Let M_0\subseteq M_1 be two real ordered fields where M_0 is dense in M_1. Then is the real closure of M_0 dense in the real closure of M_1?
1 Answer
$\begingroup$
$\endgroup$
2
Suppose there were a nonempty open set that intersected the closure of $M_1$ but not the closure of $M_0$. Take a point in that set and look at its minimal polynomial over $M_1$. A sufficiently small neighborhood of the coefficients of that polynomial should produce only polynomials that have a root in that set, since roots of polynomials are continuous. Choose a point in $M_0$ in each of those neighborhoods, look at the polynomial with those coefficients, and find a root in that open set.
-
$\begingroup$ Is it clear that the last polynomial should have a root? $\endgroup$– mosenJan 17, 2012 at 17:02
-
$\begingroup$ If the coefficients do not change too much then it should still have a root. Let $r$ be the root, and assume $a<r<b$ with, wlog $f(a)<0$ and $f(b)>0$. Then if the coefficients change so little that no point between $a$ and $b$ changes by more then $min(|f(a)|,|f(b)|)$, then there will still be a root in that interval. $\endgroup$ Jan 17, 2012 at 20:19