Let $L/K$ be a finite Galois extension, $\phi$ a valuation on $K$ and $\psi$ a valuation on $L$ extending $\phi$. I'm trying to parse a very short proof of the theorem that all extensions of $\phi$ to $L$ are conjugate under the Galois group of $L/K$. The proof starts as follows:
Let $\psi_1$ and $\psi_2$ be extensions of $\phi$ lying in different orbits of $G=\textrm{Gal}(L/K)$. Then the $G\psi_i$ are disjoint and by the approximation theorem there exists an $x\in L$ s.t. $\psi(x)<1$ for all $\psi\in G\psi_1$ and $\psi(x)>1$ for all $\psi\in G\psi_2$.
I can't see why the approximation theorem would imply this, because in this case both orbits can have more than one element. The way I've been taught the approximation theorem is that we can pick one valuation $\psi$ for which $\psi(x)<1$ and for all the others the inequality would go the other way.