this an interesting question.
– JMM
Commented Jun 8, 2023 at 18:19
• @Jephph: Apologies, I missed that hypothesis in Z Wu's question.
– Joe
Commented Jun 8, 2023 at 18:20
• Related: Souslin's problem https://en.wikipedia.org/wiki/Suslin%27s_problem Commented Jun 8, 2023 at 18:58

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.
• For a counterexample when you remove the separable condition take $[0,1]^2 \setminus \{ (0,0), (1,1)\}$ with the lexicografic order i.e. $(a,b) < (c, d)$ iif $a < c$ or $c = a \land b< d$. The box has uncountably many disjoint intervals.