Compact subgroups of $p$-adic fields and the groups $GL_n$ over them I am trying to fill in some details in a collection of materials I'm currently studying. I have a pretty poor understanding of the topology of $p$-adic fields and the general linear group over these fields.
To start with, fix a $p$-adic field $F$ (and to be honest I'm not sure if this has to be of characteristic zero for the facts I'm concerned about). I was told that the subgroup $G_j=\{x\in F:|x|_F\leqslant q^{-j}\}$ is compact, where $|\cdot|_F$ is the normalized absolute value on $F$ and $q$ the size of the residue field $\mathfrak o_F/\mathfrak p$. I read in Neukirch that the valuation rings of local fields (which seems to be the $p$-adic fields I'm referring to, i.e., local non-archimedean fields possibly of a finite characteristic) are already compact. So is it true that the sets above are just closed subsets of some compact set so they are compact?
When dealing with the locally profinite group $GL_n(F)$ where $F$ is a $p$-adic field, we have a neighborhood basis around the identity, defined by
$$K_0=\{g\in GL_n(F):g\in M_n(\mathfrak o_F),|\det g|_F=1\},\quad K_j=GL_n(\mathfrak o_F)\cap (
1+\pi^jM_n(\mathfrak o_F))$$
where $\pi$ is a prime element. It seems to me that their compactness follows from that of $M_n(\mathfrak o_F)$. I'm not sure why this is true; do we simply regard $M_n(\mathfrak o_F)$ as $\prod_i^{n^2}\mathfrak o_F$ which is simply a product of compact sets (since we give $M_n(F)$ the product topology) and thus compact?
Thanks in advance. Any help is appreciated.
Also, I sometimes see people write $K_0=GL_n(\mathfrak o_F)$. But in this case where does the condition $|\det g|_F=1$ go? (I understand that asking multiple questions in a single thread is unacceptable, but I'm not sure if this seemly minor point should be asked as a separate question; I apologize in advance if I should do the latter.)
 A: Re second paragraph:
"A $p$-adic field" almost always means a finite extension of $\mathbb Q_p$, and hence necessarily has characterstic $0$. A "local field" usually is a more general concept, in that it also includes fields like $\mathbb F_q((T))$ (actually, up to isomorphism those are the only other local fields besides p-adic ones). In both cases, the field is locally compact, and this implies what you want in your second paragraph. Cf. The ring of integers of a locally compact non-archimedean field is compact.
Re third paragraph:

"It seems to me that their compactness follows from that of $M_n(\mathfrak o_F)$."

Yes it does.

"I'm not sure why this is true; do we simply regard $M_n(\mathfrak o_F)$ as $\prod_i^{n^2}\mathfrak o_F$ which is simply a product of compact sets (since we give $M_n(F)$ the product topology) and thus compact?"

That is indeed one way to see it. Another would be to consider $M_n(F)$ as a finite-dimensional vector space over $F$, or even over $\mathbb Q_p$, and use the fact that all norms on this are equivalent i.e. induce the same topology. To see that that topology, further, is identical with the product topology, note that "the" "natural" norm on such vector spaces is the supremum norm (for any choice of basis). Cf. answers to Natural Extensions of the $p$-Adic Norm to Higher Dimensions. And then of course everything behaves well when we go to the subspaces with entries in $\mathfrak{o}_F$.
Re:

"Also, I sometimes see people write $K_0=GL_n(\mathfrak o_F)$. But in this case where does the condition $|\det g|_F=1$ go?"

$g \in GL_n(\mathfrak o_F)$ means in particular $g^{-1} \in GL_n(\mathfrak o_F)$, means that $\det g \in \mathfrak o_F^\times = \{x \in \mathfrak o_F: \lvert x \rvert =1 \}$. And conversely, for $g \in M_n(\mathfrak o_F)$ with $\lvert \det g \rvert =1$, then also $(\det g)^{-1} \in \mathfrak o_F$, and since one knows that the entries of $g^{-1}$ (which a priori is only in $GL_n(F)$) are of the form $$(\det g)^{-1} \cdot \text{ polynomials in the entries of } g$$
they are also all in $\mathfrak o_F$, i.e. $g^{-1} \in GL_n(\mathfrak o_F)$.
