How do you prove that a dense, complete, separable linear order with no end points is isomorphic to the reals as a linear order?

I have just got the hang of the back and forth construction and used it to prove that any two countable dense linear orders with no end points are isomorphic to each other.


  • $\begingroup$ Once you get the isomorphism between the countable dense subsets, show it extends uniquely. $\endgroup$ – Asaf Karagila Nov 21 '17 at 16:25

Let $(X; \preceq)$ be a dense, complete, separable linear order without endpoints.

($\mathbb Q; \le)$ is a countable dense linear order without endpoints and it's a substructure of $(\mathbb R; \le)$.

Step 1. Prove that there is a substructure $(Y; \preceq)$ of $(X; \preceq)$ which is a countable, dense and dense in $(X; \preceq)$ [sic!] linear order without endpoints. Fix an isomorphism $$ \pi \colon (Y; \preceq) \to (\mathbb Q; \le). $$

Step 2. Now use that $X$ is complete. Every element of $X$ can be approximated by elements of $Y$ and this approximation translates via $\pi$ to an approximation of some unique element, let's call it $\pi^+(x)$, of $\mathbb R$. Show that $$ \pi^+ \colon (X; \preceq) \to (\mathbb R; \le) $$ is an isomorphism. In fact, it's the unique isomorphism $\rho$ between those structures satisfying $\pi \subseteq \rho$.

  • 1
    $\begingroup$ I'm intentionally vague about the details here. Your job is to fill in the gaps and prove all the subclaims mentioned in my answer. $\endgroup$ – Stefan Mesken Nov 21 '17 at 16:29

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