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Let S be a totally ordered set such as:

-for all a in S, there exists b,c in S such as : b < a < c

-for all a < b in S there exists c in S such as a < c < b

-there exists a sequence (u_n), in S such as for all a < b in S, we have a <= u_n <= b for a certain value of n

-every non empty upper bounded subset of S have an supremum, and every non empty lower bounded subset of S have an infimum.

Show that R fits these properties. Conversely, show that if S fits these properties, there exists f: S -> R bijective and increasing.

The indication given by my teacher is to build an increasing bijection from the set of (un) and Q, then to extend it to a bijection from S to R, but i have no clue on how to do it. Any help would be appreciable.

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  • $\begingroup$ R does not fit those properties because R has not empty subsets with neither inf nor sup. $\endgroup$ – William Elliot Nov 13 '17 at 3:44
  • $\begingroup$ Thank you for mentioning this, i changed the initial conditions. $\endgroup$ – Morpheus Nov 13 '17 at 9:56
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Nowhere in your conditions does anything say that $u_n \ne u_m$ when $n \ne m$. The possibility that $u_n = u_m$ makes the construction more difficult, so I would first build a new sequence $\{\hat u_n\}$ by removing any duplicates from $\{u_n\}$. Then, note that $\{\hat u_n\}$ has exactly the same property as $\{u_n\}$ and therefore we might as well have assumed uniqueness for $\{u_n\}$ in the first place. (Thereby allowing us to drop that hat in the rest of this.)

Building your initial correspondence from $\{u_n\}$ to $\Bbb Q$ would be more difficult than necessary. I suggest building it from $\{u_n\}$ to the set $D$ of dyadic rationals instead: $D = \left\{\dfrac p{2^q} \mid p, q \in \Bbb Z\right\}$.

For $n \in \Bbb N$, let $L_n = \{ i < n \mid u_i < u_n\}$ and $U_n = \{ i < n \mid u_i > u_n\}$. Define $$m(n) = \begin{cases} \max L_n & L_n \ne \emptyset\\ -\infty & L_n = \emptyset\end{cases}$$ $$M(n) = \begin{cases} \min U_n & U_n \ne \emptyset\\\infty & U_n = \emptyset\end{cases}$$

Define $f$ on $\{u_n\}$ inductively: $f(u_0) = 0$ and if $f(u_k)$ is defined for all $k < n$, then define $$f(u_n) = \begin{cases} \max \{p \in \Bbb Z \mid \forall k < n, p < f(u_k)\} & m(n) = -\infty\\ \dfrac{u_{m(n)} + u_{M(n)}}2 & m(n) > -\infty \text{ and } M(n) < \infty\\\min \{p \in \Bbb Z \mid \forall k < n, p > f(u_k)\} & M(n) = \infty\end{cases}$$

You will need to demonstrate that $f$ is injective, increasing, and $f(\{u_n\}) = D$. Then you can use the fact that $D$ is dense in $\Bbb R$ to extend $f$ continuously to all of $S$.

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