Let $F$ be an ordered field, let $L$ be the real closure of $F$. 


Let $R \in L$ be strictly positive. Can one find a bound $M \geq 0$ and for each $x \in ]-R;R[\_L$, an element $x' \in [x-1;x+1]\_L$ and a unitary annihilating polynomial $P\_{x'}$ of $x'$ in $F[X]$ with coefficients (however numerous they may be) bounded by $M$?


If $L / F$ is finite, then $x' := x$, $P\_{x'}:= \chi\_x$ where $\chi$ denotes the characteristc polynomial works (the coefficients functions of $\chi$ are polynomial and thus lipschitz).


In the general case, I have to say I really don't know if this is possible.



Bound for annihilating polynomials
It seems that the answer is **no**, and a counterexample is found exactly in the field provided in the comment.


So, let $F=\mathbb Q(X)$ ordered by $X>\mathbb Q$, and let $L$ be its real closure. For any integer $n>1$, set $x=X^{n/(n+1)}$. Take any $\xi\in[x-1,x+1]\_L$; notice that $\xi^k=X^{kn/(n+1)}(1+o(X^{-1/(n+1)}))$.


Now consider any polynomial $P(t)=p\_st^s+p\_{s-1}t^{s-1}+\dots+p\_0\in F[t]$ annihilating $\xi$ (we have $p\_s=1$). Let $d=\max\_i \left(\frac{in}{n+1}+\deg p\_i\right)\geq \frac{sn}{n+1}$. This maximum should be attained for at least two values of $i$, otherwise the term $p\_ix^i$ has a strictly maximal degree, according to the remark above. Let $i>j$ be these two indices; then $n+1$ divides $i-j$, hence $j\leq s-n-1$, and $\deg p\_j\geq d-\frac{jn}{n+1}\geq \frac{(s-j)n}{n+1}\geq n$. Therefore, among the coefficients of $P$ there is one which is greater than $X^{n-1}$. Thus there is no required $M$.


