Quasi-connected set

Question

Let $p$ be a prime number. In $\mathbb C_p$, consider the set $E=\{x\in\mathbb C_p\mid |x|<1\}\setminus\{p^{1/n}\mid n\in\mathbb N\}$. Is this subset quasi-connected? I try to check the definition of quasi-connected. But I did not manage to do that. Recall that a subset $S$ of $\mathbb C_p$ is quasi-connected if it contains at least two points and for all $a,b\in S$, the set of circles $\mathscr C$ with center $a$ and radius $r\in[0,|a-b|]$ that such $\mathscr C\cap\complement_{\mathbb C_p}S\ne\emptyset$ is finite.

Obviously, $E$ has at least two elements but I do not manage to prove the last condition.

Thanks in advance for any help.

Presumably, $|\cdot|$ is always the $p$-adic valuation, $|\cdot|_p?$ — Thomas Andrews, Jun 30 '23 at 14:08
$\complement_{\mathbb C_p}S$ means the complement of $S$ in $\mathbb C_p$. And yes, $|.|$ is the $p$-adic norm (usually normalized by $|p|=p^{-1}$). — joaopa, Jun 30 '23 at 14:12
By "circle", do you mean a set of the form ${x : |x-a| < r}$ (disk) or set of the form ${x: |x-a| =r }$ (its boundary)? — Torsten Schoeneberg, Jul 06 '23 at 19:50

Torsten Schoeneberg · Accepted Answer · 2023-07-07T17:02:42.257

The notation $p^{1/n}$ does not well-define an element of $\mathbb C_p$; there are $n$ distinct $n$-th roots of $p$ in $\mathbb C_p$. Apparently, you make a choice for each $n$. (Compare https://math.stackexchange.com/a/4258617/96384.). However, it does not matter what choice you make, because:
(EDIT: Original erroneous answer corrected.) That being said, let $(S, d(\cdot, \cdot))$ be any ultrametric space such that some ("open") ball $\{y \in S: d(y, x_0) < 1\}$ contains a sequence $x_1, x_2, ...$ with $\lim_{n\to\infty} d(x_n, x_0) =1$. Assume the set $$E := \{y \in S : d(y,x_0) < 1\} \setminus \{x_1, x_2, ... \}$$ contains at least two elements. Then $E$ is quasi-connected. Namely, let $a,b \in E$. First of all, any circle $\{y \in S: d(a,y) =r\}$ around $a$ with $0\le r \le d(a,b)$ is, by the ultrametric inequality, fully contained in the ball $\{y \in S: d(y, x_0) < 1\}$. Hence, if it contains any elements of the complement $S \setminus E$, those must be elements of the sequence $x_1, x_2, ...$. Secondly, since $\max(d(x_0, a)), d(a,b)) <1$, there exists $N_{a,b} \in \mathbb N$ such that $d(x_n, x_0) > \max(d(x_0, a)), d(a,b))$ for all $n \ge N_{a,b}$, which implies that only the finitely many $x_n$ with $n < N_{a,b}$ can be contained in any $\{y \in S: d(a,y) =r\}$ for $0\le r \le d(a,b)$. Since for each such $x_n$ there is only one such circle (with $r_n := d(a,x_n)$), only finitely many such circles can exist.

Thank you for the answer. Just something I do not understand in it. I assume that $x_n=p^{1/n}$ in your notations. But then, $d(0,x_n)$ does not have $0$ as limit. Did I miss anything? — joaopa, Jul 06 '23 at 03:34
@joaopa: You are right, sorry, my bad, ignore part 2 then. Please see my comment under the question for clarification. I will try to answer then. — Torsten Schoeneberg, Jul 06 '23 at 19:51

Quasi-connected set

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