Does this construction yield the surreal numbers? There are two simple constructions for creating arbitrarily large non-Archimedean ordered field extensions of the reals.
First given such a field one may consider rational functions over that field with $f(x)$ declared positive if $f(x)>0$
for all sufficiently large $x$.  Secondly one can Dedekind complete the field by filling in all good/narrow cuts in the
field as detailed in 
this discussion.
I would like to know if iterating these two constructions over the ordinals produces the surreal numbers.
Specifically we define $F_0$ to be the real numbers.  For an ordinal $\lambda$ with predecessor $\lambda-1$, define $F_\lambda$ to be the
ordered field obtained by Dedekind completing the rational functions over $F_{\lambda-1}$. Otherwise if $\lambda$ is a limit ordinal,
define $F_\lambda$ to be the union of all $F_\kappa$ with $\kappa<\lambda$.
 A: I've written up an expository paper incorporating ideas from this discussion.
Here is a link:
https://drive.google.com/file/d/0B1G4mOmOYMtha1JLc2tTY1d2SnM/view
I welcome any comments, suggestions, etc.
A: No. Think about the fields as valued fields under the natural valuation induced by the order (i.e., the valuation ring consist of elements bounded by an integer). Order completion then coincides (for nonarchimedean fields) with their completion as valued fields, in particular, it preserves the value group. If $F$ has value group $\Gamma$, then $F(X)$ ordered with $X>F$ as you are doing has value group $\mathbb Z\times\Gamma$ ordered lexicographically, in particular, it contains $\Gamma$ as a convex subgroup. Thus, value groups of all $F_\alpha$ will be discretely ordered. This means none of the fields nor their union can be a real-closed field, as real-closed fields have divisible value groups. (Specifically, the $X$ from $F_1=\widehat{\mathbb R(X)}$ will have no square root in any of the fields.)
A: Edit Based on the comments above, I would like to rephrase my question.  Suppose I change the recursive procedure as follows.
To proceed from $F_\alpha$ to $F_{\alpha+1}$ I perform the constructions below in the order shown.


*

*I first fill in the good/narrow gaps in $F_\alpha$ with cut points.

*For each bad/wide gap $(A,B)$, if there is a polynomial $p(x)$ which changes sign across the gap, I algebraically
extend $F_\alpha$ by a root $X_{(A,B)}$ of $p(x)$ contained within the gap (with degree of $p(x)$ minimal).

*For each bad/wide gap $(A,B)$, such that all non-zero polynomials in $F_\alpha$ have the same sign across the gap,
I add a transcendental element $X_{(A,B)}$. I declare a rational function $f(X_{(A,B)})$ over $F_\alpha$ to be
positive if the rational function $f(x)$ is positive across the gap

*Finally I add a transcendental element $X_\infty$ with a rational function $f(X_\infty)$ being declared
positive if $f(x)>0$ for all sufficiently large $x\in F_\alpha$


Do I get the surreals if I recurse over all ordinals?
It seems to me that steps (1) and (2) have the effect of eventually real closing each $F_\alpha$ at some later stage.
Edit.  I overlooked Emil's last comment. If I understand him correctly, then my construction produces an $Ord$-saturated field.
Given sets $X<Y$, let $F_\alpha$ be the minimal field containing $X\cup Y$. Let
$$A=\{a\in F_\alpha\ |\ a\le x\in X\}$$
$$B=\{b\in F_\alpha\ |\ b\ge y\in Y\}$$
Then if $(A,B)$ is not a Dedekind cut of $F_\alpha$, then there is an element $u\in F_\alpha$ with 
$X\subset A<u<B\supset Y$. If $(A,B)$ has a cut point $c$ in $F_\alpha$ then one of $c\pm\frac{1}{X_\infty}$
produces an appropriate $u\in F_{\alpha+1}$. Otherwise my construction explicitly produces such a $u\in F_{\alpha+1}$.
Edit 2 I realize that I need to be a little more careful in formulating my construction.  I need another transfinite recursion in the passage
from $F_\alpha$ to $F_{\alpha+1}$.  First a couple of preliminary definitions.
Let $(A,B)$ be a Dedekind cut of an ordered field $F$. We say that $(A,B)$ is a gap if there is no cut point in $F$. We say that the
gap is narrow if for any $\epsilon>0$ in $F$, we can find $a\in A$, $b\in B$ with $b-a<\epsilon$.  Otherwise we say the gap is wide.
We further classify the wide gaps as follows.  If there is a polynomial $p(x)$ over $F$ which changes sign across the gap, we say the
gap is algebraic (relative to $F$). Otherwise we say the gap is transcendental (relative to $F$).
Now given an ordered field $F_\alpha$ I want to construct an ordered field $F_{\alpha+1}$ with the property that for any Dedekind
cut $(A,B)$ in $F_\alpha$ (gap or not)there is an element $u\in F_{\alpha+1}$, with $A<u<B$. I start by filling in all the narrow
gaps in $F_\alpha$ by taking the standard Dedekind completion of $F_\alpha$ as detailed in
this discussion.
I call the resulting field $F_\alpha^*$.
Now I start another transfinite recursion.  I begin by well-ordering all the wide gaps $(A_\beta,B_\beta)$ in $F_{\alpha}$. I define
$F_{\alpha,0}=F_\alpha^*$. Having defined $F_{\alpha,\beta}$, I define $F_{\alpha,\beta+1}$ as follows.
Case 1. If there is already an element $u\in F_{\alpha,\beta}$ with $A_\beta<u<B_\beta$, then I take $F_{\alpha,\beta+1}= F_{\alpha,\beta}$.
If we are not in Case 1, then we can extend the wide gap $(A_\beta,B_\beta)$ in $F_\alpha$ to a wide gap $(A'_\beta,B'_\beta)$
in $F_{\alpha,\beta}$.
Case 2. If $(A'_\beta,B'_\beta)$ is algebraic relative to $F_{\alpha,\beta}$, then I take $F_{\alpha,\beta+1}$ to be an algebraic
extension of $F_{\alpha,\beta}$ obtained by adjoining a root $X_{(A'_\beta,B'_\beta)}$ of an irreducible polynomial $p(x)$ which
changes sign across the gap. We order $F_{\alpha,\beta+1}$ so that $X_{(A'_\beta,B'_\beta)}$ lies in the gap.
Case 3. If $(A'_\beta,B'_\beta)$ is transcendental relative to $F_{\alpha,\beta}$, then I take $F_{\alpha,\beta+1}$ to be a transcendental
extension of $F_{\alpha,\beta}$ obtained by adding a transcendental element $X_{(A'_\beta,B'_\beta)}$ with a rational function
$f(X_{(A'_\beta,B'_\beta)})$ declared positive if the corresponding function $f:F_{\alpha,\beta}\to F_{\alpha,\beta}$ is positive across the
gap.
After completing the induction $F_{\alpha,\beta}$ over $\beta$, I obtain a field $\widehat{F}_\alpha$ which contains elements spanning
all gaps in $F_\alpha$. Finally I add a transcendental element $X_\infty$ to $\widehat{F}_\alpha$ with a rational function $f_(X_\infty)$
declared positive if the corresponding function $f:\widehat{F}_\alpha\to \widehat{F}_\alpha$ is positive for sufficiently large values
of the argument. I call the resulting field $F_{\alpha+1}$. This last step has the effect of adding elements between Dedekind cuts
in $F_\alpha$ which are not gaps, but rather have cut points.
