Considering the topology given by the basis $B_{\gamma}(a)=\{x\in K: v(x-a)>\gamma \}$ on (K,v) a field with valuation. I need to show that the set: $\{x\in K: v(x)=\gamma \}$ is open.
So I need to show that for every x in the set there exists a set from the basis such as $x\in B_{\gamma'}(a)\subseteq \{x\in K: v(x)=\gamma\}$ thus I need to find $\gamma'$ and show that $v(x-a)>\gamma'$ $v(x-a)\geq min\{ v(x),v(a) \}=min\{ \gamma,v(a) \}$ can I say that since a is in $B_{\gamma'}(a)$ then it must be in $\{x\in K: v(x)=\gamma\}$ then $v(a)=\gamma$ and then I just take $\gamma' < \gamma$
*I know that since I need to find $x\in B_{\gamma'}(a)$ then $B_{\gamma'}(a)=B_{\gamma'}(x)$ so maybe I just need to show that there exists $\gamma'$ such as $B_{\gamma'}(x)\subseteq \{x\in K: v(x)=\gamma\}$.
I can't really understand what is the relation between $\gamma$ and $\gamma'$ that I need to find, and I don't know how to continue from here. Should I have a different approach?
Any help would be appreciated, Thanks!
Given $a,x\in K$, if $v(a-x)>v(x)$ then $v(a)=v(x)$.
$\quad $ (This follows from $v(x)\ge \min(v(a-x),v(a))$)
So your set $E=\{ x\in K, v(x)=\gamma\}$ is open because $E=\bigcup_{x\in E} \{ a\in K, v(a-x)>\gamma\}$