Prove that the every Cauchy sequence in $CS_{\Bbb R(x)}/R=\lbrace [(u_n)]|(u_n)\in CS_{\Bbb R(x)}\rbrace$ converges. Consider the field of rational functions $\Bbb R(x)$. Let $(u_n), (v_n)$ be sequences with terms in the field of rational functions.
Define the equivalence relation $R$ by $((u_n),(v_n))\in R\Leftrightarrow lim_{n\rightarrow\infty}(u_n-v_n)=0$.
Consider the field of rational functions. 
It is $\Bbb R (x)=\lbrace \frac{f(x)}{g(x)}|f,g\in \Bbb R[x],g\ne0\rbrace$
Define $P=\lbrace p(x)|p(x)=\frac{f(x)}{g(x)}\in \Bbb R[x]\land$ the leading coefficients of $f(x),g(x)$ are of the same sign $\rbrace$.
Define for $p(x),q(x)\in \Bbb R(x), p(x)\ge q(x) \Leftrightarrow p(x)-q(x)\in P\cup \lbrace 0\rbrace$. 
For every $u=\frac{f}{g}\in \Bbb R(x)$,Define $|u|=max\lbrace u,-u\rbrace$
Define Cauchy sequence in $\Bbb R(x)$ as: A sequence $(u_n)$ is Cauchy if $(\forall \epsilon \in P)(\exists N\in \Bbb N)(n,m>N\Rightarrow |u_n-u_m|<\epsilon)$
As the way of getting $\Bbb R$ from $\Bbb Q$, extend the field of rational functions by means of equivalence classes of Cauchy sequences.
For the field of equivalence classes of Cauchy sequence in field of rational functions. Prove that every Cauchy sequence converges in the new field we get.
It seems that it is not quite same as the proof of every Cauchy sequence converges in $\Bbb R$. Could someone please give a proof? Thanks so much!
EDIT: It is seen in "conterexamples in analysis" without a proof.

 A: Virtually all of the usual proof of this fact for real numbers uses nothing special about the field $\mathbb{Q}$ (besides just that it is an ordered field).  So almost all of the steps apply immediately when you replace $\mathbb{Q}$ with the ordered field $\mathbb{R}(x)$.  The one special thing about $\mathbb{Q}$ that is used is the fact that for every rational number $q>0$, there exists a positive integer $N$ such that $q>1/N$.  This is used, for instance, in saying that you can test whether a sequence $(x_n)$ is Cauchy by just testing whether the differences $|x_n-x_m|$ eventually become smaller than $1/N$ for each $N$.  To prove every Cauchy sequence $(y_n)$ in $\mathbb{R}$ is complete, you then pick Cauchy sequences of rational numbers representing each $y_n$ and use a diagonal argument to find a limit of $(y_n)$, using the fact that you only need to make the relevant differences eventually smaller than $1/N$ for each $N$.
So to make this argument with $\mathbb{Q}$ replaced by a more general ordered field $K$, all you need to know about $K$ is that there exists a sequence $(\epsilon_N)$ of positive elements of $K$ such that for any $q>0$ in $K$, there exists $N$ such that $q>\epsilon_N$.  You can then use $\epsilon_N$ as a replacement for $1/N$ in the diagonal argument.
In the case of $K=\mathbb{R}(x)$, such a sequence is $\epsilon_N=1/x^N$.  Indeed, if $f(x)/g(x)\in K$ is a positive element, it is straightforward to verify that $f(x)/g(x)>1/x^N$ as long as $N>\deg g-\deg f$.
(In fact, if $K$ is an ordered field such that no such sequence $(\epsilon_N)$ exists, then the result is still true, for a much more trivial reason: if no such sequence $(\epsilon_N)$ exists, then every Cauchy sequence in $K$ is eventually constant.)
