Any compact subset of $\bar{\mathbb{Q}}_{p}$ is in a finite extension of $\mathbb{Q}_p$?

Question

It is well-known that any compact subgroup of $\bar{\mathbb{Q}}_{p}$ (algebraic closure of $\mathbb{Q}_p$) is in a finite extension of $\mathbb{Q}_p$. (Usually, its proof requires Baire category theorem.)

My question: does this hold for any compact set in $\bar{\mathbb{Q}}_{p}$? i.e., Any compact subset of $\bar{\mathbb{Q}}_{p}$ is in a finite extension of $\mathbb{Q}_p$?

It seems that my question is equivalent to the following: 'any compact subset of $GL_n(\bar{\mathbb{Q}}_p)$ is in $GL_n(E)$ where $E/\mathbb{Q}_p$ is a finite extension.'

Answer 1

Counterexample: $C:=\{p^n p^{1/n}\} \cup \{0\}$.

More generally, take the union of $\{0\}$ and (the set of the members of) any sequence $\alpha_n$ such that the corresponding sequence of values $(| \alpha_n |_p)_n$ has $0$ as its only accumulation point. There's plenty such sequences which are not contained in any finite extension of $\mathbb Q_p$.

Answer 2

For your implicit first question: it follows from $\Bbb{Q}_p(a+p^n b)=\Bbb{Q}_p(a, b)$ for $n$ large enough and $a,b$ algebraic which itself follows from the same proof as in the primitive element theorem

If $H\subset\overline{\Bbb{Q}_p}$ is a subgroup such that $\Bbb{Q}_p H$ is not a finite dimensional vector space then there is $h_j\in H$ such that $\dim \sum_{j=1}^J \Bbb{Q}_p h_j\to \infty$ ie. $ [\Bbb{Q}_p(h_1,\ldots,h_J):\Bbb{Q}_p]\to \infty$ thus (with $r_j$ increasing fast enough) $[\Bbb{Q}_p(\sum_{j=1}^J p^{r_j} h_j):\Bbb{Q}_p]\to \infty$ and hence $\sum_{j=1}^\infty p^{r_j} h_j$ is not algebraic ie. $(\sum_{j=1}^J p^{r_j} h_j)_J$ has no convergent subsequence and $H$ is not compact.