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.



Why is $P^n(K)$ compact, when $K$ is a local field?
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.


