How to embed a total ordering into the real field. Let $(S,<_S)$ be a total ordering with $card(S)\leq card(2^{\aleph_0})$. Does there exist a subset $A$ of the real numbers such that $(A,<_A)$, being a total ordering, is isomorphic to $(S,<_S)$?
I have tried with the following:


*

*Suppose that $S'$,a subordering of $S$, has already been mapped isomorphically into a subset of R through $U$.

*Pick an element $p$ in $S-S'$. 
2.1 If $p$ is a limit point of $S'$ in the order topology of $(S,<_S)$, there will be a sequence ${p_n}$ in $S'$ which converges to $p$ in $S$. Then $U(p_n)$ should be a converging sequence in R so that I could define $U(p)$ to be the limit of $U(p_n)$.
2.2 If $p$ is an isolated point, then simply define $U(p)$ to be the midpoint of $sup\{U(q):q<_Sp\}$ and $inf\{U(q):p<_Sq\}$.
Is my reasoning correct?If $S$ is countable then I could well-order it and use recursion to complete the construction, but I am still wondering how to deal with the limit ordinal case if $S$ is uncountable.
Thank you.
 A: There are some clear faults with your proof: the most important of which is that the result you claim is false.

If $\{ x_\alpha : \alpha < \omega_1 \}$ is a strictly increasing enumeration of some set of reals (i.e., we have an (order-)embedding $( \omega_1 , < ) \hookrightarrow ( \mathbb{R} , < )$), then for each $\alpha < \omega_1$ there must be a rational number $q_\alpha \in ( x_\alpha , x_{\alpha+1} )$. But these rational numbers must be all distinct, even though there are only countably many of them!


Looking at your proof, you seem to assume that if $p \in S \setminus S^\prime$ is a limit point of $S^\prime$, then there is some limit point of $U [ S^\prime ]$ to which we can map $p$.  But what if we're just trying to embed $\omega + 1$ into $\mathbb{R}$, and we've been a bit lazy and assumed that $U(n) = n$ for each $n < \omega$.  Then there is no limit point of $U[\omega]$ which we could map $\omega$ to.
Similarly for the isolated point case.  It could be that $\sup \{ U(q) : q \in S^\prime, q < p \} = \inf \{ U(r) : r \in S^\prime, r > p \}$, and so there is no isolated midpoint.
