Convex subring is local 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, $I \subseteq \{a : |a|<1/n \mbox{ for all } n \}$.
 A: 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.
