Ordered fields properly containing $\Bbb R$ I was reading this supplementary to Rudin's Principles of Mathematical Analysis, and I came across 'Properties of ordered fields properly containing $\Bbb R$' (on page 8). Please check if I have understood correctly:

Suppose $\Bbb F$ is  an ordered field properly containing $\Bbb R$, then for every $x\in \Bbb R$, there are infinitely many 'immediate successors' $\alpha_1,\alpha _2,\dots \in\Bbb F-\Bbb R$ such that $\alpha_1<\alpha_2<\dots$ and infinitely many 'immediate predecessors' $\beta_1,\beta _2,\dots \in\Bbb F-\Bbb R$ such that $\beta_1<\beta_2<\dots$ .

Here, I use this definition: 
Let $a\in \Bbb R$ and $\gamma\in \Bbb F$ and $a<\gamma$. If set $\{z\in \Bbb R\ : a<z<\gamma\}$ is empty, then we say that $\gamma$ is immediate successor of $a$.

Also, Please recommend some reading about  ordered fields properly containing $\Bbb R$.

 A: (1). Preliminary. Let $F,G$ be sub-fields of $\Bbb R$ which are isomorphic as fields. 
Theorem: If $\sqrt x\;\in F$ whenever $0<x\in F$ then the only field-isomorphism $h:F\to G$ is $id_F$ and $F=G.$ 
Proof: Let $h:F\to G$ be a field-isomorphism. If $x,y\in F$ with $x>y$ then $$0\ne h(x-y)=h((\sqrt {x-y}\;)^2)=(h(\sqrt {x-y}\;))^2$$ so $h(x)>h(y).$ So $h$ preserves order. Since $\Bbb Q\subset F$ and $h|_{\Bbb Q}=id_{\Bbb Q}$ we have $$\forall x\in F\;\forall q\in \Bbb Q\;(q<x\iff q=h(q)<h(x)\;).$$ So for $x\in F$ we have $\{q\in \Bbb Q: q<x\}=\{q\in \Bbb Q:q<h(x)\},$ which implies $x=h(x).$
Corollary: If $\Bbb R$ is a proper sub-field of an ordered field $H$ then $H$ is not field-isomorphic to $\Bbb R.$ 
Proof: If not,  let $g:H\to \Bbb R$ be a field-isomorphism. Applying the above theorem with $F=g(\Bbb R)$ and $G=\Bbb R=g(H)$ we have $F=G.$ That is, $g(\Bbb R)=g(H),$ so $\Bbb R=g^{-1}(g(\Bbb R))=g^{-1}(g(H))=H.$
(2). Main argument. Let $\Bbb R$ be a proper sub-field of the ordered field $H$ where the order on $\Bbb R$ is inherited from the order on $H.$  Then $H$ is not order-complete because any complete ordered field is field-isomorphic to $\Bbb R,$ but by the corollary above, $H$ is not field-isomorphic to $\Bbb R.$ 
So let $\phi\ne S\subset H$ where $S$ has an upper bound but lub $S$ does not exist. 
(i). If no $x\in \Bbb R$ is an upper bound for $S,$ let $t$ be any upper bound for $S.$ Then $0<1/t<x$ for every $x\in \Bbb R^+$.
(ii). If some $r\in \Bbb R$ is an upper bound for $S$ then let $r_0$ be the glb in $\Bbb R$ ($not$ the glb in $H$) of the set of real-number upper bounds for $S.$ We split now to two sub-cases: 
Case (ii-a).... If there exists $s\in S$ with $s>r_0$ then $s\not \in \Bbb R$ (Otherwise the def'n of $r_0$ implies $r_0\geq s$, but we have $r_0<s$). So $s-r_0>0.$ 
Now $0<s-r_0<x$ for every $x\in \Bbb R^+.$ Because if $x\in \Bbb R$ and $0< x\leq s-r_0$ then  $r_0<x+r_0\leq s ,$ but then any real number $r$ that is an upper bound for $S$ would satisfy $r\geq x+r_0\in \Bbb R,$ contrary to the def'n of $r_0.$ 
Case (ii-b).... If $s\leq r_0 $ for every $s\in S$ then  $r_0$ is an upper bound for $S$ but $r_0\ne$  \lub $S$. So let $t$ be an upper bound for $S$ with $t<r_0.$ Now $0<r_0-t<x$ for every $x\in \Bbb R^+.$ Because if  $x\in \Bbb R$ and  $0<x\leq r_0-t$ then $r_0>r_0-x\geq t,$ but then the  real number $r_0-x$ is an upper bound for S, and is less than $r_0,$ contrary to the def'n of $r_0.$
Conclusion: In each of (i), (ii-a), and (ii-b) we obtain some $y\in H^+$ such that $\forall x\in \Bbb R^+\;(y<x).$ So for any $x\in \Bbb R$  we have, for every $n\in \Bbb N$,    $$x-(n+1)y^2<x-ny^2 <x<ny^2<(n+1)y^2$$  $$\text { and }\quad \Bbb R\cap (x-ny^2,x+ny^2)=\phi.$$
