Let $K$ be an ordered field, $R\subseteq K$ a convex subring - that is, for all $a \in K$, $a \in R$ if $x \leq a \leq y$ for some $x,y \in R$. Define $I = \{ x \in R: x^{-1} \notin R\}$. It is clear that if $I$ is an ideal, then $R$ is local. However, I am having some trouble to understand why it is a convex ideal of $R$. I know that, if $a>1$ and $a \in R$, then $0 < a^{-1}< 1$, which implies that $a \notin I$. I am also unsure of the hypothesis one should assume about $K$ (for example, $K$ may be real closed - or even a model of the theory of the reals as an ordered field). Can someone help me?
Edit: For $a \in R$, assume that $|a|\geq 1/n$ for some $n \in \mathbb{N}$. If $a>0$, then $0<a^{-1}\leq n$ which implies $a \notin I$. If $a<0$, $-n \leq a^{-1} < 0$ - from which $a^{-1} \notin I$. Thus, $I \subseteq \{a : |a|<1/n \mbox{ for all } n \}$.
Let $K$ be an ordered field.
Call a subset $G$ of $K$ convex if for all $x,z\in G$ with $x < z$, we have $\{y\in K{\,\mid\,}x < y < z\}\subset G$.
Let $R$ be a convex subring of $K$, and let $I=\{a\in R{\,\mid\,}a^{-1} \not\in R\}$
Claim:$\;I$ is a convex ideal of $R$.
Proof:
It's immediate that $0\in I$.
Next let $a\in I$ and $r\in R$.
If $a=0$ or $r=0$, then $ra=0$, so $ra\in I$.
Suppose $a,r\ne 0$, and suppose $ra\not\in I$. \begin{align*} \text{Then}\;\;& ra\not\in I \\[4pt] \implies\;& \frac{1}{ra}\in R \\[4pt] \implies\;& \frac{1}{ra}=s\;\text{for some $s\in R$} \\[4pt] \implies\;& \frac{1}{a}=rs \\[4pt] \implies\;& \frac{1}{a}\in R \\[4pt] \end{align*} contrary to $a\in I$.
It follows that $ra\in I$ for all $a,r$ with $a\in I$ and $r\in R$.
As an immediate consequence, if $a\in I$, then $-a=(-1)a\in I$, so $I$ is closed under negation.
Next let $a,b\in I$, and suppose $a+b\not\in I$.
Then $a,b,a+b\ne 0$.
Without loss of generality, assume $|a|\le |b|$. \begin{align*} \text{Then}\;\;& a+b\not\in I \\[4pt] \implies\;& |a+b|\not\in I \\[4pt] \implies\;& \frac{1}{|a+b|}\in R \\[4pt] \implies\;& \frac{1}{|a+b|}=r\;\text{for some $r\in R$ with $r > 0$} \\[4pt] \implies\;& |a+b|=\frac{1}{r} \\[4pt] \implies\;& |a|+|b|\ge\frac{1}{r} \\[4pt] \implies\;& 2|b|\ge\frac{1}{r} \\[4pt] \implies\;& |b|\ge\frac{1}{2r} \\[4pt] \implies\;& \frac{1}{|b|}\le 2r \\[4pt] \implies\;& 0 < \frac{1}{|b|} < 3r \\[4pt] \implies\;& \frac{1}{|b|}\in R \\[4pt] \implies\;& |b|\not\in I \\[4pt] \implies\;& b\not\in I \\[4pt] \end{align*} contradiction, hence we must have $a+b\in I$.
It follows that $I$ is an ideal of $R$.
It remains to show that $I$ is convex.
Thus suppose $a < b < c$, where $a,c\in I$ and $b\in K$.
Our goal is to show $b\in I$.
Since $R$ is convex, it's immediate that $b\in R$.
Suppose $b\not\in I$.
Then $b\ne 0$ and ${\Large{\frac{1}{b}}}\in R$, hence also ${\Large{\frac{1}{|b|}}}\in R$.
Letting $x=\max(|a|,|c|)$, we have $x\in I$ and $0 < |b| < x$. \begin{align*} \text{Then}\;\;& 0 < |b| < x \\[4pt] \implies\;& 0 < \frac{1}{x} < \frac{1}{|b|} \\[4pt] \implies\;& \frac{1}{x}\in R \\[4pt] \end{align*} contrary to $x\in I$.
It follows that $b\in I$, hence $I$ is convex.
This completes the proof.