Why is $P^n(K)$ compact, when $K$ is a local field? In Silverman's book AEC, question 7.6 asks to prove $E_0(K)$ has finite index in $E(K)$ for $K$ a local field. For part (a), I know the topology on $P^{n}(K)$ is the quotient topology on $K^{n+1}$, and the topology on $K^{n+1}$ is induced by the absolute value. However, I do not know how to prove the compactness of $P^{n}(K)$ in this topology.
 A: Because the quotient mapping $K^{n+1}  - \{\mathbf 0\} \to {\mathbf P}^n(K)$ (note  $\mathbf 0$ is not in the domain) is continuous by the definition of the quotient topology, it suffices to show there is a compact subset $C$ of $K^{n+1}  - \{\mathbf 0\}$ such that every element of ${\mathbf P}^n(K)$ is hit by some element of $C$, as that makes the natural mapping $C \to {\mathbf P}^n(K)$ continuous and surjective, so the target is compact . We want to show every line through the origin in $K^{n+1}$ contains an element of $C$. What can $C$ be?
When $K = \mathbf R$ (not a local field,  but a good warm-up example), for each $\mathbf v \in \mathbf R^{n+1} - \{\mathbf 0\}$ the vector $\mathbf v/||\mathbf v||$ is equal to $\mathbf v$ in ${\mathbf P}^n(\mathbf R)$, so as $C$ we can use the ordinary unit sphere
$$
S^n = \{\mathbf v \in \mathbf R^{n+1} : ||\mathbf v|| = 1\},
$$
which is compact.
When $K$ is a local field, let $||\cdot||_\infty$ be the sup-norm on $K^{n+1}$, so its values are in $|K|$: for each $\mathbf v \in \mathbf K^{n+1}$ there is $c \in K$ such that $||\mathbf v||_\infty = |c|$.
When $\mathbf v \not= \mathbf 0$, $\mathbf v/c$ is on the same line through the origin as $\mathbf v$ and $||\mathbf v/c||_\infty = 1$, so as $C$ we want to take the "sup-norm unit sphere"
$$
\{\mathbf w \in K^{n+1} : ||\mathbf w||_\infty = 1\}.
$$
Note this is a subset of $K^{n+1}- \{\mathbf 0\}$, and superficially its definition resembles the unit sphere in $\mathbf R^{n+1}$, but it is different in one sense: it is an open subset. (Its subspace topology in $K^{n+1}$ or $K^{n+1} - \{\mathbf 0\}$ is the same.) Why is it compact?
Method 1: the sup-norm on $K^{n+1}$ is continuous, so the above set is closed (inverse image of a point) and bounded in $K^{n+1}$, and thus is compact because $K$ is locally compact.
Method 2: the above set is the set-theoretic difference
$$
\{\mathbf w \in K^{n+1} : ||\mathbf w||_\infty \leq 1\} - 
\{\mathbf w \in K^{n+1} : ||\mathbf w||_\infty < 1\}
$$
where the bigger set is compact in $K^{n+1}$ (here we use that $K$ is a local field) and the set being removed is open in $K^{n+1}$, so its complement in $K^{n+1}$ is closed. Thus this set-theoretic difference is closed inside a compact set and thus is compact.
A: (Expanded version of my initial comment) Here I use that $K$ is a normed field in which the closed 1-ball is compact.
Let $\pi$ be the projection $K^{n+1}\smallsetminus\{0\}\to \mathbf{P}^n(K)$. Consider the compact subset $$W=\big\{x\in K^{n+1}:(\forall i:|x_i|\le 1) \wedge \sup_j|x_j|=1\big\}\subseteq K^{n+1}\smallsetminus\{0\}.$$
Then $\pi(W)=\mathbf{P}^n(K)$. So $\mathbf{P}^n(K)$ is compact (at least once it's granted $P^n(K)$ is Hausdorff).
To check Hausdorff of the quotient topology, to separate $\pi(y)\neq\pi(z)$, after a linear transformation we can suppose $y=e_1$, $z=e_2$, and then they are separated by the saturated open subsets $\{x:|x_1|>|x_2|\}$ and $\{x:|x_1|<|x_2|\}$.
