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Let $K$ be a discrete valued (with discrete valuation $v$) complete local division ring with ring of valuation $V$. Let $F$ be a compact subset of $V$. Suppose that for all $x\in F$ and for all $y\in V^\times$, one has $x^y:=yxy^{-1}\in F$. Can one find a finite number of elements $b_1,\dotsc,b_r\in F$ and an $e\in\mathbb N$, denoting $B_i:=\{x\in V\mid v(b_i-x)\ge e\}$, such that $B_i\cap B_j=\emptyset$ and $F\subset \bigcup_{i=1}^rB_i$ and for all $1\le i\le r$ and for all $y\in V^\times$, $yB_iy^{-1}\subset B_i$.

That's clearly true in the commutative case, but I do not if it is true when $K$ is not commutative.

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    $\begingroup$ Your condition forces $v(b_i - y b_i y^{-1}) \ge e$ for all $i$ and $y$, which (I think) forces $v(b_i) \ge e - 1$ for all $i$ and hence $v(x) \ge e - 1$ for all $x \in F$. This seems like a strong restriction: either $F$ must be very small, or the balls must be very big. $\endgroup$
    – LSpice
    Commented Dec 19, 2018 at 3:39
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    $\begingroup$ That being said, expanding @LSpice's comment: Let $d:= min\lbrace v(x-x^y) : x \in V, y \in V^\times \rbrace$, and let $x,y$ be such that they realise that minimum (they exist because $v$ is discrete). Then even for the singleton $F = \lbrace x \rbrace$, we need $e \le d$. On the other hand, if I don't miss something, that criterion seems to be sufficent as well. So the task is reduced to computing $d$, and I have a feeling that $d=1$ for $v(K^\times) = \Bbb Z$ and $K$ non-commutative. $\endgroup$
    – Torsten Schoeneberg
    Commented Jan 8, 2019 at 22:23

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