Consider $0<s<1$. Is true the following statement?
$(\forall 0<\delta<1)(\exists \gamma>0)$ such that $(\forall x>0)(\exists n\in\mathbb{Z})$ such that $\gamma<s^nx<\delta. \hspace{6mm} (*)$
Context:
Let $K$ be a valued field and let $X$ be a vector space over a $K$. I am trying to prove that:
"Two norms $p_1,p_2$ on $X$ induce the same topology iff there are constants $c_1$ and $c_2$ such that $c_1p_1\leq p_2\leq c_2p_1$" $\hspace{6mm} (**)$.
I already proved $(**)$ when the valuation is dense. (For definitions and proof see my answer here).
If the statement $(*)$ could be proven, then using the same proof of the link we could prove $(**)$ when the valuation is discrete and non-trivial.
Regarding to the statement $(*)$, I don't have a clear approach yet.
With the way the quantifiers are written, $\gamma$ can depend on $\delta$ and $s$ while $n$ can depend on $s,\delta$ and $x$. That is, you want functions $\gamma,n$ so that
$$\gamma(s,\delta)<s^{n(s,\delta,x)} x<\delta.$$
To do that, identify $n$ so that $s^n x$ is as close to $\delta$ as possible while still being smaller. Since $0<s<1$, this is $n=\lceil \log_s(\delta/x) \rceil$. Then $n<\log_s(\delta/x)+1$ so $s^n x>s\delta$. So you can take $\gamma=s\delta$.
Strictly speaking, this permits the possibility $s^n x = \delta$. I will leave it as an exercise to see how to shift things to remove that possibility.