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.

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

1 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.

  • $\begingroup$ 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. $\endgroup$
    – FZan
    Jun 8 at 18:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .